Chapter 9

Random walks, large deviations, and martingales

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Problem 9.1

Consider the simple random walk {Sn;n≥1}\{ S_{n}; n \geq 1\} with Sn=X1+⋯+XnS_{n} = X_{1}+\cdots +X_{n} and P(Xi=1)=p\mathbb {P}\left(X_{i} = 1\right) = p, P(Xi=−1)=1−p\mathbb {P}\left(X_{i} = -1\right) = 1-p; assume that p≤1/2p \leq 1/2.

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(a)

Show that P(⋃n≥1{Sn≥k})=[P(⋃n≥1{Sn≥1})]k\mathbb {P}\left(\bigcup_{n \geq 1} \left\{ S_{n} \geq k\right\} \right) = \left[\mathbb {P}\left(\bigcup_{n \geq 1} \left\{ S_{n} \geq 1\right\} \right)\right]^{k} for any positive integer kk.

(b)

Find a quadratic equation for y=P(⋃n≥1{Sn≥1})y = \mathbb {P}\left(\bigcup_{n \geq 1} \left\{ S_{n} \geq 1\right\} \right).

(c)

For p<1/2p<1/2, show that the two roots of this quadratic equation are p/(1−p)p/(1-p) and 11. Argue that P(⋃n≥1{Sn≥1})\mathbb {P}\left(\bigcup_{n \geq 1} \left\{ S_{n} \geq 1\right\} \right) cannot be 11 and thus must be p/(1−p)p/(1-p).

(d)

For p=1/2p = 1/2, show that the quadratic equation in (c) has a double root at 11, and thus P(⋃n≥1{Sn≥1})=1\mathbb {P}\left(\bigcup_{n \geq 1} \left\{ S_{n} \geq 1\right\} \right) = 1.

(e)

Let r∗r^{*} be the unique positive root of g(r)=1\mathsf{g}(r) = 1 where g(r)=E[erX]\mathsf{g}(r) = \mathbb {E}\left[e^{rX}\right]. Show that p/(1−p)=exp⁡(−r∗)p/(1-p) = \exp (-r^{*}).

Problem 9.2

Consider a G/G/1 queue with IID arrivals {Xi;i≥1}\{ X_{i}; i \geq 1\}, IID FCFS service times {Yi;i≥0}\{ Y_{i}; i \geq 0\}, and an initial arrival to an empty system at time 00. Define Ui=Xi−Yi−1U_{i} = X_{i}-Y_{i-1} for i≥1i \geq 1. Consider a sample path where (u1,…,u6)=(1,−2,2,−1,3,−2)(u_{1}, \ldots , u_{6}) = (1, -2, 2, -1, 3, -2).

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(a)

Let Zi6=U6+U6−1+⋯+U6−i+1Z_{i}^{6} = U_{6}+U_{6-1}+\cdots +U_{6-i+1}. Find the queueing delay for customer 66 as the maximum of the `backward' random walk with elements 0,Z16,Z26,…,Z660, Z_{1}^{6}, Z_{2}^{6}, \ldots , Z_{6}^{6}; sketch this random walk.

(b)

Find the queueing delay for customers 11 to 55.

(c)

Which customers start a busy period (i.e., arrive when the queue and server are both empty)? Verify that if Zi6Z_{i}^{6} maximizes the random walk in (a), then a busy period starts with arrival 6−i6-i.

(d)

Now consider a forward random walk Vn=U1+⋯+UnV_{n} = U_{1}+\cdots +U_{n}. Sketch this walk for the sample path above and show that the queueing delay for each customer is the difference between two appropriately chosen values of this walk.

Problem 9.3

A G/G/1 queue has a deterministic service time of 22 and interarrival times that are 33 with probability p<1/2p<1/2 and 11 with probability 1−p1-p.

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(a)

Find the distribution of W1W_{1}, the wait in queue of the first arrival after the beginning of a busy period.

(b)

Find the distribution of W∞W_{\infty }, the steady-state wait in queue.

(c)

Repeat (a) and (b) assuming the service times and interarrival times are exponentially distributed with rates μ\mu and λ\lambda respectively.

Problem 9.4

Let U=V1+⋯+VnU = V_{1}+\cdots +V_{n}, where V1,…,VnV_{1}, \ldots , V_{n} are IID rv s with the MGF gV(s)\mathsf{g}_{V}(s). Show that gU(s)=[gV(s)]n\mathsf{g}_{U}(s) = \left[\mathsf{g}_{V}(s)\right]^{n}.

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Problem 9.5

Let {an;n≥1}\{ a_{n}; n \geq 1\} and {αn;n≥1}\{ \alpha_{n}; n \geq 1\} be sequences of numbers. For some bb, assume that an≤αne−bna_{n} \leq \alpha_{n} e^{-bn} for all n≥1n \geq 1. For each of the following choices for αn\alpha_{n} and arbitrary bb, determine whether the bound is exponentially tight. You may assume b≥0b \geq 0 if you wish, but it really does not matter.

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(a)

αn=∑j=1kcjn−j\alpha_{n} = \sum_{j=1}^{k} c_{j} n^{-j}.

(b)

αn=exp⁡(−n)\alpha_{n} = \exp (-\sqrt{n}).

(c)

αn=e−n2\alpha_{n} = e^{-n^{2}}.

What you are intended to learn from this is that an exponentially tight bound is not necessarily a reasonable approximation for large nn. It is intended for sequences that are essentially exponentially decreasing in nn and where one is satisfied with knowing the exponent without concern for the non-exponential coefficients.

Problem 9.6

Define γ(r)\gamma (r) as ln⁡[g(r)]\ln \left[\mathsf{g}(r)\right], where g(r)=E[exp⁡(rX)]\mathsf{g}(r) = \mathbb {E}\left[\exp (rX)\right]. Assume that XX is discrete with possible outcomes {ai;i≥1}\{ a_{i}; i \geq 1\}, let pip_{i} denote P(X=ai)\mathbb {P}\left(X=a_{i}\right), and assume that g(r)\mathsf{g}(r) exists in some open interval (r−,r+)(r_{-}, r_{+}) containing r=0r=0. For any given rr, r−<r<r+r_{-}<r<r_{+}, define a rv XrX_{r} with the same set of possible outcomes {ai;i≥1}\{ a_{i}; i \geq 1\} as XX, but with a PMF qi=P(Xr=ai)=piexp⁡[air−γ(r)]q_{i} = \mathbb {P}\left(X_{r}=a_{i}\right) = p_{i} \exp \left[a_{i} r-\gamma (r)\right]. Note that XrX_{r} is not a function of XX, and is not even to be viewed as in the same probability space as XX; it is of interest simply because of the behavior of its defined probability mass function. It is called a tilted rv relative to XX, and this exercise, along with Exercise 9.11, will justify our interest in it.

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(a)

Verify that ∑iqi=1\sum_{i} q_{i} = 1.

(b)

Verify that E[Xr]=∑iaiqi\mathbb {E}\left[X_{r}\right] = \sum_{i} a_{i} q_{i} is equal to γ′(r)\gamma '(r).

(c)

Verify that Var⁡[Xr]=∑iai2qi−(E[Xr])2\operatorname {Var}\left[X_{r}\right] = \sum_{i} a_{i}^{2} q_{i}-(\mathbb {E}\left[X_{r}\right])^{2} is equal to γ′′(r)\gamma ''(r).

(d)

Argue that γ′′(r)≥0\gamma ''(r) \geq 0 for all rr such that g(r)\mathsf{g}(r) exists, and that γ′′(r)>0\gamma ''(r)>0 if γ′′(0)>0\gamma ''(0)>0.

(e)

Give a similar definition of XrX_{r} for a rv XX with a density, and modify (a)--(d) accordingly.

Problem 9.7
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(a)

Suppose ZZ is uniformly distributed over [−b,+b][-b, +b]. Find gZ(r)\mathsf{g}_{Z}(r), gZ′(r)\mathsf{g}_{Z}'(r), and γ′(r)\gamma '(r) as a function of rr.

(b)

Show that the interval over which gZ(r)\mathsf{g}_{Z}(r) exists is the real line, i.e., r+=∞r_{+}=\infty and r−=−∞r_{-}=-\infty. Show that lim⁡r→∞γ′(r)=b\lim_{r \rightarrow \infty } \gamma '(r) = b.

(c)

Show that if a>ba>b, then inf⁡r∈I(X)γ(r)−ra=−∞\inf_{r \in I(X)} \gamma (r)-ra = -\infty, so that the optimized Chernoff bound says that P(Sn>na)=0\mathbb {P}\left(S_{n}>na\right) = 0. Explain without any mathematics why P(Sn>na)=0\mathbb {P}\left(S_{n}>na\right)=0 must be valid. Explain (with as little mathematical verbiage as possible) why an infimum rather than a minimum must be used in the optimized Chernoff bound.

(d)

Show that for an arbitrary rv XX if r+=∞r_{+}=\infty and lim⁡r→∞γ′(r)=b\lim_{r \rightarrow \infty } \gamma '(r) = b, then the optimized Chernoff bound says that P(Sn>na)=0\mathbb {P}\left(S_{n}>na\right)=0 for a>ba>b.

Problem 9.8

Note that the MGF of the non-negative exponential rv with density e−xe^{-x} is (1−r)−1(1-r)^{-1} for r<r+=1r<r_{+}=1. It can be seen from this that g(r+)\mathsf{g}(r_{+}) does not exist (i.e., it is infinite) and both lim⁡r→r+g(r)=∞\lim_{r \rightarrow r_{+}} \mathsf{g}(r) = \infty and lim⁡r→r+g′(r)=∞\lim_{r \rightarrow r_{+}} \mathsf{g}'(r) = \infty, where the limit is over r<r+r<r_{+}. In this exercise, you are first to assume an arbitrary rv XX for which r+<∞r_{+}<\infty and g(r+)=∞\mathsf{g}(r_{+}) = \infty and show that both lim⁡r→r+g(r)=∞\lim_{r \rightarrow r_{+}} \mathsf{g}(r) = \infty and lim⁡r→r+g′(r)=∞\lim_{r \rightarrow r_{+}} \mathsf{g}'(r) = \infty. You will then use this to show that if sup⁡r<r+g′(r)<∞\sup_{r<r_{+}} \mathsf{g}'(r)<\infty, then γ(r+)<∞\gamma (r_{+})<\infty and the optimized Chernoff exponent μ(a)\mu (a) is given by μ(a)=γ(r+)−r+a\mu (a) = \gamma (r_{+})-r_{+} a for a>sup⁡r<r+γ′(r)a>\sup_{r<r_{+}} \gamma '(r).

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(a)

For XX such that r+<∞r_{+}<\infty and g(r+)=∞\mathsf{g}(r_{+})=\infty, explain why

lim⁡A→∞∫0Aexr+ dF(x)=∞. \lim _{A \rightarrow \infty } \int _{0}^{A} e^{x r_{+}} \, d\mathsf{F}(x) = \infty .
(b)

Show that for any ϵ>0\epsilon >0 and any A>0A>0,

g(r+−ϵ)≥e−ϵA∫0Aexr+ dF(x). \mathsf{g}(r_{+}-\epsilon ) \geq e^{-\epsilon A} \int _{0}^{A} e^{x r_{+}} \, d\mathsf{F}(x).
(c)

Choose A=1/ϵA=1/\epsilon and show that

lim⁡ϵ→0g(r+−ϵ)=∞. \lim _{\epsilon \rightarrow 0} \mathsf{g}(r_{+}-\epsilon ) = \infty .
(d)

Show that lim⁡ϵ→0g′(r+−ϵ)=∞\lim_{\epsilon \rightarrow 0} \mathsf{g}'(r_{+}-\epsilon ) = \infty.

(e)

Use (a)--(d) to show that if r+<∞r_{+}<\infty and sup⁡r<r+g′(r)<∞\sup_{r<r_{+}} \mathsf{g}'(r)<\infty, then g(r+)<∞\mathsf{g}(r_{+})<\infty.

(f)

Show that if r+<∞r_{+}<\infty and sup⁡r<r+g′(r)<∞\sup_{r<r_{+}} \mathsf{g}'(r)<\infty, then μ(a)=γ(r+)−r+a\mu (a) = \gamma (r_{+})-r_{+} a for a>sup⁡r<r+γ′(r)a>\sup_{r<r_{+}} \gamma '(r).

Problem 9.9
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(a)

Show that the two appearances of ϵ\epsilon in the exponential lower bound established in the text for P(Sn≥nγ′(r))\mathbb {P}\left(S_{n} \geq n \gamma '(r)\right) can be replaced with two independent arbitrary positive quantities ϵ1\epsilon_{1} and ϵ2\epsilon_{2}, getting

P(Sn≥n(γ′(r)−ϵ1))≥(1−δ)exp⁡[−n(rγ′(r)+rϵ2−γ(r))]. \mathbb {P}\left(S_{n} \geq n(\gamma '(r)-\epsilon _{1})\right) \geq (1-\delta ) \exp \left[-n(r \gamma '(r)+r \epsilon _{2}-\gamma (r))\right].

Show that if this equation is valid for ϵ1\epsilon_{1} and ϵ2\epsilon_{2}, then it is valid for all larger values of ϵ1\epsilon_{1} and ϵ2\epsilon_{2}.

(b)

Show that by increasing the required value of non_{o} such that this equation is valid for all n≥non \geq n_{o}, the factor of (1−δ)(1-\delta ) can be eliminated above.

(c)

For any r∈(0,r+)r \in (0, r_{+}), let δ1\delta_{1} be an arbitrary number in (0,r+−r)(0, r_{+}-r), let r1=r+δ1r_{1} = r+\delta_{1}, and let ϵ1=γ′(r1)−γ′(r)\epsilon_{1} = \gamma '(r_{1})-\gamma '(r). Show that there is an mm such that for all n≥mn \geq m,

P(Sn≥nγ′(r))≥exp⁡{−n[(r+δ1)γ′(r+δ1)+(r+δ1)ϵ2−γ(r+δ1)]}.(9.164) \mathbb {P}\left(S_{n} \geq n \gamma '(r)\right) \geq \exp \left\{ -n\left[(r+\delta _{1}) \gamma '(r+\delta _{1})+(r+\delta _{1}) \epsilon _{2}-\gamma (r+\delta _{1})\right]\right\} . \tag {9.164}

Using the continuity of γ\gamma and its derivatives, show that for any ϵ>0\epsilon >0, there is a δ1>0\delta_{1}>0 so that the right-hand side of (9.164) is greater than or equal to exp⁡[−n(γ′(r)−rγ(r)+ϵ)]\exp \left[-n(\gamma '(r)-r \gamma (r)+\epsilon )\right].

Problem 9.10

In this exercise, we show that the optimized Chernoff bound is tight for the third case of the large-deviations exponent (the case a>sup⁡r<r+γ′(r)a>\sup_{r<r_{+}} \gamma '(r), with r+<∞r_{+}<\infty) as well as the first case. That is, we show that if r+<∞r_{+}<\infty and a>sup⁡r<r+γ′(r)a>\sup_{r<r_{+}} \gamma '(r), then, for any ϵ>0\epsilon >0, P(Sn≥na)≥exp⁡{n[γ(r+)−r+a−ϵ]}\mathbb {P}\left(S_{n} \geq na\right) \geq \exp \left\{ n\left[\gamma (r_{+})-r_{+} a-\epsilon \right]\right\} for all large enough nn.

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(a)

Let YiY_{i} be the truncated version of XiX_{i}, truncated for some given bb to Yi=XiY_{i}=X_{i} for Xi≤bX_{i} \leq b and Yi=bY_{i}=b otherwise. Let Wn=Y1+⋯+YnW_{n} = Y_{1}+\cdots +Y_{n}. Show that P(Sn≥na)≥P(Wn≥na)\mathbb {P}\left(S_{n} \geq na\right) \geq \mathbb {P}\left(W_{n} \geq na\right).

(b)

Let gb(r)\mathsf{g}_{b}(r) be the MGF of YY. Show that gb(r)<∞\mathsf{g}_{b}(r)<\infty and that gb(r)\mathsf{g}_{b}(r) is non-decreasing in bb for all r<∞r<\infty.

(c)

Show that lim⁡b→∞gb(r)=∞\lim_{b \rightarrow \infty } \mathsf{g}_{b}(r) = \infty for all r>r+r>r_{+} and that lim⁡b→∞gb(r)=g(r)\lim_{b \rightarrow \infty } \mathsf{g}_{b}(r) = \mathsf{g}(r) for all r≤r+r \leq r_{+}.

(d)

Let γb(r)=ln⁡gb(r)\gamma_{b}(r) = \ln \mathsf{g}_{b}(r). Show that γb(r)<∞\gamma_{b}(r)<\infty for all r<∞r<\infty. Also show that lim⁡b→∞γb(r)=∞\lim_{b \rightarrow \infty } \gamma_{b}(r) = \infty for r>r+r>r_{+} and lim⁡b→∞γb(r)=γ(r)\lim_{b \rightarrow \infty } \gamma_{b}(r) = \gamma (r) for r≤r+r \leq r_{+}.

(e)

Let γb′(r)=(∂/∂r)γb(r)\gamma_{b}'(r) = (\partial /\partial r) \gamma_{b}(r) and let δ>0\delta >0 be arbitrary. Show that for all large enough bb, γb′(r++δ)>a\gamma_{b}'(r_{+}+\delta )>a.

(f)

Show that the optimized Chernoff bound for P(Wn≥na)\mathbb {P}\left(W_{n} \geq na\right) is exponentially tight for the values of bb in (e). Show that the optimizing rr is less than r++δr_{+}+\delta.

(g)

Show that for any ϵ>0\epsilon >0 and all sufficiently large bb,

γb(r)−ra≥γ(r)−ra−ϵ≥γ(r+)−r+a−ϵfor 0<r≤r+. \gamma _{b}(r)-ra \geq \gamma (r)-ra-\epsilon \geq \gamma (r_{+})-r_{+} a-\epsilon \qquad \text{for } 0<r \leq r_{+}.
(h)

Show that for arbitrary ϵ>0\epsilon >0, there exists δ>0\delta >0 and bob_{o} such that

γb(r)−ra≥γ(r+)−r+a−ϵfor r+<r<r++δ and b≥bo. \gamma _{b}(r)-ra \geq \gamma (r_{+})-r_{+} a-\epsilon \qquad \text{for } r_{+}<r<r_{+}+\delta \text{ and } b \geq b_{o}.
(i)

Note that if we put (g), (h) and (d) together, and use the δ\delta of (h) in (d), then we have shown that the optimized exponent in the Chernoff bound for P(Wn≥na)\mathbb {P}\left(W_{n} \geq na\right) satisfies μb(a)≥γ(r+)−r+a−ϵ\mu_{b}(a) \geq \gamma (r_{+})-r_{+} a-\epsilon for sufficiently large bb. Show that this means that P(Sn≥na)≥γ(r+)−r+a−2ϵ\mathbb {P}\left(S_{n} \geq na\right) \geq \gamma (r_{+})-r_{+} a-2 \epsilon for sufficiently large nn.

Problem 9.11

Assume that XX is discrete, with possible values {ai;i≥1}\{ a_{i}; i \geq 1\} and probabilities P(X=ai)=pi\mathbb {P}\left(X=a_{i}\right) = p_{i}. Let XrX_{r} be the corresponding tilted rv as defined in Exercise 9.6. Let Sn=X1+⋯+XnS_{n} = X_{1}+\cdots +X_{n} be the sum of nn IID rv s with the distribution of XX, and let Sn,r=X1,r+⋯+Xn,rS_{n,r} = X_{1,r}+\cdots +X_{n,r} be the sum of nn IID tilted rv s with the distribution of XrX_{r}. Assume that X‾<0\overline{X}<0 and that r>0r>0 is such that γ(r)\gamma (r) exists.

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(a)

Show that P(Sn,r=s)=P(Sn=s)exp⁡[sr−nγ(r)]\mathbb {P}\left(S_{n,r}=s\right) = \mathbb {P}\left(S_{n}=s\right) \exp \left[sr-n \gamma (r)\right].

(b)

Find the mean and variance of Sn,rS_{n,r} in terms of γ(r)\gamma (r).

(c)

Define a=γ′(r)a = \gamma '(r) and σr2=γ′′(r)\sigma_{r}^{2} = \gamma ''(r). Show that P(∣Sn,r−na∣≤2nσr)>1/2\mathbb {P}\left(\left|S_{n,r}-na\right| \leq \sqrt{2n} \sigma_{r}\right)>1/2. Use this to show that

P(∣Sn−na∣≤2nσr)>(1/2)exp⁡[−r(an+2nσr)+nγ(r)]. \mathbb {P}\left(\left|S_{n}-na\right| \leq \sqrt{2n} \sigma _{r}\right) > (1/2) \exp \left[-r(an+\sqrt{2n} \sigma _{r})+n \gamma (r)\right].
(d)

Use this to show that for any ϵ\epsilon and for all sufficiently large nn,

P(Sn≥n(γ′(r)−ϵ))>12exp⁡[−rn(γ′(r)+ϵ)+nγ(r)]. \mathbb {P}\left(S_{n} \geq n(\gamma '(r)-\epsilon )\right) > \frac{1}{2} \exp \left[-rn(\gamma '(r)+\epsilon )+n \gamma (r)\right].
Problem 9.12

Let p→=(p1,…,pM)\overrightarrow {p} = (p_{1}, \ldots , p_{\mathsf{M}}) and p~→=(p~1,…,p~M)\overrightarrow {\tilde{p}} = (\tilde{p}_{1}, \ldots , \tilde{p}_{\mathsf{M}}) be strictly positive probability vectors. The Kullback--Leibler divergence between p~→\overrightarrow {\tilde{p}} and p→\overrightarrow {p} is defined by

D(p~→∥p→)=∑jp~jln⁡(p~jpj). D(\overrightarrow {\tilde{p}} \| \overrightarrow {p}) = \sum _{j} \tilde{p}_{j} \ln \left(\frac{\tilde{p}_{j}}{p_{j}}\right).
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(a)

Show that the divergence D(p~→∥p→)≥0D(\overrightarrow {\tilde{p}} \| \overrightarrow {p}) \geq 0 is 00 if p~→=p→\overrightarrow {\tilde{p}} = \overrightarrow {p}.

(b)

Let u→=u1,…,uM\overrightarrow {u} = u_{1}, \ldots , u_{\mathsf{M}} satisfy ∑juj=0\sum_{j} u_{j} = 0 and show that p~→+ϵu→\overrightarrow {\tilde{p}} + \epsilon \overrightarrow {u} is a probability vector over a sufficiently small region of ϵ\epsilon around 00. Show that

∂∂ϵD(p~→+ϵu→∥p→)∣ϵ=0=0. \frac{\partial }{\partial \epsilon } D(\overrightarrow {\tilde{p}} + \epsilon \overrightarrow {u} \| \overrightarrow {p})\Big|_{\epsilon = 0} = 0.
(c)

Show that

∂2∂ϵ2D(p~→+ϵu→∥p→)≥0, \frac{\partial ^{2}}{\partial \epsilon ^{2}} D(\overrightarrow {\tilde{p}} + \epsilon \overrightarrow {u} \| \overrightarrow {p}) \geq 0,

where p~→+ϵu→\overrightarrow {\tilde{p}} + \epsilon \overrightarrow {u} is a probability vector with non-zero components. Explain why this implies that DD is convex in p~→\overrightarrow {\tilde{p}} and non-negative over the region of probability vectors.

Problem 9.13

Consider a random walk {Sn;n≥1}\left\{ S_{n}; n \geq 1\right\}, where Sn=X1+⋯+XnS_{n} = X_{1} + \cdots + X_{n} and {Xi;i≥1}\left\{ X_{i}; i \geq 1\right\} is a sequence of IID exponential rv s with the PDF f(x)=λe−λx\mathsf{f}(x) = \lambda e^{-\lambda x} for x≥0x \geq 0. In other words, the random walk is the sequence of arrival epochs in a Poisson process.

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(a)

Show that for λa>1\lambda a > 1, the optimized Chernoff bound for P(Sn≥na)\mathbb {P}\left(S_{n} \geq na\right) is given by

P(Sn≥na)≤(aλ)ne−n(aλ−1). \mathbb {P}\left(S_{n} \geq na\right) \leq (a \lambda )^{n} e^{-n(a \lambda - 1)}.
(b)

Show that the exact value of P(Sn≥na)\mathbb {P}\left(S_{n} \geq na\right) is given by

P(Sn≥na)=∑i=0n−1(naλ)ie−naλi!. \mathbb {P}\left(S_{n} \geq na\right) = \sum _{i=0}^{n-1} \frac{(na \lambda )^{i} e^{-na \lambda }}{i!}.
(c)

By upper and lower bounding the quantity on the right above, show that

(naλ)ne−naλn! aλ≤P(Sn≥na)≤(naλ)ne−naλn!(aλ−1). \frac{(na \lambda )^{n} e^{-na \lambda }}{n! \, a \lambda } \leq \mathbb {P}\left(S_{n} \geq na\right) \leq \frac{(na \lambda )^{n} e^{-na \lambda }}{n! (a \lambda - 1)}.
(d)

Use the Stirling bounds on n!n! to show that

(aλ)ne−n(aλ−1)2πn aλexp⁡(1/12n)≤P(Sn≥na)≤(aλ)ne−n(aλ−1)2πn (aλ−1). \frac{(a \lambda )^{n} e^{-n(a \lambda - 1)}}{\sqrt{2 \pi n} \, a \lambda \exp (1 / 12 n)} \leq \mathbb {P}\left(S_{n} \geq na\right) \leq \frac{(a \lambda )^{n} e^{-n(a \lambda - 1)}}{\sqrt{2 \pi n} \, (a \lambda - 1)}.
Remark.
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The point of this exercise is to demonstrate that the Chernoff bound is not only exponentially tight for this example but captures the important factors in the behavior of P(Sn≥na)\mathbb {P}\left(S_{n} \geq na\right).

Problem 9.14

Consider a random walk with thresholds α>0\alpha > 0, β<0\beta < 0. We wish to find P(SJ≥α)\mathbb {P}\left(S_{J} \geq \alpha \right) in the absence of a lower threshold. Use the upper bound P(SJ≥α)≤exp⁡(−r∗α)\mathbb {P}\left(S_{J} \geq \alpha \right) \leq \exp (-r^{*}\alpha ) (eq. (9.46), Corollary 9.4.4) for the probability that the random walk crosses α\alpha before β\beta.

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(a)

Given that the random walk crosses β\beta first, find an upper bound to the probability that α\alpha is crossed before a yet lower threshold at 2β2\beta is crossed.

(b)

Given that 2β2\beta is crossed before α\alpha, upper bound the probability that α\alpha is crossed before a threshold at 3β3\beta. Extending this argument to successively lower thresholds, find an upper bound to each successive term, and find an upper bound on the overall probability that α\alpha is crossed. By observing that β\beta is arbitrary, show that (9.46) is valid with no lower threshold.

Problem 9.15

This exercise verifies that Corollary 9.4.4 holds in the situation where γ(r+)<0\gamma (r_{+}) < 0. We have defined r∗=r+r^{*} = r_{+} in this case.

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(a)

Use the Wald identity at r=r+r = r_{+} to show that

P(SJ≥α)E[exp⁡(r+SJ−Jγ(r+))∣SJ≥α]≤1. \mathbb {P}\left(S_{J} \geq \alpha \right)\mathbb {E}\left[\exp (r_{+}S_{J} - J\gamma (r_{+})) \mid S_{J} \geq \alpha \right] \leq 1.
(b)

Show that

P(SJ≥α)exp⁡[r+α−γ(r+)]≤1. \mathbb {P}\left(S_{J} \geq \alpha \right)\exp \left[r_{+}\alpha - \gamma (r_{+})\right] \leq 1.
(c)

Show that

P(SJ≥α)≤exp⁡[−r+α+γ(r+)]andP(SJ≥α)≤exp⁡[−r∗α]. \mathbb {P}\left(S_{J} \geq \alpha \right) \leq \exp \left[-r_{+}\alpha + \gamma (r_{+})\right] \qquad \text{and} \qquad \mathbb {P}\left(S_{J} \geq \alpha \right) \leq \exp \left[-r^{*}\alpha \right].
Remark.
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Note that the first bound above is a little stronger than the second, and note that lower bounding JJ by 11 is not necessarily a very weak bound, since this is the case where if α\alpha is crossed, it tends to be crossed for small JJ.

Problem 9.16
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(a)

Use Wald's equality to show that if X‾=0\overline{X} = 0, then E[SJ]=0\mathbb {E}\left[S_{J}\right] = 0, where JJ is the time of threshold crossing with one threshold at α>0\alpha > 0 and another at β<0\beta < 0.

(b)

Obtain an expression for P(SJ≥α)\mathbb {P}\left(S_{J} \geq \alpha \right). Your expression should involve the expected value of SJS_{J} conditional on crossing the individual thresholds (you need not try to calculate these expected values).

(c)

Evaluate your expression for the case of a simple random walk.

(d)

Evaluate your expression when XX has an exponential density, fX(x)=a1e−λx\mathsf{f}_{X}(x) = a_{1}e^{-\lambda x} for x≥0x \geq 0 and fX(x)=a2eμx\mathsf{f}_{X}(x) = a_{2}e^{\mu x} for x<0x < 0 and where a1a_{1} and a2a_{2} are chosen so that X‾=0\overline{X} = 0.

Problem 9.17

A random walk {Sn;n≥1}\left\{ S_{n}; n \geq 1\right\}, with Sn=∑i=1nXiS_{n} = \sum_{i=1}^{n} X_{i}, has the following probability density for XiX_{i}:

fX(x)={e−xe−e−1,−1≤x≤1,0,elsewhere. \mathsf{f}_{X}(x) = \begin{cases} \dfrac {e^{-x}}{e - e^{-1}}, & -1 \leq x \leq 1, \\ 0, & \text{elsewhere.} \end{cases}
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(a)

Find the values of rr for which g(r)=E[exp⁡(rX)]=1\mathsf{g}(r) = \mathbb {E}\left[\exp (rX)\right] = 1.

(b)

Let PαP_{\alpha } be the probability that the random walk ever crosses a threshold at α\alpha for some α>0\alpha > 0. Find an upper bound to PαP_{\alpha } of the form Pα≤e−αAP_{\alpha } \leq e^{-\alpha A}, where AA is a constant that does not depend on α\alpha; evaluate AA.

(c)

Find a lower bound to PαP_{\alpha } of the form Pα≥Be−αAP_{\alpha } \geq Be^{-\alpha A}, where AA is the same as in (b) and BB is a constant that does not depend on α\alpha.

Problem 9.18

Let {Xn;n≥1}\left\{ X_{n}; n \geq 1\right\} be a sequence of IID integer-valued rv s with the PMF pX(k)=Qk\mathsf{p}_{X}(k) = Q_{k}. Assume that Qk>0Q_{k} > 0 for ∣k∣≤10\left|k\right| \leq 10 and Qk=0Q_{k} = 0 for ∣k∣>10\left|k\right| > 10. Let {Sn;n≥1}\left\{ S_{n}; n \geq 1\right\} be a random walk with Sn=X1+⋯+XnS_{n} = X_{1} + \cdots + X_{n}. Let α>0\alpha > 0 and β<0\beta < 0 be integer-valued thresholds, let JJ be the smallest value of nn for which either Sn≥αS_{n} \geq \alpha or Sn≤βS_{n} \leq \beta. Let {Sn∗;n≥1}\left\{ S_{n}^{*}; n \geq 1\right\} be the stopped random walk; i.e., Sn∗=SnS_{n}^{*} = S_{n} for n≤Jn \leq J and Sn∗=SJS_{n}^{*} = S_{J} for n>Jn > J. Let πi∗=P(SJ=i)\pi_{i}^{*} = \mathbb {P}\left(S_{J} = i\right).

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(a)

Consider a Markov chain in which this stopped random walk is run repeatedly until the point of stopping. That is, the Markov chain transition probabilities are given by Pij=Qj−iP_{ij} = Q_{j-i} for β<i<α\beta < i < \alpha and Pi0=1P_{i0} = 1 for i≤βi \leq \beta and i≥αi \geq \alpha. All other transition probabilities are 00 and the set of states is the set of integers [−9+β,9+α][-9+\beta , 9+\alpha ]. Show that this Markov chain is ergodic.

(b)

Let {πi}\left\{ \pi_{i}\right\} be the set of steady-state probabilities for this Markov chain. Find the set of probabilities {πi∗}\left\{ \pi_{i}^{*}\right\} for the stopping states of the stopped random walk in terms of {πi}\left\{ \pi_{i}\right\}.

(c)

Find E[SJ]\mathbb {E}\left[S_{J}\right] and E[J]\mathbb {E}\left[J\right] in terms of {πi}\left\{ \pi_{i}\right\}.

Problem 9.19

Consider the following binary hypothesis testing problem: there are two hypotheses X=0,X=1X=0, X=1, and, conditional on each hypothesis, IID observations Y1,…,YnY_{1}, \ldots , Y_{n} are drawn with conditional density fY∣X(y∣0)\mathsf{f}_{Y|X}(y \mid 0) or fY∣X(y∣1)\mathsf{f}_{Y|X}(y \mid 1) respectively.

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(a)

Conditional on X=0X = 0 for the hypothesis testing problem, consider the rv s Zi=ln⁡[f(yi∣X=1)/f(yi∣X=0)]Z_{i} = \ln \left[\mathsf{f}(y_{i} \mid X=1)/\mathsf{f}(y_{i} \mid X=0)\right]. Show that r∗r^{*}, the positive solution to g(r)=1\mathsf{g}(r) = 1, where g(r)=E[exp⁡(rZi)]\mathsf{g}(r) = \mathbb {E}\left[\exp (rZ_{i})\right], is given by r∗=1r^{*} = 1.

(b)

Assuming that YY is a discrete rv (under each hypothesis), show that the tilted rv ZrZ_{r} with r=1r = 1 has the PMF pY(y∣X=1)\mathsf{p}_{Y}(y \mid X=1).

Problem 9.20

Recall the bounds from Theorem 9.5.1: for a threshold test at η=enγ0(r)\eta = e^{n\gamma_{0}(r)},

P(e(n)∣X=0)≤exp⁡(n[γ0(r)−rγ0′(r)])(9.61) \mathbb {P}\left(e(n) \mid X=0\right) \leq \exp \left(n\left[\gamma _{0}(r) - r\gamma _{0}'(r)\right]\right) \tag {9.61}

and

P(e(n)∣X=1)≤exp⁡(n[γ0(r)+(1−r)γ0′(r)]),(9.62) \mathbb {P}\left(e(n) \mid X=1\right) \leq \exp \left(n\left[\gamma _{0}(r) + (1-r)\gamma _{0}'(r)\right]\right), \tag {9.62}

where γ0(r)\gamma_{0}(r) is the semi-invariant MGF of the log-likelihood ratio ZZ conditional on X=0X = 0.

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(a)

Use a power series expansion around r=0r = 0 to show that γ0(r)−rγ0′(r)=−r2γ0′′(0)/2+o(r2)\gamma_{0}(r) - r\gamma_{0}'(r) = -r^{2}\gamma_{0}''(0)/2 + o(r^{2}).

(b)

Choose r=n−1/3r = n^{-1/3} in (9.61) and (9.62). Show that (9.61) can then be rewritten as

P(e(n)∣X=0)≤exp⁡{n[−r2γ0′′(0)/2+o(r2)]}=exp⁡{−n1/3γ0′′(0)/2+o(n1/3)}. \begin{aligned} \mathbb {P}\left(e(n) \mid X=0\right) & \leq \exp \left\{ n\left[-r^{2}\gamma _{0}''(0)/2 + o(r^{2})\right]\right\} \\ & = \exp \left\{ -n^{1/3}\gamma _{0}''(0)/2 + o(n^{1/3})\right\} . \end{aligned}

Thus P(e(n)∣X=0)\mathbb {P}\left(e(n) \mid X=0\right) approaches 00 with increasing nn exponentially in n1/3n^{1/3}.

(c)

Using the same power series expansion for rr in (9.62), show that

P(e(n)∣X=1)≤exp⁡{n[r2γ0′′(0)/2+γ0′(0)+o(r)]}=exp⁡{nγ0′(0)+o(n2/3)}. \begin{aligned} \mathbb {P}\left(e(n) \mid X=1\right) & \leq \exp \left\{ n\left[r^{2}\gamma _{0}''(0)/2 + \gamma _{0}'(0) + o(r)\right]\right\} \\ & = \exp \left\{ n\gamma _{0}'(0) + o(n^{2/3})\right\} . \end{aligned}

Thus lim⁡n(1/n)ln⁡P(e(n)∣X=1)=γ0′(0)=E[Z∣X=0]\lim_{n}(1/n)\ln \mathbb {P}\left(e(n) \mid X=1\right) = \gamma_{0}'(0) = \mathbb {E}\left[Z \mid X=0\right].

(d)

Explain why the bound in (9.62) is exponentially tight given that P(e(n)∣X=0)→0\mathbb {P}\left(e(n) \mid X=0\right) \rightarrow 0 as n→∞n \rightarrow \infty.

Problem 9.21

This illustrates a very different type of sequential decision problem from those of Section 9.5. Alice is looking for a spouse and dates a set of nn suitors sequentially, one per week. For simplicity, assume that Alice must make her decision to accept a suitor for marriage immediately after dating that suitor; she cannot come back later to accept a formerly rejected suitor. Her decision must be based only on whether the current suitor is more suitable than all previous suitors. Mathematically, we can view the dating as continuing for all nn weeks, but the choice at week mm is a stopping rule. Assume that each suitor's suitability is represented by a real number and that all nn numbers are different. Alice does not observe the suitability numbers, but only observes at each time mm whether suitor mm has the highest suitability so far. The suitors are randomly permuted before Alice dates them.

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(a)

A reasonable algorithm for Alice is to reject the first kk suitors (for some kk to be optimized later) and then to choose the first suitor m>km > k that is more suitable than all the previous m−1m - 1 suitors. Find the probability qkq_{k} that Alice chooses the most suitable of all nn when she uses this algorithm for given kk.

(b)

Approximating ∑i=1j1/i\sum_{i=1}^{j} 1/i by ln⁡j\ln j, show that for nn and kk large

qk≈knln⁡(n/k). q_{k} \approx \frac{k}{n}\ln (n/k).

Ignoring the constraint that nn and kk are integers, show that the right-hand side above is maximized at k/n=ek/n = e and that max⁡kqk≈1/e\max_{k} q_{k} \approx 1/e.

(c)

(Optional) Show that the algorithm of (a), optimized over kk, is optimal over all algorithms (given the constraints of the problem). Let pmp_{m} be the maximum probability of choosing the optimal suitor given that no choice has been made before time mm. Show that

pm=max⁡[1n+m−1mpm+1,  pm+1]; p_{m} = \max \left[\frac{1}{n} + \frac{m-1}{m}p_{m+1}, \; p_{m+1}\right];

part of the problem here is understanding exactly what pmp_{m} means.

(d)

Explain why this is a poor model for choosing a spouse (or for making a best choice in a wide variety of similar problems).

Remark.
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Note: If no suitor is selected by the algorithm in part (a), it makes no difference whether suitor nn is chosen at the end or no suitor is chosen.

Problem 9.22
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(a)

Suppose {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} is a martingale. Verify

E[Zn]=E[Z1]for n>1.(9.83) \mathbb {E}\left[Z_{n}\right] = \mathbb {E}\left[Z_{1}\right] \quad \text{for } n > 1. \tag {9.83}
(b)

If {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} is a submartingale, verify

E[Zn]≥E[Zi]for all i,1≤i<n,(9.88) \mathbb {E}\left[Z_{n}\right] \geq \mathbb {E}\left[Z_{i}\right] \quad \text{for all } i, 1 \leq i < n, \tag {9.88}

and if a supermartingale, verify

E[Zn]≤E[Zi]for all i,1≤i<n.(9.89) \mathbb {E}\left[Z_{n}\right] \leq \mathbb {E}\left[Z_{i}\right] \quad \text{for all } i, 1 \leq i < n. \tag {9.89}
Problem 9.23

Suppose {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} is a martingale. Show that

E[Zm∣Zni,Zni−1,…,Zn1]=Znifor all 0<n1<n2<⋯<ni<m. \mathbb {E}\left[Z_{m} \mid Z_{n_{i}}, Z_{n_{i-1}}, \ldots , Z_{n_{1}}\right] = Z_{n_{i}} \quad \text{for all } 0 < n_{1} < n_{2} < \cdots < n_{i} < m.
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Problem 9.24
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(a)

Assume that {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} is a submartingale. Show that

E[Zm∣Zn,Zn−1,…,Z1]≥Znfor all n<m. \mathbb {E}\left[Z_{m} \mid Z_{n}, Z_{n-1}, \ldots , Z_{1}\right] \geq Z_{n} \quad \text{for all } n < m.
(b)

Show that

E[Zm∣Zni,Zni−1,…,Zn1]≥Znifor all 0<n1<n2<⋯<ni<m. \mathbb {E}\left[Z_{m} \mid Z_{n_{i}}, Z_{n_{i-1}}, \ldots , Z_{n_{1}}\right] \geq Z_{n_{i}} \quad \text{for all } 0 < n_{1} < n_{2} < \cdots < n_{i} < m.
(c)

Assume now that {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} is a supermartingale. Show that (a) and (b) still hold with ≥\geq replaced by ≤\leq.

Problem 9.25

Let {Zn=exp⁡(rSn−nγ(r));n≥1}\left\{ Z_{n} = \exp \left(rS_{n} - n\gamma (r)\right); n \geq 1\right\} be the generating function martingale of (9.77), where Sn=X1+⋯+XnS_{n} = X_{1} + \cdots + X_{n} and X1,…,XnX_{1}, \ldots , X_{n} are IID with mean X‾<0\overline{X} < 0. Let JJ be the possibly-defective stopping trial for which the process stops after crossing a threshold at α>0\alpha > 0 (there is no negative threshold). Show that exp⁡(r∗α)\exp \left(r^{*}\alpha \right) is an upper bound to the probability of threshold crossing by considering the stopped process {Zn∗;n≥1}\left\{ Z_{n}^{*}; n \geq 1\right\}.

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Remark.
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The purpose of this exercise is to illustrate that the stopped process can yield useful upper bounds even when the stopping trial is defective.

Problem 9.26

This exercise uses a martingale to find the expected number of successive trials E[J]\mathbb {E}\left[J\right] until some fixed pattern, a1,a2,…,aka_{1}, a_{2}, \ldots , a_{k}, of successive binary digits occurs within a sequence of IID binary rv's X1,X2,…X_{1}, X_{2}, \ldots (see Example 4.5.1 and Exercise 5.35 for alternative approaches). We take the stopping time JJ to be the smallest nn for which (Xn−k+1,…,Xn)=(a1,…,ak)(X_{n-k+1}, \ldots , X_{n}) = (a_{1}, \ldots , a_{k}). A mythical casino and sequence of gamblers who follow a prescribed strategy will be used to determine E[J]\mathbb {E}\left[J\right]. The outcomes of the plays (trials), {Xn;n≥1}\left\{ X_{n}; n \geq 1\right\} at the casino is a binary IID sequence for which P(Xn=i)=pi\mathbb {P}\left(X_{n} = i\right) = p_{i} for i∈{0,1}i \in \left\{ 0,1\right\}.

If a gambler places a bet ss on 1 at play nn, the return is s/p1s/p_{1} if Xn=1X_{n} = 1 and 00 otherwise. With a bet ss on 0, the return is s/p0s/p_{0} if Xn=0X_{n} = 0 and 00 otherwise; i.e., the game is fair.

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(a)

Assume an arbitrary choice of bets on 0 and 1 by the various gamblers on the various trials. Let YnY_{n} be the net gain of the casino on trial nn. Show that E[Yn]=0\mathbb {E}\left[Y_{n}\right] = 0. Let Zn=Y1+Y2+⋯+YnZ_{n} = Y_{1} + Y_{2} + \cdots + Y_{n} be the aggregate gain of the casino over nn trials. Show that for any given pattern of bets, {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} is a martingale.

(b)

In order to determine E[J]\mathbb {E}\left[J\right] for a given pattern a1,a2,…,aka_{1}, a_{2}, \ldots , a_{k}, we program our gamblers to bet as follows:

(i) Gambler 1 has an initial capital of 1 which is bet on a1a_{1} at trial 1. If X1=a1X_{1} = a_{1}, the capital grows to 1/pa11/p_{a_{1}}, all of which is bet on a2a_{2} at trial 2. If X2=a2X_{2} = a_{2}, the capital grows to 1/(pa1pa2)1/(p_{a_{1}}p_{a_{2}}), all of which is bet on a3a_{3} at trial 3. Gambler 1 continues in this way until either losing at some trial (in which case he leaves with no money) or winning on kk successive trials (in which case he leaves with 1/[pa1⋯pak]1/\left[p_{a_{1}} \cdots p_{a_{k}}\right]).

(ii) Gambler ℓ\ell, for each ℓ>1\ell > 1, follows the same strategy, but starts at trial ℓ\ell. Note that if the pattern a1,…,aka_{1}, \ldots , a_{k} appears for the first time at trials n−k+1,n−k+2,…,nn-k+1, n-k+2, \ldots , n, i.e., if J=nJ = n, then gambler n−k+1n-k+1 leaves at time nn with capital 1/[p(a1)⋯p(ak)]1/\left[p(a_{1}) \cdots p(a_{k})\right] and gamblers j<n−k+1j < n-k+1 have all lost their capital. We will come back later to investigate the capital at time nn for gamblers n−k+2n-k+2 to nn.

First consider the string a1=0,a2=1a_{1}=0, a_{2}=1 with k=2k = 2. Find the sample values of Z1,Z2,Z3Z_{1}, Z_{2}, Z_{3} for the sample sequence X1=1,X2=0,X3=1,…X_{1} = 1, X_{2} = 0, X_{3} = 1, \ldots. Note that gamblers 1 and 3 have lost their capital, but gambler 2 now has capital 1/p0p11/p_{0}p_{1}. Show that the sample value of the stopping time for this case is J=3J = 3. Given an arbitrary sample value n≥2n \geq 2 for JJ, show that Zn=n−1/p0p1Z_{n} = n - 1/p_{0}p_{1}.

(c)

Find E[ZJ]\mathbb {E}\left[Z_{J}\right] from (a). Use this plus (b) to find E[J]\mathbb {E}\left[J\right].

(d)

Repeat (b) and (c) using the string (a1,…,ak)=(1,1)(a_{1}, \ldots , a_{k}) = (1,1) and initially assuming (X1X2X3)=(011)(X_{1}X_{2}X_{3}) = (011). Be careful about gambler 3 for J=3J = 3. Show that E[J]=1/p12+1/p1\mathbb {E}\left[J\right] = 1/p_{1}^{2} + 1/p_{1}.

(e)

Repeat (b) and (c) for (a1,…,ak)=(1,1,1,0,1,1)(a_{1}, \ldots , a_{k}) = (1,1,1,0,1,1).

(f)

Consider an arbitrary binary string a1,…,aka_{1}, \ldots , a_{k} and condition on J=nJ = n for some n≥kn \geq k. Show that the sample capital of gambler ℓ\ell is then equal to

  • 00 for ℓ<n−k\ell < n-k;

  • 1/[pa1pa2⋯pak]1/\left[p_{a_{1}}p_{a_{2}} \cdots p_{a_{k}}\right] for ℓ=n−k+1\ell = n-k+1;

  • 1/[pa1pa2⋯pai]1/\left[p_{a_{1}}p_{a_{2}} \cdots p_{a_{i}}\right] for ℓ=n−i+1\ell = n-i+1, 1≤i<k1 \leq i < k if (a1,…,ai)=(ak−i+1,…,ak)(a_{1}, \ldots , a_{i}) = (a_{k-i+1}, \ldots , a_{k});

  • 00 for ℓ=n−i+1\ell = n-i+1, 1≤i<k1 \leq i < k if (a1,…,ai)≠(ak−i+1,…,ak)(a_{1}, \ldots , a_{i}) \neq (a_{k-i+1}, \ldots , a_{k}).

Verify that this general formula agrees with (b), (d), and (e).

(g)

For a given binary string (a1,…,ak)(a_{1}, \ldots , a_{k}), and each jj, 1≤j≤k1 \leq j \leq k let Ij=1\mathbb {I}_{j} = 1 if (a1,…,aj)=(ak−j+1,…,ak)(a_{1}, \ldots , a_{j}) = (a_{k-j+1}, \ldots , a_{k}) and let Ij=0\mathbb {I}_{j} = 0 otherwise. Show that

E[J]=∑i=1kIi∏m=1ipam. \mathbb {E}\left[J\right] = \sum _{i=1}^{k} \frac{\mathbb {I}_{i}}{\prod _{m=1}^{i} p_{a_{m}}}.

Note that this is the same as the final result in Exercise 4.28. The argument is shorter here, but more motivated and insightful there. Both approaches are useful and lead to simple generalizations.

Problem 9.27
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(a)

This exercise shows why the condition E[∣ZJ∣]<∞\mathbb {E}\left[\left|Z_{J}\right|\right] < \infty is required in Lemma 9.8.4. Let Z1=−2Z_{1} = -2 and, for n≥1n \geq 1, let Zn+1=Zn[1+Xn(3n+1)/(n+1)]Z_{n+1} = Z_{n}\left[1 + X_{n}(3n+1)/(n+1)\right] where X1,X2,…X_{1}, X_{2}, \ldots are IID and take on the values +1+1 and −1-1 with probability 1/21/2 each. Show that {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} is a martingale.

(b)

Consider the stopping trial JJ such that JJ is the smallest value of n>1n > 1 for which ZnZ_{n} and Zn−1Z_{n-1} have the same sign. Show that, conditional on n<Jn < J, Zn=(−2)n/nZ_{n} = (-2)^{n}/n and, conditional on n=Jn = J, ZJ=−(−2)n(2n−1)/(n2−n)Z_{J} = -(-2)^{n}(2n-1)/(n^{2}-n).

(c)

Show that E[∣ZJ∣]\mathbb {E}\left[\left|Z_{J}\right|\right] is infinite, so that E[ZJ]\mathbb {E}\left[Z_{J}\right] does not exist according to the definition of expectation, and show that lim⁡n→∞E[Zn∣J>n]P(J>n)=0\lim_{n \rightarrow \infty } \mathbb {E}\left[Z_{n} \mid J > n\right] \mathbb {P}\left(J > n\right) = 0.

Problem 9.28

This exercise shows why the supremum of a martingale can behave markedly differently from the maximum of an arbitrarily large number of the variables. More precisely, it shows that P(sup⁡n≥1Zn≥a)\mathbb {P}\left(\sup_{n \geq 1} Z_{n} \geq a\right) can be unequal to P(⋃n≥1{Zn≥a})\mathbb {P}\left(\bigcup_{n \geq 1}\left\{ Z_{n} \geq a\right\} \right).

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(a)

Consider a martingale where ZnZ_{n} can take on only the values 2−n−12^{-n-1} and 1−2−n−11-2^{-n-1}, each with probability 1/21/2. Given that ZnZ_{n}, conditional on Zn−1Z_{n-1}, is independent of Z1,…Zn−2Z_{1}, \ldots Z_{n-2}, find P(Zn∣Zn−1)\mathbb {P}\left(Z_{n} \mid Z_{n-1}\right) for each nn so that the martingale condition is satisfied.

(b)

Show that P(sup⁡n≥1Zn≥1)=1/2\mathbb {P}\left(\sup_{n \geq 1} Z_{n} \geq 1\right) = 1/2 and show that P(⋃n{Zn≥1})=0\mathbb {P}\left(\bigcup_{n}\left\{ Z_{n} \geq 1\right\} \right) = 0.

(c)

Show that for every ϵ>0\epsilon > 0, P(sup⁡n≥1Zn≥a)≤E[Z1]/(a−ϵ)\mathbb {P}\left(\sup_{n \geq 1} Z_{n} \geq a\right) \leq \mathbb {E}\left[Z_{1}\right]/(a-\epsilon ).

(d)

Use (c) to establish

P(sup⁡n≥1Zn≥a)≤E[Z1]afor all a>0.(9.111) \mathbb {P}\left(\sup _{n \geq 1} Z_{n} \geq a\right) \leq \frac{\mathbb {E}\left[Z_{1}\right]}{a} \quad \text{for all } a > 0. \tag {9.111}
Problem 9.29

Theorem 9.7.4 states: assume that hh is a convex function from R\mathbb {R} to R\mathbb {R}, that {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} is a martingale, and that E[∣h(Zn)∣]<∞\mathbb {E}\left[\left|h(Z_{n})\right|\right] < \infty for all nn. Then {h(Zn);n≥1}\left\{ h(Z_{n}); n \geq 1\right\} is a submartingale.

Show that Theorem 9.7.4 is also valid for martingales relative to a joint process. That is, show that if hh is a convex function of a real variable and if {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} is a martingale relative to a joint process {Zn,Xn;n≥1}\left\{ Z_{n}, X_{n}; n \geq 1\right\}, then {h(Zn);n≥1}\left\{ h(Z_{n}); n \geq 1\right\} is a submartingale relative to {h(Zn),Xn;n≥1}\left\{ h(Z_{n}), X_{n}; n \geq 1\right\}.

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Problem 9.30

Show that if {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} is a martingale (submartingale or supermartingale) relative to a joint process {Zn,Xn;n≥1}\left\{ Z_{n}, X_{n}; n \geq 1\right\} and if JJ is a stopping trial for {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} relative to {Zn,Xn;n≥1}\left\{ Z_{n}, X_{n}; n \geq 1\right\}, then the stopped process is a martingale (submartingale or supermartingale) respectively relative to the joint process.

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Problem 9.31

Prove Corollaries 9.9.3 to 9.9.5, i.e., prove the following three statements.

  1. Let {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} be a martingale with E[Zn2]<∞\mathbb {E}\left[Z_{n}^{2}\right] < \infty for all n≥1n \geq 1. Then

    P(max⁡1≤n≤m∣Zn∣≥b)≤E[Zm2]b2for all integer m≥2, and all b>0. \mathbb {P}\left(\max _{1 \leq n \leq m} \left|Z_{n}\right| \geq b\right) \leq \frac{\mathbb {E}\left[Z_{m}^{2}\right]}{b^{2}} \quad \text{for all integer } m \geq 2, \text{ and all } b > 0.
  2. (Kolmogorov’s random walk inequality) Let {Sn;n≥1}\left\{ S_{n}; n \geq 1\right\} be a random walk with Sn=X1+⋯+XnS_{n} = X_{1} + \cdots + X_{n}, where {Xi;i≥i}\left\{ X_{i}; i \geq i\right\} is a set of IID rv’s with mean X‾\overline{X} and variance σ2\sigma^{2}. Then for any positive integer mm and any ϵ>0\epsilon > 0,

    P(max⁡1≤n≤m∣Sn−nX‾∣≥mϵ)≤σ2mϵ2. \mathbb {P}\left(\max _{1 \leq n \leq m} \left|S_{n} - n\overline{X}\right| \geq m\epsilon \right) \leq \frac{\sigma ^{2}}{m\epsilon ^{2}}.
  3. Let {Sn;n≥1}\left\{ S_{n}; n \geq 1\right\} be a random walk, Sn=X1+⋯+XnS_{n} = X_{1} + \cdots + X_{n}, where each XiX_{i} has mean X‾<0\overline{X} < 0 and semi-invariant MGF γ(r)\gamma (r). For any r>0r > 0 such that 0<γ(r)<∞0 < \gamma (r) < \infty, and for any α>0\alpha > 0,

    P(max⁡1≤i≤nSi≥α)≤exp⁡{−rα+nγ(r)}. \mathbb {P}\left(\max _{1 \leq i \leq n} S_{i} \geq \alpha \right) \leq \exp \left\{ -r\alpha + n\gamma (r)\right\} .
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Problem 9.32

Let {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} be the scaled branching process of Section 9.6.2: {Xn;n≥0}\left\{ X_{n}; n \geq 0\right\} is a branching process where XnX_{n} is the aggregate number of elements of generation nn, each element ii of generation nn (for 1≤i≤Xn1 \leq i \leq X_{n}) has a number of offspring Yi,nY_{i,n} collectively constituting generation n+1n+1 (i.e. Xn+1=∑i=1XnYi,nX_{n+1} = \sum_{i=1}^{X_{n}} Y_{i,n}), and the rv s Yi,nY_{i,n} are IID over both ii and nn with mean Y‾:=E[Yi,n]\overline{Y} := \mathbb {E}\left[Y_{i,n}\right]; the scaled process is Zn:=Xn/Y‾nZ_{n} := X_{n}/\overline{Y}^{n}.

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(a)

Assume that YY, the number of offspring of each element, has finite mean Y‾\overline{Y} and finite variance σ2\sigma^{2}. Assume that the population at time 0 is 1. Show that

E[Zn2]=1+σ2[Y‾−2+Y‾−3+⋯+Y‾−n−1]. \mathbb {E}\left[Z_{n}^{2}\right] = 1+\sigma ^{2}\left[\overline{Y}^{-2}+\overline{Y}^{-3}+\cdots +\overline{Y}^{-n-1}\right].
(b)

Assume that Y‾>1\overline{Y} > 1 and find lim⁡n→∞E[Zn2]\lim_{n \rightarrow \infty } \mathbb {E}\left[Z_{n}^{2}\right]. Show from this that the conditions for the martingale convergence theorem, Theorem 9.9.8, are satisfied.

Remark.
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Theorem 9.9.8 (Martingale convergence theorem): let {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} be a martingale and assume that there is some finite MM such that E[Zn2]≤M\mathbb {E}\left[Z_{n}^{2}\right] \leq M for all nn. Then there is a rv ZZ such that, for all sample sequences except a set of probability 0, lim⁡n→∞Zn=Z\lim_{n \rightarrow \infty } Z_{n} = Z.

Note: this condition is not satisfied if Y‾≤1\overline{Y} \leq 1. The general martingale convergence theorem requires only a bounded first absolute moment, so it holds for Y‾≤1\overline{Y} \leq 1.

Problem 9.33

Show that if {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} is a martingale (submartingale or supermartingale) relative to a joint process {Zn,Xn;n≥1}\left\{ Z_{n}, X_{n}; n \geq 1\right\} and if JJ is a stopping trial for {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} relative to {Zn,Xn;n≥1}\left\{ Z_{n}, X_{n}; n \geq 1\right\}, then the stopped process satisfies (9.98), (9.99), or (9.100) respectively.

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Problem 9.34

Show that if {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} is a martingale relative to a joint process {Zn,Xn;n≥1}\left\{ Z_{n}, X_{n}; n \geq 1\right\} and if JJ is a stopping trial for {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} relative to {Zn,Xn;n≥1}\left\{ Z_{n}, X_{n}; n \geq 1\right\}, then E[ZJ]=E[Z1]\mathbb {E}\left[Z_{J}\right] = \mathbb {E}\left[Z_{1}\right] if and only if (9.104) is satisfied.

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Problem 9.35

Consider an investment example similar to that of Example 9.10.1 in which there is only one investment other than cash. The ratio XX of that investment value at the end of an epoch to that at the beginning is either 2 or 1/41/4, each with equal probability. Thus P(X=2)=1/2\mathbb {P}\left(X=2\right) = 1/2 and P(X=1/4)=1/2\mathbb {P}\left(X=1/4\right) = 1/2. Successive ratios, X1,X2,…X_{1}, X_{2}, \ldots, are IID.

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(a)

In (a)--(c), assume a fixed allocation strategy where a fraction λ\lambda is kept in the double-or-quarter investment and 1−λ1-\lambda is kept in cash. Find the expected wealth, Wn(λ)W_{n}(\lambda ) and the expected log wealth E[Ln(λ)]\mathbb {E}\left[L_{n}(\lambda )\right] as a function of a constant λ∈[0,1]\lambda \in \left[0,1\right] and n≥1n \geq 1. Assume unit initial wealth.

(b)

For λ=1\lambda =1, find the PMF for W4(1)W_{4}(1) and give a brief explanation of why E[Wn(1)]\mathbb {E}\left[W_{n}(1)\right] is growing exponentially with nn but E[Ln(1)]\mathbb {E}\left[L_{n}(1)\right] is decreasing linearly toward −∞-\infty.

(c)

Using the same approach as in Example 9.10.1, find the value of λ∗\lambda^{*} that maximizes E[Ln(λ)]\mathbb {E}\left[L_{n}(\lambda )\right]. Show that your solution satisfies the optimality conditions in (9.136).

(d)

Find the range of λ\lambda over which E[Ln(λ)]\mathbb {E}\left[L_{n}(\lambda )\right] is positive.

(e)

Find the PMF of the rv Zn/Zn−1Z_{n}/Z_{n-1} as given in (9.138) for any given λn\lambda_{n}.

Problem 9.36

An interesting special case of this simple theory of investment is the horse-racing game due to J. Kelly. There are ℓ−1\ell -1 horses in a race and each horse jj wins with some probability p(j)>0p(j) > 0. One and only one horse wins, and if jj wins, the gambler receives r(j)>0r(j) > 0 for each dollar bet on jj and nothing for the bets on other horses. In other words, the price-relative X(j)X(j) for each jj, 1≤j≤ℓ−11 \leq j \leq \ell -1, is r(j)r(j) with probability p(j)p(j) and 0 otherwise. For cash, X(ℓ)=1X(\ell )=1.

The gambler's allocation of wealth on the race is denoted by λ(j)\lambda (j) on each horse jj with λ(ℓ)\lambda (\ell ) kept in cash. As usual, ∑jλ(j)=1\sum_{j} \lambda (j)=1 and λ(j)≥0\lambda (j) \geq 0 for 1≤j≤ℓ1 \leq j \leq \ell. Note that X(1),…,X(ℓ−1)X(1), \ldots , X(\ell -1) are highly dependent, since only one has a nonzero sample value in any race.

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(a)

For any given allocation λ→\overrightarrow {\lambda } find the expected wealth and the expected log wealth at the end of a race for unit initial wealth.

(b)

Assume that a statistically identical sequence of races is run, i.e., X→1,X→2,…\overrightarrow {X}_{1}, \overrightarrow {X}_{2}, \ldots, are IID, where each X→n=(Xn(1),…,Xn(ℓ))T\overrightarrow {X}_{n} = \left(X_{n}(1), \ldots , X_{n}(\ell )\right)^{\mathsf{T}}. Assuming a constant allocation on each race and unit initial wealth, find the expected log wealth E[Ln(λ→)]\mathbb {E}\left[L_{n}(\overrightarrow {\lambda })\right] at the end of the nnth race and express it as nE[Y(λ→)]n\mathbb {E}\left[Y(\overrightarrow {\lambda })\right].

(c)

Let λ→∗\overrightarrow {\lambda }^{*} maximize E[Y(λ→)]\mathbb {E}\left[Y(\overrightarrow {\lambda })\right]. Use the necessary and sufficient condition (9.136) on λ→∗\overrightarrow {\lambda }^{*} for horse jj, 1≤j<ℓ1 \leq j < \ell to show that λ∗(j)\lambda^{*}(j) can be expressed in the following two equivalent ways; each uniquely specifies each λ∗(j)\lambda^{*}(j) in terms of λ∗(ℓ)\lambda^{*}(\ell ).

λ∗(j)≥p(j)−λ∗(ℓ)r(j);with equality if λ∗(j)>0.λ∗(j)=max⁡{p(j)−λ∗(ℓ)r(j),0}. \begin{align} \lambda ^{*}(j) & \geq p(j)-\frac{\lambda ^{*}(\ell )}{r(j)}; \qquad \text{with equality if } \lambda ^{*}(j) > 0. \\ \lambda ^{*}(j) & = \max \left\{ p(j)-\frac{\lambda ^{*}(\ell )}{r(j)}, 0\right\} . \end{align}

Solving for λ∗(ℓ)\lambda^{*}(\ell ) (which in turn specifies the other components of λ→∗\overrightarrow {\lambda }^{*}) breaks into 3 special cases which are treated below in (d), (e), and (f) respectively. The first case, in (d), shows that if ∑j<ℓ1/r(j)<1\sum_{j<\ell } 1/r(j) < 1, then λ∗(ℓ)=0\lambda^{*}(\ell )=0. The second case, in (e), shows that if ∑j<ℓ1/r(j)>1\sum_{j<\ell } 1/r(j) > 1, then λ∗(ℓ)>min⁡j(p(j)r(j))\lambda^{*}(\ell ) > \min_{j}(p(j)r(j)), with the specific value specified by the unique solution to

∑j<ℓp(j)max⁡[p(j)r(j),  λ∗(ℓ)]≤1;with equality if λ∗(ℓ)>0. \sum _{j<\ell } \frac{p(j)}{\max \left[p(j)r(j), \; \lambda ^{*}(\ell )\right]} \leq 1; \qquad \text{with equality if } \lambda ^{*}(\ell ) > 0.

The third case, in (f), shows that if ∑j<ℓ1/r(j)=1\sum_{j<\ell } 1/r(j)=1, then λ∗(ℓ)\lambda^{*}(\ell ) is nonunique, and its set of possible values occupy the range [0,min⁡j(p(j)r(j))]\left[0, \min_{j}\left(p(j)r(j)\right)\right].

(d)

Sum the first inequality of (c) over jj to show that if λ∗(ℓ)>0\lambda^{*}(\ell ) > 0, then ∑j<ℓ1/r(j)≥1\sum_{j<\ell } 1/r(j) \geq 1. Note that the logical obverse of this is that ∑j<ℓ1/r(j)<1\sum_{j<\ell } 1/r(j) < 1 implies that λ∗(ℓ)=0\lambda^{*}(\ell )=0 and thus that λ∗(j)=p(j)\lambda^{*}(j)=p(j) for each horse jj.

(e)

In (c), λ∗(j)\lambda^{*}(j) for each j<ℓj < \ell was specified in terms of λ∗(ℓ)\lambda^{*}(\ell ); here you are to use the necessary and sufficient condition (9.136) on cash to specify λ∗(ℓ)\lambda^{*}(\ell ). More specifically, you are to show that λ∗(ℓ)\lambda^{*}(\ell ) satisfies each of the following two equivalent inequalities:

∑j<ℓp(j)λ∗(j)r(j)+λ∗(ℓ)≤1;with equality if λ∗(ℓ)>0.∑j<ℓp(j)max⁡[p(j)r(j),  λ∗(ℓ)]≤1;with equality if λ∗(ℓ)>0. \begin{align} \sum _{j<\ell } \frac{p(j)}{\lambda ^{*}(j)r(j)+\lambda ^{*}(\ell )} & \leq 1; \qquad \text{with equality if } \lambda ^{*}(\ell ) > 0. \\ \sum _{j<\ell } \frac{p(j)}{\max \left[p(j)r(j), \; \lambda ^{*}(\ell )\right]} & \leq 1; \qquad \text{with equality if } \lambda ^{*}(\ell ) > 0. \end{align}

Show from the second inequality above that if λ∗(ℓ)≤p(j)r(j)\lambda^{*}(\ell ) \leq p(j)r(j) for each jj, then ∑j1/r(j)≤1\sum_{j} 1/r(j) \leq 1. Point out that the logical obverse of this is that if ∑j1/r(j)>1\sum_{j} 1/r(j) > 1, then λ∗(ℓ)>min⁡j(p(j)r(j))\lambda^{*}(\ell ) > \min_{j}\left(p(j)r(j)\right). Explain why this second inequality has a unique solution for λ∗(ℓ)\lambda^{*}(\ell ) in this case. Note that λ∗(j)=0\lambda^{*}(j)=0 for each jj such that p(j)r(j)<λ∗(ℓ)p(j)r(j) < \lambda^{*}(\ell ).

(f)

Now consider the case in which ∑j<ℓ1/r(j)=1\sum_{j<\ell } 1/r(j)=1. Show that the second inequality of (e) is satisfied with equality for each choice of λ∗(ℓ)\lambda^{*}(\ell ), 0≤λ∗(ℓ)≤min⁡j<ℓ(p(j)r(j))0 \leq \lambda^{*}(\ell ) \leq \min_{j<\ell }\left(p(j)r(j)\right).

(g)

Consider the special case of a race with only two horses. Let p(1)=p(2)=1/2p(1)=p(2)=1/2. Assume that r(1)r(1) and r(2)r(2) are large enough to satisfy 1/r(1)+1/r(2)<11/r(1)+1/r(2) < 1; thus no cash allotment is used in maximizing E[Y(λ→)]\mathbb {E}\left[Y(\overrightarrow {\lambda })\right]. With λ(3)=0\lambda (3)=0, we have

E[Y(λ→)]=12ln⁡[λ(1)r(1)]+12ln⁡[λ(2)r(2)]=12ln⁡[λ(1)r(1)(1−λ(1))r(2)]. \mathbb {E}\left[Y(\overrightarrow {\lambda })\right] = \frac{1}{2}\ln \left[\lambda (1)r(1)\right]+\frac{1}{2}\ln \left[\lambda (2)r(2)\right] = \frac{1}{2}\ln \left[\lambda (1)r(1)\left(1-\lambda (1)\right)r(2)\right].

Use this equation to give an intuitive explanation for why λ∗(1)=1/2\lambda^{*}(1)=1/2, independent of r(1)r(1) and r(2)r(2).

(h)

Again consider the special case of two horses with p(1)=p(2)=1/2p(1)=p(2)=1/2, but let r(1)=3r(1)=3 and r(2)=3/2r(2)=3/2. Show that λ→∗\overrightarrow {\lambda }^{*} is nonunique with (1/2,1/2,0)(1/2,1/2,0) and (1/4,0,3/4)(1/4,0,3/4) as possible values. Show that if r(2)>3/2r(2) > 3/2, then the first solution above is unique, and if r(2)<3/2r(2) < 3/2, then the second solution is unique, assuming p(1)=1/2p(1)=1/2 and r(1)=3r(1)=3 throughout.

(i)

For the case where ∑j<ℓ1/r(j)=1\sum_{j<\ell } 1/r(j)=1, define q(j)=1/r(j)q(j)=1/r(j) as a PMF on {1,…,ℓ−1}\left\{ 1, \ldots , \ell -1\right\}. Show that E[Y(λ→∗)]=D(p∥q)\mathbb {E}\left[Y(\overrightarrow {\lambda }^{*})\right] = D(\mathsf{p} \parallel \mathsf{q}) for the conditions of (f).

Remark.
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Note (part (h)): the case 3/2<r(2)<23/2 < r(2) < 2 provides an example where an investment is used to maximize log-wealth even though E[X(2)]=p(2)r(2)<1\mathbb {E}\left[X(2)\right] = p(2)r(2) < 1, i.e., horse 2 is a poor investment, but is preferable to cash in this case as a hedge against horse 1 losing.

Note (part (i)): to interpret this, we could view a horse race where each horse jj has probability q(j)q(j) of winning the reward r(j)=1/q(j)r(j)=1/q(j) as a fair game'. Our gambler, who knows that the true probabilities are $\left\{ p(j); 1 \leq j < \ell \right\}$, then has inside information' if p(j)≠q(j)p(j) \neq q(j), and can establish a positive rate of return equal to D(p∥q)D(\mathsf{p} \parallel \mathsf{q}).

Problem 9.37

Let the function hϵ(u)h_{\epsilon }(u) have the value 0 for −ϵ<E[Y(λ→)]<ϵ-\epsilon < \mathbb {E}\left[Y(\overrightarrow {\lambda })\right] < \epsilon and the value 1 elsewhere.

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(a)

Show that lim⁡n→∞hϵ(Ln(λ→))=0\lim_{n \rightarrow \infty } h_{\epsilon }\left(L_{n}(\overrightarrow {\lambda })\right) = 0 WP1.

(b)

Show that for all ϵ>0,δ>0\epsilon > 0, \delta > 0, there is an non_{o} such that

P(⋃n≥no{∣1nLn(λ→)−E[Y(λ→)]∣>ϵ})≤δ. \mathbb {P}\left(\bigcup _{n \geq n_{o}} \left\{ \left|\frac{1}{n}L_{n}(\overrightarrow {\lambda })-\mathbb {E}\left[Y(\overrightarrow {\lambda })\right]\right| > \epsilon \right\} \right) \leq \delta .
(c)

Show that the probability of the complementary event is

P(⋂n≥no{E[Y(λ→)]−ϵ≤1nLn(λ→)≤E[Y(λ→)]+ϵ})>1−δ. \mathbb {P}\left(\bigcap _{n \geq n_{o}} \left\{ \mathbb {E}\left[Y(\overrightarrow {\lambda })\right]-\epsilon \leq \frac{1}{n}L_{n}(\overrightarrow {\lambda }) \leq \mathbb {E}\left[Y(\overrightarrow {\lambda })\right]+\epsilon \right\} \right) > 1-\delta .
(d)

Show that

P(⋂n≥no{exp⁡[n(E[Y(λ→)]−ϵ)]≤Wn(λ→)≤exp⁡[n(E[Y(λ→)]+ϵ)]})>1−δ. \mathbb {P}\left(\bigcap _{n \geq n_{o}} \left\{ \exp \left[n\left(\mathbb {E}\left[Y(\overrightarrow {\lambda })\right]-\epsilon \right)\right] \leq W_{n}(\overrightarrow {\lambda }) \leq \exp \left[n\left(\mathbb {E}\left[Y(\overrightarrow {\lambda })\right]+\epsilon \right)\right]\right\} \right) > 1-\delta .
Remark.
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Theorem 9.10.2: assume that E[∣Y(λ→)∣]<∞\mathbb {E}\left[\left|Y(\overrightarrow {\lambda })\right|\right] < \infty. Then for every ϵ,δ>0\epsilon , \delta > 0, there exists an non_{o} such that

P(⋂n≥no{exp⁡[n(E[Y(λ→)]−ϵ)]≤Wn(λ→)≤exp⁡[n(E[Y(λ→)]+ϵ)]})>1−δ. \mathbb {P}\left(\bigcap _{n \geq n_{o}} \left\{ \exp \left[n\left(\mathbb {E}\left[Y(\overrightarrow {\lambda })\right]-\epsilon \right)\right] \leq W_{n}(\overrightarrow {\lambda }) \leq \exp \left[n\left(\mathbb {E}\left[Y(\overrightarrow {\lambda })\right]+\epsilon \right)\right]\right\} \right) > 1-\delta .
Problem 9.38

In this exercise, you are to express the buy-and-hold strategy and the two-pile strategy of (9.137) as special cases of a time-varying allocation.

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(a)

Suppose the buy-and-hold strategy starts with an allocation λ→1\overrightarrow {\lambda }_{1}. Let Wn(k)W_{n}(k) be the investor's wealth in investment kk at time nn. Show that

Wn(k)=λ1(k)∏m=1nXm(k). W_{n}(k) = \lambda _{1}(k) \prod _{m=1}^{n} X_{m}(k).
(b)

Show that λn(k)\lambda_{n}(k) can be expressed in each of the following ways:

λn(k)=Wn−1(k)∑jWn−1(j)=λn−1(k)Xn−1(k)∑jλn−1(j)Xn−1(j). \lambda _{n}(k) = \frac{W_{n-1}(k)}{\sum _{j} W_{n-1}(j)} = \frac{\lambda _{n-1}(k)X_{n-1}(k)}{\sum _{j} \lambda _{n-1}(j)X_{n-1}(j)}.
(c)

For the two pile strategy of (9.137), verify (9.137) and find an expression for λn(k)\lambda_{n}(k).

Problem 9.39
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(a)

Consider the martingale {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} where Zn=Wn/Wn∗Z_{n} = W_{n}/W_{n}^{*} and WnW_{n} is the wealth at time nn for the pure triple or nothing strategy of Example 9.10.1 and Wn∗W_{n}^{*} is the wealth using the constant allocation λ→∗\overrightarrow {\lambda }^{*}. Find the PMF of ZZ where Z=lim⁡n→∞ZnZ = \lim_{n \rightarrow \infty } Z_{n} WP1.

(b)

Now consider the two-pile strategy of (9.137) and find the PMF of the limiting rv ZZ for this strategy.

(c)

Now consider a `tempt-the-fates' strategy where one uses the pure triple or nothing strategy for the first three epochs and uses the constant allocation λ→∗\overrightarrow {\lambda }^{*} thereafter. Again find the PMF of the limiting rv ZZ.

Problem 9.40

Consider the Markov modulated random walk in the figure below. The rv s YnY_{n} in this example take on only a single value for each transition, that value being 1 for all transitions from state 1, 10 for all transitions from state 2, and 0 otherwise. ϵ>0\epsilon > 0 is a very small number, say ϵ<10−6\epsilon < 10^{-6}.

 {#fig-1 width="50%"}

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(a)

Show that the steady-state gain per transition is 5.5/(1+ϵ)5.5/(1+\epsilon ). Show that the relative-gain vector is w→=(0,(ϵ−4.5)/[ϵ(1+ϵ)],(10ϵ+4.5)/[ϵ(1+ϵ)])\overrightarrow {w} = \left(0, (\epsilon -4.5)/\left[\epsilon (1+\epsilon )\right], (10\epsilon +4.5)/\left[\epsilon (1+\epsilon )\right]\right).

(b)

Let Sn=Y0+Y1+⋯+Yn−1S_{n} = Y_{0}+Y_{1}+\cdots +Y_{n-1} and take the starting state X0X_{0} to be 0. Let JJ be the smallest value of nn for which Sn≥100S_{n} \geq 100. Find P(J=11)\mathbb {P}\left(J=11\right) and P(J=101)\mathbb {P}\left(J=101\right). Find an estimate of E[J]\mathbb {E}\left[J\right] that is exact in the limit ϵ→0\epsilon \rightarrow 0.

(c)

Show that P(XJ=1)=(1−45ϵ+o(ϵ))/2\mathbb {P}\left(X_{J}=1\right) = (1-45\epsilon +o(\epsilon ))/2 and that P(XJ=2)=(1+45ϵ+o(ϵ))/2\mathbb {P}\left(X_{J}=2\right) = (1+45\epsilon +o(\epsilon ))/2. Verify, to first order in ϵ\epsilon, that (9.159) is satisfied.

Problem 9.41

Show that (9.159) results from taking the derivative of (9.163) and evaluating it at r=0r=0.

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Problem 9.42

Let {Zn;n≥1}\left\{ Z_{n}; n \geq 1\right\} be a martingale, and for some integer mm, let Yn=Zn+m−ZmY_{n} = Z_{n+m}-Z_{m}.

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(a)

Show that E[Yn∣Zn+m−1=zn+m−1,Zn+m−2=zn+m−2,…,Zm=zm,…,Z1=z1]=zn+m−1−zm\mathbb {E}\left[Y_{n} \mid Z_{n+m-1}=z_{n+m-1}, Z_{n+m-2}=z_{n+m-2}, \ldots , Z_{m}=z_{m}, \ldots , Z_{1}=z_{1}\right] = z_{n+m-1}-z_{m}.

(b)

Show that E[Yn∣Yn−1=yn−1,…,Y1=y1]=yn−1\mathbb {E}\left[Y_{n} \mid Y_{n-1}=y_{n-1}, \ldots , Y_{1}=y_{1}\right] = y_{n-1}.

(c)

Show that E[∣Yn∣]<∞\mathbb {E}\left[\left|Y_{n}\right|\right] < \infty. Note that (b) and (c) show that {Yn;n≥1}\left\{ Y_{n}; n \geq 1\right\} is a martingale.

Problem 9.43

This exercise views the continuous-time branching process of Exercise 7.15 as a stopped random walk. Recall that the process was specified there as a Markov process such that for each state jj, j≥0j \geq 0, the transition rate to j+1j+1 is jλj\lambda and to j−1j-1 is jμj\mu. There are no other transitions, and in particular, there are no transitions out of state 0, so that the Markov process is reducible. Recall that the embedded Markov chain is the same as the embedded chain of an M/M/1\mathrm{M}/\mathrm{M}/1 queue except that there is no transition from state 0 to state 1.

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(a)

To model the possible extinction of the population, convert the embedded Markov chain above to a stopped random walk, {Sn;n≥0}\left\{ S_{n}; n \geq 0\right\}. The stopped random walk starts at S0=0S_{0}=0 and stops on reaching a threshold at −1-1. Before stopping, it moves up by 1 with probability λ/(λ+μ)\lambda /(\lambda +\mu ) and downward by 1 with probability μ/(λ+μ)\mu /(\lambda +\mu ) at each step. Give the (very simple) relationship between the state XnX_{n} of the Markov chain and the state SnS_{n} of the stopped random walk for each n≥0n \geq 0.

(b)

Find the probability that the population eventually dies out as a function of λ\lambda and μ\mu. Be sure to consider all three cases λ>μ\lambda > \mu, λ<μ\lambda < \mu, and λ=μ\lambda =\mu.