Chapter IV

Martingales. Discrete and Continuous Time

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Problem IV.1

Let a random variable ξ∈L2(Ω,F,P)\xi \in L^{2}(\Omega , \mathscr {F}, \mathbb {P}) and a σ\sigma-algebra A⊂F\mathscr {A} \subset \mathscr {F} be given. Denote by Pr⁡H\operatorname {Pr}_{H} the operator of orthogonal projection onto the subspace H=L2(Ω,A,P)⊂L2(Ω,F,P)H = L^{2}(\Omega , \mathscr {A}, \mathbb {P}) \subset L^{2}(\Omega , \mathscr {F}, \mathbb {P}). Show that E[ξ∣A]=Pr⁡Hξ\mathbb {E}\left[\xi \mid \mathscr {A}\right] = \operatorname {Pr}_{H} \xi.

Note that since the space L2(Ω,F,P)L^{2}(\Omega , \mathscr {F}, \mathbb {P}) is dense in the space L1(Ω,F,P)L^{1}(\Omega , \mathscr {F}, \mathbb {P}), the operator Pr⁡H\operatorname {Pr}_{H} can be extended by continuity to L1(Ω,F,P)L^{1}(\Omega , \mathscr {F}, \mathbb {P}).

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Problem IV.2

Let (Ft)t⩾0\left(\mathscr {F}_{t}\right)_{t \geqslant 0} be a filtration. Prove that τ\tau is an optional time with respect to the filtration (Ft)t⩾0\left(\mathscr {F}_{t}\right)_{t \geqslant 0} if and only if τ\tau is a Markov time with respect to the filtration (Ft+)t⩾0(\mathscr {F}_{t+})_{t \geqslant 0}, where Ft+=⋂s>tFs\mathscr {F}_{t+} = \bigcap_{s > t} \mathscr {F}_{s}, t⩾0t \geqslant 0.

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Problem IV.3

Let (Ft)t⩾0\left(\mathscr {F}_{t}\right)_{t \geqslant 0} be a filtration and τ\tau an optional time with respect to it. Introduce the σ\sigma-algebra

Fτ+={A∈F:A∩{τ<t}∈Ft for every t⩾0}. \mathscr {F}_{\tau +} = \left\{ A \in \mathscr {F}: A \cap \left\{ \tau < t\right\} \in \mathscr {F}_{t} \text{ for every } t \geqslant 0\right\} .

Can one assert that in relation (51) the σ\sigma-algebra G=Fσ+\mathscr {G} = \mathscr {F}_{\sigma +}?

Relation (51) arises in the proof of Lemma 7. Lemma 7: let X=(Xt,Ft)t⩾0X = \left(X_{t}, \mathscr {F}_{t}\right)_{t \geqslant 0} be a martingale with a.s. right-continuous sample paths, and let σ\sigma and τ\tau be optional times (with respect to the filtration F=(Ft)t⩾0\mathbb {F} = \left(\mathscr {F}_{t}\right)_{t \geqslant 0}) such that σ⩽τ⩽c\sigma \leqslant \tau \leqslant c a.s., where cc is a positive constant; then E[Xτ]=E[Xσ]\mathbb {E}\left[X_{\tau }\right] = \mathbb {E}\left[X_{\sigma }\right]. In the proof one sets σ(n)=2−n[2nσ+1]\sigma (n) = 2^{-n}\left[2^{n} \sigma +1\right], n∈Nn \in \mathbb {N}, where [⋅][\cdot ] denotes the integer part of a number, and

G=⋂n=1∞F‾σ(n)(51) \mathscr {G} = \bigcap _{n = 1}^{\infty } \overline{\mathscr {F}}_{\sigma (n)} \tag {51}

where A‾=σ{A,N}\overline{\mathscr {A}} = \sigma \left\{ \mathscr {A}, \mathscr {N}\right\} for a σ\sigma-algebra A\mathscr {A}, and N\mathscr {N} is the class of P\mathbb {P}-null sets.

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Problem IV.4

Let {ξn}n⩾1\left\{ \xi_{n}\right\}_{n \geqslant 1} be a sequence of integrable random variables, Fn=σ{ξ1,…,ξn},Xn=∑k=1nξk,n∈N\mathscr {F}_{n} = \sigma \left\{ \xi_{1}, \ldots , \xi_{n}\right\} , X_{n} = \sum_{k = 1}^{n} \xi_{k}, n \in \mathbb {N}. Prove that (Xn,Fn)n⩾1\left(X_{n}, \mathscr {F}_{n}\right)_{n \geqslant 1} is a martingale if and only if

E[ξn+1fn(ξ1,…,ξn)]=0 \mathbb {E}\left[\xi _{n+1} f_{n}\left(\xi _{1}, \ldots , \xi _{n}\right)\right] = 0

for every bounded Borel function fn:Rn→Rf_{n}: \mathbb {R}^{n} \rightarrow \mathbb {R} and all n⩾1n \geqslant 1.

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Problem IV.5

Prove that if (ξn,Fn)n⩾1\left(\xi_{n}, \mathscr {F}_{n}\right)_{n \geqslant 1} is a sequence of independent centered random variables and Fn=σ{ξ1,…,ξn},n⩾1\mathscr {F}_{n} = \sigma \left\{ \xi_{1}, \ldots , \xi_{n}\right\} , n \geqslant 1, then the process (Sn,m,Fn)n⩾m\left(S_{n, m}, \mathscr {F}_{n}\right)_{n \geqslant m} of the form

Sn,m=∑1⩽i1<⋯<im⩽nξi1⋯ξim,n⩾m S_{n, m} = \sum _{1 \leqslant i_{1} < \cdots < i_{m} \leqslant n} \xi _{i_{1}} \cdots \xi _{i_{m}}, \quad n \geqslant m

is a martingale with zero mean for each m⩾1m \geqslant 1.

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Problem IV.6

Let an urn contain aa white and bb black balls. Balls are drawn from the urn successively at random, one at a time; each drawn ball is returned to the urn together with dd additional balls of the same color as the drawn one. Denote by Xn(n⩾1)X_{n}(n \geqslant 1) the ratio of the number of white balls to the total number of balls in the urn after the nn-th drawing of a ball and its return to the urn together with dd balls of the corresponding color (X0:=a/(a+b))\left(X_{0}: = a /(a+b)\right). Prove that X={Xn,n⩾0}X = \left\{ X_{n}, n \geqslant 0\right\} is a martingale.

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Problem IV.7

Let (Ω,F,P)=([0,1],B[0,1],mes)(\Omega , \mathscr {F}, \mathbb {P}) = ([0,1], \mathscr {B}[0,1], \mathrm{mes}), where mes is the Lebesgue measure. Let {Tn,n⩾1}\left\{ T_{n}, n \geqslant 1\right\} be a sequence of refining partitions of the segment [0,1][0,1], i.e., TnT_{n} consists of the points 0=tn,0<⋯<tn,mn=10 = t_{n, 0} < \cdots < t_{n, m_{n}} = 1 and Tn⊂Tn+1,n⩾1,mn⩾2T_{n} \subset T_{n+1}, n \geqslant 1, m_{n} \geqslant 2. Introduce the σ\sigma-algebras Fn\mathscr {F}_{n} on [0,1][0,1] generated by the sets Δn,1=[0,tn,0]\Delta_{n, 1} = \left[0, t_{n, 0}\right], Δn,k=(tn,k−1,tn,k],2⩽k⩽mn,n⩾1\Delta_{n, k} = \left(t_{n, k-1}, t_{n, k}\right], 2 \leqslant k \leqslant m_{n}, n \geqslant 1. For a function f:[0,1]→Rf:[0,1] \rightarrow \mathbb {R} set

Xn(ω)=∑k=1mnf(tn,k)−f(tn,k−1)tn,k−tn,k−11Δn,k(ω),n⩾1 X_{n}(\omega ) = \sum _{k = 1}^{m_{n}} \frac{f\left(t_{n, k}\right)-f\left(t_{n, k-1}\right)}{t_{n, k}-t_{n, k-1}} \mathbf{1}_{\Delta _{n, k}}(\omega ), \quad n \geqslant 1

Prove that (Xn,Fn)n⩾1\left(X_{n}, \mathscr {F}_{n}\right)_{n \geqslant 1} is a martingale.

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Problem IV.8

Under the conditions of the previous exercise, let the function ff satisfy the Lipschitz condition, i.e., ∣f(x)−f(y)∣⩽L∣x−y∣\left|f(x)-f(y)\right| \leqslant L\left|x-y\right| for x,y∈[0,1]x, y \in [0,1] and some constant L>0L > 0. Let the partitions TnT_{n} be such that max⁡0⩽k⩽mn(tn,k−tn,k−1)→0\max_{0 \leqslant k \leqslant m_{n}}\left(t_{n, k}-t_{n, k-1}\right) \rightarrow 0, n→∞n \rightarrow \infty. Prove that ⋁n=1∞Fn(:=σ{Fn,n∈N})=B([0,1])\bigvee_{n = 1}^{\infty } \mathscr {F}_{n}\left(: = \sigma \left\{ \mathscr {F}_{n}, n \in \mathbb {N}\right\} \right) = \mathscr {B}([0,1]) and the family {Xn,n⩾1}\left\{ X_{n}, n \geqslant 1\right\} is uniformly integrable. Then by Theorem 7, Xn→X∞X_{n} \rightarrow X_{\infty } a.s. and in L1L^{1} as n→∞n \rightarrow \infty, where X∞∈B([0,1])∣B(R)X_{\infty } \in \mathscr {B}([0,1]) \mid \mathscr {B}(\mathbb {R}).

Theorem 7. Let X=(Xn,Fn)n⩾1X = \left(X_{n}, \mathscr {F}_{n}\right)_{n \geqslant 1} be a martingale. The following conditions are equivalent:

  1. Xn=E[ξ∣Fn]X_{n} = \mathbb {E}\left[\xi \mid \mathscr {F}_{n}\right], n⩾1n \geqslant 1, where ξ\xi is some integrable random variable;

  2. the family {Xn,n⩾1}\left\{ X_{n}, n \geqslant 1\right\} is uniformly integrable;

  3. there exists an integrable random variable X∞X_{\infty } such that convergence in L1L^{1} holds:

    E[∣X∞−Xn∣]→0asn→∞; \mathbb {E}\left[\left|X_{\infty }-X_{n}\right|\right] \rightarrow 0 \quad \text{as} \quad n \rightarrow \infty ;
  4. sup⁡nE[∣Xn∣]<∞\sup_{n} \mathbb {E}\left[\left|X_{n}\right|\right] < \infty and Xk=E[X∞∣Fk]X_{k} = \mathbb {E}\left[X_{\infty } \mid \mathscr {F}_{k}\right] with X∞=lim⁡n→∞XnX_{\infty } = \lim_{n \rightarrow \infty } X_{n} (the limit exists a.s. and is integrable by Theorem 6), k⩾1k \geqslant 1.

Consequently, for any [a,b]⊂[0,1][a, b] \subset [0,1]

∫abXndt=E[Xn1[a,b]]→E[X∞1[a,b]]=∫abX∞dt,n→∞(63) \int _{a}^{b} X_{n} d t = \mathbb {E}\left[X_{n} \mathbf{1}_{[a, b]}\right] \rightarrow \mathbb {E}\left[X_{\infty } \mathbf{1}_{[a, b]}\right] = \int _{a}^{b} X_{\infty } d t, \quad n \rightarrow \infty \tag {63}

Since also ∫abXndt→f(b)−f(a)\int_{a}^{b} X_{n} d t \rightarrow f(b)-f(a) as n→∞n \rightarrow \infty (explain why), together with (63) we conclude from this that the function ff is absolutely continuous and its density (with respect to Lebesgue measure) can be found as the limit of the variables Xn,n→∞X_{n}, n \rightarrow \infty.

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Problem IV.9

Let {Xn}n⩾1\left\{ X_{n}\right\}_{n \geqslant 1} be a sequence of real random variables defined on a probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}), and let fn0(x1,…,xn)f_{n}^{0}\left(x_{1}, \ldots , x_{n}\right) and fn1(x1,…,xn)f_{n}^{1}\left(x_{1}, \ldots , x_{n}\right) be some probability densities with respect to Lebesgue measure on (Rn,B(Rn)),n⩾1\left(\mathbb {R}^{n}, \mathscr {B}\left(\mathbb {R}^{n}\right)\right), n \geqslant 1. Consider the likelihood ratio

Ln(ω)=fn1(X1(ω),…,Xn(ω))fn0(X1(ω),…,Xn(ω)),n⩾1 L_{n}(\omega ) = \frac{f_{n}^{1}\left(X_{1}(\omega ), \ldots , X_{n}(\omega )\right)}{f_{n}^{0}\left(X_{1}(\omega ), \ldots , X_{n}(\omega )\right)}, \quad n \geqslant 1

assuming that fn1(z)=0f_{n}^{1}(z) = 0 if fn0(z)=0f_{n}^{0}(z) = 0 (in which case Ln(z):=0L_{n}(z): = 0 ), z∈Rnz \in \mathbb {R}^{n}. Show that (Ln,Fn)n⩾1\left(L_{n}, \mathscr {F}_{n}\right)_{n \geqslant 1} is a martingale, where Fn=σ{X1,…,Xn},n⩾1\mathscr {F}_{n} = \sigma \left\{ X_{1}, \ldots , X_{n}\right\} , n \geqslant 1.

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Problem IV.10

Let T⊂R+T \subset \mathbb {R}_{+} and let (Xt,Ft)t∈T\left(X_{t}, \mathscr {F}_{t}\right)_{t \in T} be a complex-valued martingale, i.e., ((Re⁡Xt,Im⁡Xt),Ft)t∈T((\operatorname {Re} X_{t}, \operatorname {Im} X_{t}), \mathscr {F}_{t})_{t \in T} is a martingale with values in R2\mathbb {R}^{2}, each component of which is (by definition) a martingale. Then E[Xt∣Fs]=Xs\mathbb {E}\left[X_{t} \mid \mathscr {F}_{s}\right] = X_{s} a.s. for s⩽ts \leqslant t, s,t∈Ts, t \in T (here E[Y+iZ∣A]:=E[Y∣A]+iE[Z∣A]\mathbb {E}\left[Y+i Z \mid \mathscr {A}\right]: = \mathbb {E}\left[Y \mid \mathscr {A}\right]+i \mathbb {E}\left[Z \mid \mathscr {A}\right] for integrable Y,ZY, Z and a σ\sigma-algebra A⊂F)\mathscr {A} \subset \mathscr {F}). Let E[∣Xt∣2]<∞\mathbb {E}\left[\left|X_{t}\right|^{2}\right] < \infty for t∈Tt \in T. Prove that X={Xt,t∈T}X = \left\{ X_{t}, t \in T\right\} is a process with uncorrelated increments, i.e., E[(Xt−Xs)(Xu−Xv‾)]=0\mathbb {E}\left[(X_{t}-X_{s})\left(\overline{X_{u}-X_{v}}\right)\right] = 0 for v<u⩽s<t,v,u,s,t∈Tv < u \leqslant s < t, v, u, s, t \in T. (Thus, square-integrable martingales occupy an intermediate position between processes with independent increments having constant mean and processes with uncorrelated increments.)

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Problem IV.11

Let {ξn}n⩾1\left\{ \xi_{n}\right\}_{n \geqslant 1} be a sequence of independent real random variables. Set Sn=∑k=1nξk,n⩾1S_{n} = \sum_{k = 1}^{n} \xi_{k}, n \geqslant 1. Using martingale methods, prove that the following statements about the series ∑k=1∞ξk\sum_{k = 1}^{\infty } \xi_{k} are equivalent:

  1. the series converges a.s.,

  2. the series converges in probability,

  3. the series converges in distribution.

Note that by Kolmogorov's zero-one law (Theorem 6 of Chapter III) this series converges or diverges with probability one.

Theorem 6 of Chapter III (zero-one law). Let a family {Xt,t∈T}\left\{ X_{t}, t \in T\right\}, T⊂RT \subset \mathbb {R}, of independent random elements with values in measurable spaces (St,Bt)(\mathrm{S}_{t}, \mathscr {B}_{t}), i.e. Xt:Ω→StX_{t}: \Omega \rightarrow \mathrm{S}_{t}, Xt∈F∣BtX_{t} \in \mathscr {F} \mid \mathscr {B}_{t}, t∈Tt \in T, be given on a probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}). Denote

F⩾tX=σ{Xs,s∈T∩[t,∞)} \mathscr {F}_{\geqslant t}^{X} = \sigma \left\{ X_{s}, s \in T \cap [t, \infty )\right\}

Then the "tail" σ\sigma-algebra, i.e. the σ\sigma-algebra G=⋂t∈TF⩾tX\mathscr {G} = \bigcap_{t \in T} \mathscr {F}_{\geqslant t}^{X}, is degenerate. In other words, for every A∈GA \in \mathscr {G} the probability P(A)\mathbb {P}\left(A\right) equals zero or one (if T∩[t,∞)=∅T \cap [t, \infty ) = \varnothing, then F⩾tX:=∅\mathscr {F}_{\geqslant t}^{X}: = \varnothing).

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Problem IV.12

Let Y={Yt,t∈T},T⊂RY = \left\{ Y_{t}, t \in T\right\} , T \subset \mathbb {R}, be a (complex-valued) process with independent increments such that E[∣Yt∣2]<∞,t∈T\mathbb {E}\left[\left|Y_{t}\right|^{2}\right] < \infty , t \in T. Prove that there exists a nonrandom function h:T→Rh: T \rightarrow \mathbb {R} such that Xt=∣Yt∣2−h(t)X_{t} = \left|Y_{t}\right|^{2}-h(t) is a martingale with respect to the filtration Ft=σ{Ys,s⩽t,s∈T}=σ{Re⁡Ys,Im⁡Ys,s⩽t,s∈T},t∈T\mathscr {F}_{t} = \sigma \left\{ Y_{s}, s \leqslant t, s \in T\right\} = \sigma \left\{ \operatorname {Re} Y_{s}, \operatorname {Im} Y_{s}, s \leqslant t, s \in T\right\} , t \in T, if and only if

E[∣Yt−Ys∣2]=h(t)−h(s),s⩽t,s,t∈T \mathbb {E}\left[\left|Y_{t}-Y_{s}\right|^{2}\right] = h(t)-h(s), \quad s \leqslant t, \quad s, t \in T
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Problem IV.13

Let W={Wt,t⩾0}W = \left\{ W_{t}, t \geqslant 0\right\} be a Brownian motion with the natural filtration F=(Ft)t⩾0\mathbb {F} = \left(\mathscr {F}_{t}\right)_{t \geqslant 0}.

Find all α,β∈R\alpha , \beta \in \mathbb {R} for which the process

X={exp(αWt+βt),t⩾0} X = \left\{ \mathrm{exp}\left(\alpha W_{t}+\beta t\right), t \geqslant 0\right\}

is a martingale.

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Problem IV.14

Let W={Wt,t⩾0}W = \left\{ W_{t}, t \geqslant 0\right\} be a Brownian motion with the natural filtration F=(Ft)t⩾0\mathbb {F} = \left(\mathscr {F}_{t}\right)_{t \geqslant 0}.

Let τ\tau be a stopping time (with respect to the filtration F\mathbb {F} ). Prove that (Wt∧τ,Ft)t⩾0\left(W_{t \wedge \tau }, \mathscr {F}_{t}\right)_{t \geqslant 0} is a martingale.

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Problem IV.15

Let W={Wt,t⩾0}W = \left\{ W_{t}, t \geqslant 0\right\} be a Brownian motion with the natural filtration F=(Ft)t⩾0\mathbb {F} = \left(\mathscr {F}_{t}\right)_{t \geqslant 0}.

For α,x∈R\alpha , x \in \mathbb {R} and t⩾0t \geqslant 0 define the function

h(α,x,t)=exp(αx−α2t/2). h(\alpha , x, t) = \mathrm{exp}\left(\alpha x-\alpha ^{2} t / 2\right) .

Set

hk(x,t)=∂kh(α,x,t)∂αk∣α=0,k⩾1 h_{k}(x, t) = \left.\frac{\partial ^{k} h(\alpha , x, t)}{\partial \alpha ^{k}}\right|_{\alpha = 0}, \quad k \geqslant 1

Prove that (hk(Wt,t),Ft)t⩾0\left(h_{k}\left(W_{t}, t\right), \mathscr {F}_{t}\right)_{t \geqslant 0} is a martingale for each k⩾1k \geqslant 1 (note also that, by Exercise 13, the process {h(α,Wt,t),t⩾0}\left\{ h\left(\alpha , W_{t}, t\right), t \geqslant 0\right\} is a martingale for each α∈R\alpha \in \mathbb {R} ). Thus the processes Wt,Wt2−t,Wt3−3tWtW_{t}, W_{t}^{2}-t, W_{t}^{3}-3 t W_{t}, Wt4−6tWt2+3t2,…W_{t}^{4}-6 t W_{t}^{2}+3 t^{2}, \ldots are martingales (t⩾0)(t \geqslant 0).

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Problem IV.16

Let W={Wt,t⩾0}W = \left\{ W_{t}, t \geqslant 0\right\} be a Brownian motion with the natural filtration F=(Ft)t⩾0\mathbb {F} = \left(\mathscr {F}_{t}\right)_{t \geqslant 0}.

Prove that E[τa2]=5a4/3\mathbb {E}\left[\tau_{a}^{2}\right] = 5 a^{4} / 3 for τa=inf⁡{t⩾0:∣Wt∣=a},a>0\tau_{a} = \inf \left\{ t \geqslant 0:\left|W_{t}\right| = a\right\} , a > 0.

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Problem IV.17

Let W={Wt,t⩾0}W = \left\{ W_{t}, t \geqslant 0\right\} be a Brownian motion with the natural filtration F=(Ft)t⩾0\mathbb {F} = \left(\mathscr {F}_{t}\right)_{t \geqslant 0}.

Let Xt=γt+σWt,t⩾0X_{t} = \gamma t+\sigma W_{t}, t \geqslant 0, and Ra=inf⁡{t:Xt=a}R_{a} = \inf \left\{ t: X_{t} = a\right\}. Prove that if γ>0,σ≠0\gamma > 0, \sigma \neq 0 and a<0a < 0, then

P(Ra<∞)=e2γa/σ2 \mathbb {P}\left(R_{a} < \infty \right) = e^{2 \gamma a / \sigma ^{2}}
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Problem IV.18

Does condition (9) hold if

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(a)

Y={Yt,t⩾0}Y = \left\{ Y_{t}, t \geqslant 0\right\} is a Wiener process,

(b)

Y={Yt,t⩾0}Y = \left\{ Y_{t}, t \geqslant 0\right\} is a Poisson process?

Recall condition (9) (from Lemma 1 of the main text of the chapter). Let Y={Yt,t⩾0}Y = \left\{ Y_{t}, t \geqslant 0\right\} be a real-valued process with independent increments such that for some α∈R\alpha \in \mathbb {R} and all s,t⩾0s, t \geqslant 0 one has E[eαYt]<∞\mathbb {E}\left[e^{\alpha Y_{t}}\right] < \infty and E[eα(Yt−Ys)]<∞\mathbb {E}\left[e^{\alpha \left(Y_{t}-Y_{s}\right)}\right] < \infty. Set Ft=σ{Ys,s⩽t}\mathscr {F}_{t} = \sigma \left\{ Y_{s}, s \leqslant t\right\} and Zt=eαYt/E[eαYt]Z_{t} = e^{\alpha Y_{t}} / \mathbb {E}\left[e^{\alpha Y_{t}}\right], t⩾0t \geqslant 0; then Z={Zt,t⩾0}Z = \left\{ Z_{t}, t \geqslant 0\right\} is a martingale with respect to (Ft)t⩾0\left(\mathscr {F}_{t}\right)_{t \geqslant 0} if and only if for all s⩽ts \leqslant t

E[eαYteαYs]=E[eαYt]E[eαYs].(9) \mathbb {E}\left[\frac{e^{\alpha Y_{t}}}{e^{\alpha Y_{s}}}\right] = \frac{\mathbb {E}\left[e^{\alpha Y_{t}}\right]}{\mathbb {E}\left[e^{\alpha Y_{s}}\right]} . \tag {9}
Problem IV.19

(see [12; p. 166]). Let W={Wt,t⩾0}W = \left\{ W_{t}, t \geqslant 0\right\} be an mm-dimensional Brownian motion and f(x)f(x) a continuous superharmonic function in Rm\mathbb {R}^{m}, i.e., f(x)f(x), for any x∈Rmx \in \mathbb {R}^{m}, is not less than the integral of this function over any sphere centered at xx. Prove that the process X={Xt=f(Wt),t⩾0}X = \left\{ X_{t} = f\left(W_{t}\right), t \geqslant 0\right\} is a supermartingale with respect to the natural filtration of the Brownian motion if E[∣Xt∣]<∞\mathbb {E}\left[\left|X_{t}\right|\right] < \infty for all t⩾0t \geqslant 0.

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Problem IV.20

(Krickeberg decomposition). A martingale X=(Xn,Fn)n⩾0X = \left(X_{n}, \mathscr {F}_{n}\right)_{n \geqslant 0} has the property sup⁡E[∣Xn∣]<∞\sup \mathbb {E}\left[\left|X_{n}\right|\right] < \infty if and only if there exist nonnegative martingales Y=(Yn,Fn)n⩾0Y = \left(Y_{n}, \mathscr {F}_{n}\right)_{n \geqslant 0} and Z=(Zn,Fn)n⩾0Z = \left(Z_{n}, \mathscr {F}_{n}\right)_{n \geqslant 0} such that Xn=Yn−Zn,n⩾0X_{n} = Y_{n}-Z_{n}, n \geqslant 0.

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Problem IV.21

(Riesz decomposition). Let X=(Xn,Fn)n⩾0X = \left(X_{n}, \mathscr {F}_{n}\right)_{n \geqslant 0} be a uniformly integrable supermartingale. Then there exists a representation of the form Xn=Mn+RnX_{n} = M_{n}+R_{n}, where M=(Mn,Fn)n⩾0M = \left(M_{n}, \mathscr {F}_{n}\right)_{n \geqslant 0} is a uniformly integrable martingale and R=(Rn,Fn)n⩾0R = \left(R_{n}, \mathscr {F}_{n}\right)_{n \geqslant 0} is a potential, i.e., a uniformly integrable nonnegative supermartingale such that Rn→0R_{n} \rightarrow 0 a.s. as n→∞n \rightarrow \infty.

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Problem IV.22

Let h:R→R+h: \mathbb {R} \rightarrow \mathbb {R}_{+} be a nondecreasing convex function and (Xn,Fn)1⩽n⩽N\left(X_{n}, \mathscr {F}_{n}\right)_{1 \leqslant n \leqslant N} a submartingale. Show that then for any u∈Ru \in \mathbb {R} and t>0t > 0

P(max⁡1⩽n⩽NXn⩾u)⩽E[h(tXN)]/h(tu). \mathbb {P}\left(\max _{1 \leqslant n \leqslant N} X_{n} \geqslant u\right) \leqslant \mathbb {E}\left[h\left(t X_{N}\right)\right] / h(t u) .
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Problem IV.23

Let (Xn,Fn)1⩽n⩽N\left(X_{n}, \mathscr {F}_{n}\right)_{1 \leqslant n \leqslant N} be a martingale or a nonnegative submartingale. Prove that

E[max⁡1⩽n⩽N∣Xn∣]⩽ee−1(1+E[∣XN∣log⁡+∣XN∣]) \mathbb {E}\left[\max _{1 \leqslant n \leqslant N}\left|X_{n}\right|\right] \leqslant \frac{e}{e-1}\left(1+\mathbb {E}\left[\left|X_{N}\right| \log ^{+}\left|X_{N}\right|\right]\right)

where log⁡+x=log⁡x\log^{+} x = \log x for x⩾1x \geqslant 1 and log⁡+x=0\log^{+} x = 0 for x<1x < 1.

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Problem IV.24

Construct an example (see [85; vol. 2, p. 678]) showing that for p=1p = 1 the double inequality (68) need not hold.

Theorem 11 (Burkholder). Let X=(Xn,Fn)n⩾0X = \left(X_{n}, \mathscr {F}_{n}\right)_{n \geqslant 0} be a martingale with X0=0X_{0} = 0. Then for each p>1p > 1 there exist constants ApA_{p} and BpB_{p}, not depending on XX, such that

Ap∥[X]n∥p⩽∥Xn∥p⩽Bp∥[X]n∥p,n⩾0(68) A_{p}\left\| \sqrt{[X]_{n}}\right\| _{p} \leqslant \left\| X_{n}\right\| _{p} \leqslant B_{p}\left\| \sqrt{[X]_{n}}\right\| _{p}, \quad n \geqslant 0 \tag {68}

where [X]n[X]_{n} is the quadratic variation of XX (see (22)).

Definition 5. The quadratic variation of a process X={Xn,n⩾0}X = \left\{ X_{n}, n \geqslant 0\right\} is the process [X]=([X]n)n⩾0[X] = \left([X]_{n}\right)_{n \geqslant 0} given by the formula

[X]n=∑k=1n(ΔXk)2,n⩾1 and [X]0=0.(22) [X]_{n} = \sum _{k = 1}^{n}\left(\Delta X_{k}\right)^{2}, \quad n \geqslant 1 \quad \text{ and } \quad [X]_{0} = 0 . \tag {22}
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Problem IV.25

Let W={Wt,t⩾0}W = \left\{ W_{t}, t \geqslant 0\right\} be an mm-dimensional Brownian motion. Prove that (∥Wt∥,Ft)t⩾0\left(\left\| W_{t}\right\| , \mathscr {F}_{t}\right)_{t \geqslant 0} is a submartingale (with a.s. continuous sample paths), where ∥⋅∥\left\| \cdot \right\| is the Euclidean norm in Rm\mathbb {R}^{m}. Using Exercise 22 (with the function h(x)=etxh(x) = e^{t x} and a suitable value t>0t > 0 ), prove that for every s>0s > 0 and all x>msx > \sqrt{m s} the following inequality holds

P(sup⁡t∈[0,s]∥Wt∥⩾x)⩽(sdex2)−m/2e−x2/(2s) \mathbb {P}\left(\sup _{t \in [0, s]}\left\| W_{t}\right\| \geqslant x\right) \leqslant \left(\frac{s d}{e x^{2}}\right)^{-m / 2} e^{-x^{2} /(2 s)}

(The process (∥Wt∥,Ft)t⩾0\left(\left\| W_{t}\right\| , \mathscr {F}_{t}\right)_{t \geqslant 0} is called the Bessel process of dimension mm. )

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Problem IV.26

(see [12; p. 182]). Let W={Wt,t⩾0}W = \left\{ W_{t}, t \geqslant 0\right\} be an mm-dimensional Brownian motion, where m⩾3m \geqslant 3. Prove that ∥Wt∥→∞\left\| W_{t}\right\| \rightarrow \infty a.s. as t→∞t \rightarrow \infty. Prove that with probability 1 the process WW does not return to the point 0 in any finite time. Explain why for m=1m = 1 these statements are false. What does the corresponding result look like in the case m=2m = 2 ?

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Problem IV.27

Definition 7. Let (Gt)t∈T\left(\mathscr {G}_{t}\right)_{t \in T} be a decreasing family of σ\sigma-algebras in F\mathscr {F}, i.e., Gt⊂Gs⊂F\mathscr {G}_{t} \subset \mathscr {G}_{s} \subset \mathscr {F} for s<t;s,t∈T⊂Rs < t ; s, t \in T \subset \mathbb {R}. Let X={Xt,t∈T}X = \left\{ X_{t}, t \in T\right\} be a real-valued stochastic process adapted to the filtration (Gt)t∈T\left(\mathscr {G}_{t}\right)_{t \in T}. The process X←=(Xt,Gt)t∈T\overleftarrow {X} = \left(X_{t}, \mathscr {G}_{t}\right)_{t \in T} is called a reversed martingale (reversed submartingale, reversed supermartingale) if the (reversed) process (Xu,Gu)u∈U\left(X_{u}, \mathscr {G}_{u}\right)_{u \in U} with U=−T={−t,t∈T}U = -T = \left\{ -t, t \in T\right\} is a martingale (submartingale, supermartingale).

In particular, a stochastic sequence X←=(Xn,Gn)n⩾0\overleftarrow {X} = \left(X_{n}, \mathscr {G}_{n}\right)_{n \geqslant 0} is a reversed martingale if E[Xn∣Gn+1]=Xn+1,n⩾0\mathbb {E}\left[X_{n} \mid \mathscr {G}_{n+1}\right] = X_{n+1}, n \geqslant 0. Replacing " = " in the last equality by the sign " ⩾\geqslant ", we obtain the definition of a reversed submartingale (with " ⩽\leqslant ", that of a reversed supermartingale). The notion of a reversed martingale also extends to processes with values in Rm(m⩾1)\mathbb {R}^{m}(m \geqslant 1).

Let ξ1,…,ξN\xi_{1}, \ldots , \xi_{N} be independent identically distributed vectors in Rm(m⩾1)\mathbb {R}^{m} (m \geqslant 1) with E[∥ξ1∥]<∞\mathbb {E}\left[\left\| \xi_{1}\right\| \right] < \infty, where ∥⋅∥\left\| \cdot \right\| is the Euclidean norm. Set Xn=(1/n)∑k=1nξk,Gn,N=σ{Xn,…,XN}X_{n} = (1 / n) \sum_{k = 1}^{n} \xi_{k}, \mathscr {G}_{n, N} = \sigma \left\{ X_{n}, \ldots , X_{N}\right\}. Prove that (Xn,Gn,N)1⩽n⩽N\left(X_{n}, \mathscr {G}_{n, N}\right)_{1 \leqslant n \leqslant N} is a reversed martingale.

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Problem IV.28

Definition 7. Let (Gt)t∈T\left(\mathscr {G}_{t}\right)_{t \in T} be a decreasing family of σ\sigma-algebras in F\mathscr {F}, i.e., Gt⊂Gs⊂F\mathscr {G}_{t} \subset \mathscr {G}_{s} \subset \mathscr {F} for s<t;s,t∈T⊂Rs < t ; s, t \in T \subset \mathbb {R}. Let X={Xt,t∈T}X = \left\{ X_{t}, t \in T\right\} be a real-valued stochastic process adapted to the filtration (Gt)t∈T\left(\mathscr {G}_{t}\right)_{t \in T}. The process X←=(Xt,Gt)t∈T\overleftarrow {X} = \left(X_{t}, \mathscr {G}_{t}\right)_{t \in T} is called a reversed martingale (reversed submartingale, reversed supermartingale) if the (reversed) process (Xu,Gu)u∈U\left(X_{u}, \mathscr {G}_{u}\right)_{u \in U} with U=−T={−t,t∈T}U = -T = \left\{ -t, t \in T\right\} is a martingale (submartingale, supermartingale).

In particular, a stochastic sequence X←=(Xn,Gn)n⩾0\overleftarrow {X} = \left(X_{n}, \mathscr {G}_{n}\right)_{n \geqslant 0} is a reversed martingale if E[Xn∣Gn+1]=Xn+1,n⩾0\mathbb {E}\left[X_{n} \mid \mathscr {G}_{n+1}\right] = X_{n+1}, n \geqslant 0. Replacing " = " in the last equality by the sign " ⩾\geqslant ", we obtain the definition of a reversed submartingale (with " ⩽\leqslant ", that of a reversed supermartingale). The notion of a reversed martingale also extends to processes with values in Rm(m⩾1)\mathbb {R}^{m}(m \geqslant 1).

Definition 8. A sequence of real random variables (ξn)n⩾1\left(\xi_{n}\right)_{n \geqslant 1} is called exchangeable if (ξi1,…,ξiN)=D(ξ1,…,ξN)\left(\xi_{i_{1}}, \ldots , \xi_{i_{N}}\right) \stackrel{\mathscr {D}}{ = }\left(\xi_{1}, \ldots , \xi_{N}\right) for every N⩾1N \geqslant 1 and every permutation (i1,…,iN)\left(i_{1}, \ldots , i_{N}\right) of the set {1,…,N}\left\{ 1, \ldots , N\right\}.

Let m⩾1m \geqslant 1 and let g:Rm→Rg: \mathbb {R}^{m} \rightarrow \mathbb {R} be a Borel function. Consider the sequence of UU-statistics of the form

Un,m=(Cnm)−1∑1⩽i1<⋯<im⩽ng(ξi1,…,ξim),n⩾m U_{n, m} = \left(C_{n}^{m}\right)^{-1} \sum _{1 \leqslant i_{1} < \cdots < i_{m} \leqslant n} g\left(\xi _{i_{1}}, \ldots , \xi _{i_{m}}\right), \quad n \geqslant m

where Cnm=n!/(m!(n−m)!)C_{n}^{m} = n ! /(m !(n-m) !). Set Gn,m=σ{Uk,m,k⩾n},n⩾m\mathscr {G}_{n, m} = \sigma \left\{ U_{k, m}, k \geqslant n\right\} , n \geqslant m.

Prove that for the sequence of UU-statistics introduced above, (Un,m,Gn,m)n⩾m\left(U_{n, m}, \mathscr {G}_{n, m}\right)_{n \geqslant m} is a reversed martingale.

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Problem IV.29

Let E[∣X∣]<∞\mathbb {E}\left[\left|X\right|\right] < \infty and let (Gn)n⩾1\left(\mathscr {G}_{n}\right)_{n \geqslant 1} be a nonincreasing sequence of σ\sigma-algebras in (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}). Set G∞=⋂n=1∞Gn\mathscr {G}_{\infty } = \bigcap_{n = 1}^{\infty } \mathscr {G}_{n}. Then as n→∞n \rightarrow \infty

E[X∣Gn]→E[X∣G∞] a.s. and in L1(Ω,F,P). \mathbb {E}\left[X \mid \mathscr {G}_{n}\right] \rightarrow \mathbb {E}\left[X \mid \mathscr {G}_{\infty }\right] \text{ a.s. and in } L^{1}(\Omega , \mathscr {F}, \mathbb {P}).
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Problem IV.30

Let m⩾1m \geqslant 1 and let g:Rm→Rg: \mathbb {R}^{m} \rightarrow \mathbb {R} be a Borel function. Consider the sequence of UU-statistics of the form

Un,m=(Cnm)−1∑1⩽i1<⋯<im⩽ng(ξi1,…,ξim),n⩾m U_{n, m} = \left(C_{n}^{m}\right)^{-1} \sum _{1 \leqslant i_{1} < \cdots < i_{m} \leqslant n} g\left(\xi _{i_{1}}, \ldots , \xi _{i_{m}}\right), \quad n \geqslant m

where Cnm=n!/(m!(n−m)!)C_{n}^{m} = n ! /(m !(n-m) !). Set Gn,m=σ{Uk,m,k⩾n},n⩾m\mathscr {G}_{n, m} = \sigma \left\{ U_{k, m}, k \geqslant n\right\} , n \geqslant m.

(continuation of Exercise 28). By Theorem 14 we obtain that

Un,m→U∞,m a.s. as n→∞.  U_{n, m} \rightarrow U_{\infty , m} \text{ a.s. as } n \rightarrow \infty \text{. }

Definition 9. A one-to-one mapping π:N→N\pi : \mathbb {N} \rightarrow \mathbb {N} is called a finite permutation (we write π∈Π(N)\pi \in \Pi (\mathbb {N}) ) if π(n)≠n\pi (n) \neq n for at most finitely many (depending on π\pi ) values n∈Nn \in \mathbb {N}. For a sequence of real random variables {ξk}k⩾1\left\{ \xi_{k}\right\}_{k \geqslant 1} defined on a probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}) (i.e., for a random element ξ=(ξ1,ξ2,…)∈F∣B(R∞)\xi = \left(\xi_{1}, \xi_{2}, \ldots \right) \in \mathscr {F} \mid \mathscr {B}\left(\mathbb {R}^{\infty }\right) ), set πξ=(ξπ1,ξπ2,…)\pi \xi = \left(\xi_{\pi_{1}}, \xi_{\pi_{2}}, \ldots \right), where π=(π1,π2,…)\pi = \left(\pi_{1}, \pi_{2}, \ldots \right) is a permutation of the set N\mathbb {N}, and define in F\mathscr {F} the σ\sigma-algebra of exchangeable events

G={ξ−1(B),B∈B(R∞):ξ−1(B)=(πξ)−1(B) for all π∈Π(N)} \mathscr {G} = \left\{ \xi ^{-1}(B), B \in \mathscr {B}\left(\mathbb {R}^{\infty }\right): \xi ^{-1}(B) = (\pi \xi )^{-1}(B) \text{ for all } \pi \in \Pi (\mathbb {N})\right\}

Theorem 14 (zero-one law; Hewitt-Savage). Let {ξk}k⩾1\left\{ \xi_{k}\right\}_{k \geqslant 1} be a sequence of independent identically distributed random variables. Then the σ\sigma-algebra G\mathscr {G} of exchangeable events is degenerate, i.e., it contains only events of probability 0 and 1.

Using Theorem 13, prove that if {ξk}k⩾1\left\{ \xi_{k}\right\}_{k \geqslant 1} is a sequence of independent identically distributed random variables, then U∞,m=E[g(ξ1,…,ξm)]U_{\infty , m} = \mathbb {E}\left[g\left(\xi_{1}, \ldots , \xi_{m}\right)\right].

Theorem 13. Let (Xn,Gn)n⩾0\left(X_{n}, \mathscr {G}_{n}\right)_{n \geqslant 0} be a reversed submartingale. Then a.s. there exists lim⁡n→∞Xn=X∞∈[−∞,∞)\lim_{n \rightarrow \infty } X_{n} = X_{\infty } \in [-\infty , \infty ). If, in addition, there exists

lim⁡n→∞E[Xn]=c>−∞(71) \lim _{n \rightarrow \infty } \mathbb {E}\left[X_{n}\right] = c > -\infty \tag {71}

then X∞∈(−∞,∞)X_{\infty } \in (-\infty , \infty ) and Xn→X∞X_{n} \rightarrow X_{\infty } in the space L1(Ω,F,P)L^{1}(\Omega , \mathscr {F}, \mathbb {P}) as n→∞n \rightarrow \infty. If Xn⩾0X_{n} \geqslant 0 a.s. for n⩾0n \geqslant 0 and E[X0p]<∞\mathbb {E}\left[X_{0}^{p}\right] < \infty for some p∈(1,∞)p \in (1, \infty ), then Xn→X∞X_{n} \rightarrow X_{\infty } in Lp(Ω,F,P)L^{p}(\Omega , \mathscr {F}, \mathbb {P}) as n→∞n \rightarrow \infty.

Definition 7. Let (Gt)t∈T\left(\mathscr {G}_{t}\right)_{t \in T} be a decreasing family of σ\sigma-algebras in F\mathscr {F}, i.e., Gt⊂Gs⊂F\mathscr {G}_{t} \subset \mathscr {G}_{s} \subset \mathscr {F} for s<t;s,t∈T⊂Rs < t ; s, t \in T \subset \mathbb {R}. Let X={Xt,t∈T}X = \left\{ X_{t}, t \in T\right\} be a real-valued stochastic process adapted to the filtration (Gt)t∈T\left(\mathscr {G}_{t}\right)_{t \in T}. The process X←=(Xt,Gt)t∈T\overleftarrow {X} = \left(X_{t}, \mathscr {G}_{t}\right)_{t \in T} is called a reversed martingale (reversed submartingale, reversed supermartingale) if the (reversed) process (Xu,Gu)u∈U\left(X_{u}, \mathscr {G}_{u}\right)_{u \in U} with U=−T={−t,t∈T}U = -T = \left\{ -t, t \in T\right\} is a martingale (submartingale, supermartingale).

In particular, a stochastic sequence X←=(Xn,Gn)n⩾0\overleftarrow {X} = \left(X_{n}, \mathscr {G}_{n}\right)_{n \geqslant 0} is a reversed martingale if E[Xn∣Gn+1]=Xn+1,n⩾0\mathbb {E}\left[X_{n} \mid \mathscr {G}_{n+1}\right] = X_{n+1}, n \geqslant 0. Replacing " = " in the last equality by the sign " ⩾\geqslant ", we obtain the definition of a reversed submartingale (with " ⩽\leqslant ", that of a reversed supermartingale). The notion of a reversed martingale also extends to processes with values in Rm(m⩾1)\mathbb {R}^{m}(m \geqslant 1).

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Problem IV.31

Definition 7. Let (Gt)t∈T\left(\mathscr {G}_{t}\right)_{t \in T} be a decreasing family of σ\sigma-algebras in F\mathscr {F}, i.e., Gt⊂Gs⊂F\mathscr {G}_{t} \subset \mathscr {G}_{s} \subset \mathscr {F} for s<t;s,t∈T⊂Rs < t ; s, t \in T \subset \mathbb {R}. Let X={Xt,t∈T}X = \left\{ X_{t}, t \in T\right\} be a real-valued stochastic process adapted to the filtration (Gt)t∈T\left(\mathscr {G}_{t}\right)_{t \in T}. The process X←=(Xt,Gt)t∈T\overleftarrow {X} = \left(X_{t}, \mathscr {G}_{t}\right)_{t \in T} is called a reversed martingale (reversed submartingale, reversed supermartingale) if the (reversed) process (Xu,Gu)u∈U\left(X_{u}, \mathscr {G}_{u}\right)_{u \in U} with U=−T={−t,t∈T}U = -T = \left\{ -t, t \in T\right\} is a martingale (submartingale, supermartingale).

In particular, a stochastic sequence X←=(Xn,Gn)n⩾0\overleftarrow {X} = \left(X_{n}, \mathscr {G}_{n}\right)_{n \geqslant 0} is a reversed martingale if E[Xn∣Gn+1]=Xn+1,n⩾0\mathbb {E}\left[X_{n} \mid \mathscr {G}_{n+1}\right] = X_{n+1}, n \geqslant 0. Replacing " = " in the last equality by the sign " ⩾\geqslant ", we obtain the definition of a reversed submartingale (with " ⩽\leqslant ", that of a reversed supermartingale). The notion of a reversed martingale also extends to processes with values in Rm(m⩾1)\mathbb {R}^{m}(m \geqslant 1).

Let (Xn,Gn)n⩾1\left(X_{n}, \mathscr {G}_{n}\right)_{n \geqslant 1} be a reversed submartingale, and suppose condition (71) holds. Prove that X∞⩽E[Xm∣G∞],m⩾1X_{\infty } \leqslant \mathbb {E}\left[X_{m} \mid \mathscr {G}_{\infty }\right], m \geqslant 1, where G∞=⋂n=1∞Gn\mathscr {G}_{\infty } = \bigcap_{n = 1}^{\infty } \mathscr {G}_{n} (recall that X∞=lim⁡n→∞XnX_{\infty } = \lim_{n \rightarrow \infty } X_{n} exists a.s. and in L1(Ω,F,P)L^{1}(\Omega , \mathscr {F}, \mathbb {P}) by Theorem 13). In other words, the "closed" process (Xn,Gn)1⩽n⩽∞\left(X_{n}, \mathscr {G}_{n}\right)_{1 \leqslant n \leqslant \infty } is a reversed submartingale. Show that if (Xn,Gn)n⩾1\left(X_{n}, \mathscr {G}_{n}\right)_{n \geqslant 1} is a reversed martingale, then (Xn,Gn)1⩽n⩽∞\left(X_{n}, \mathscr {G}_{n}\right)_{1 \leqslant n \leqslant \infty } is also a reversed martingale.

Theorem 13. Let (Xn,Gn)n⩾0\left(X_{n}, \mathscr {G}_{n}\right)_{n \geqslant 0} be a reversed submartingale. Then a.s. there exists lim⁡n→∞Xn=X∞∈[−∞,∞)\lim_{n \rightarrow \infty } X_{n} = X_{\infty } \in [-\infty , \infty ). If, in addition, there exists

lim⁡n→∞E[Xn]=c>−∞(71) \lim _{n \rightarrow \infty } \mathbb {E}\left[X_{n}\right] = c > -\infty \tag {71}

then X∞∈(−∞,∞)X_{\infty } \in (-\infty , \infty ) and Xn→X∞X_{n} \rightarrow X_{\infty } in the space L1(Ω,F,P)L^{1}(\Omega , \mathscr {F}, \mathbb {P}) as n→∞n \rightarrow \infty. If Xn⩾0X_{n} \geqslant 0 a.s. for n⩾0n \geqslant 0 and E[X0p]<∞\mathbb {E}\left[X_{0}^{p}\right] < \infty for some p∈(1,∞)p \in (1, \infty ), then Xn→X∞X_{n} \rightarrow X_{\infty } in Lp(Ω,F,P)L^{p}(\Omega , \mathscr {F}, \mathbb {P}) as n→∞n \rightarrow \infty.

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Problem IV.32

Can condition (71) be dropped in the statement of Theorem 13?

Theorem 13. Let (Xn,Gn)n⩾0\left(X_{n}, \mathscr {G}_{n}\right)_{n \geqslant 0} be a reversed submartingale. Then a.s. there exists lim⁡n→∞Xn=X∞∈[−∞,∞)\lim_{n \rightarrow \infty } X_{n} = X_{\infty } \in [-\infty , \infty ). If, in addition, there exists

lim⁡n→∞E[Xn]=c>−∞(71) \lim _{n \rightarrow \infty } \mathbb {E}\left[X_{n}\right] = c > -\infty \tag {71}

then X∞∈(−∞,∞)X_{\infty } \in (-\infty , \infty ) and Xn→X∞X_{n} \rightarrow X_{\infty } in the space L1(Ω,F,P)L^{1}(\Omega , \mathscr {F}, \mathbb {P}) as n→∞n \rightarrow \infty. If Xn⩾0X_{n} \geqslant 0 a.s. for n⩾0n \geqslant 0 and E[X0p]<∞\mathbb {E}\left[X_{0}^{p}\right] < \infty for some p∈(1,∞)p \in (1, \infty ), then Xn→X∞X_{n} \rightarrow X_{\infty } in Lp(Ω,F,P)L^{p}(\Omega , \mathscr {F}, \mathbb {P}) as n→∞n \rightarrow \infty.

Definition 7. Let (Gt)t∈T\left(\mathscr {G}_{t}\right)_{t \in T} be a decreasing family of σ\sigma-algebras in F\mathscr {F}, i.e., Gt⊂Gs⊂F\mathscr {G}_{t} \subset \mathscr {G}_{s} \subset \mathscr {F} for s<t;s,t∈T⊂Rs < t ; s, t \in T \subset \mathbb {R}. Let X={Xt,t∈T}X = \left\{ X_{t}, t \in T\right\} be a real-valued stochastic process adapted to the filtration (Gt)t∈T\left(\mathscr {G}_{t}\right)_{t \in T}. The process X←=(Xt,Gt)t∈T\overleftarrow {X} = \left(X_{t}, \mathscr {G}_{t}\right)_{t \in T} is called a reversed martingale (reversed submartingale, reversed supermartingale) if the (reversed) process (Xu,Gu)u∈U\left(X_{u}, \mathscr {G}_{u}\right)_{u \in U} with U=−T={−t,t∈T}U = -T = \left\{ -t, t \in T\right\} is a martingale (submartingale, supermartingale).

In particular, a stochastic sequence X←=(Xn,Gn)n⩾0\overleftarrow {X} = \left(X_{n}, \mathscr {G}_{n}\right)_{n \geqslant 0} is a reversed martingale if E[Xn∣Gn+1]=Xn+1,n⩾0\mathbb {E}\left[X_{n} \mid \mathscr {G}_{n+1}\right] = X_{n+1}, n \geqslant 0. Replacing " = " in the last equality by the sign " ⩾\geqslant ", we obtain the definition of a reversed submartingale (with " ⩽\leqslant ", that of a reversed supermartingale). The notion of a reversed martingale also extends to processes with values in Rm(m⩾1)\mathbb {R}^{m}(m \geqslant 1).

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Problem IV.33

Let (Ft)t⩾0\left(\mathscr {F}_{t}\right)_{t \geqslant 0} be a nondecreasing flow of σ\sigma-algebras (not necessarily complete) in a complete probability space (Ω,F,P)(\Omega , \mathscr {F}, \mathbb {P}). Then F‾t+=Ft+‾\overline{\mathscr {F}}_{t+} = \overline{\mathscr {F}_{t+}}, i.e. the operations of augmentation by the class of P\mathbb {P}-null sets and of right-continuous closure of the filtration commute (see Exercise 2).

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Problem IV.34

For a filtration (Ft)t⩾0\left(\mathscr {F}_{t}\right)_{t \geqslant 0} set Ft+=⋂s>tFs\mathscr {F}_{t+} = \bigcap_{s > t} \mathscr {F}_{s}, t⩾0t \geqslant 0.

Is it true that for the Wiener process and its natural filtration (Ft)t⩾0\left(\mathscr {F}_{t}\right)_{t \geqslant 0} the relation Ft=Ft+\mathscr {F}_{t} = \mathscr {F}_{t+} holds for all t⩾0t \geqslant 0 ?

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Problem IV.35

For a filtration (Ft)t⩾0\left(\mathscr {F}_{t}\right)_{t \geqslant 0} set F‾t=σ{Ft,N}\overline{\mathscr {F}}_{t} = \sigma \left\{ \mathscr {F}_{t}, \mathscr {N}\right\}, where N\mathscr {N} is the class of P\mathbb {P}-null sets, and F‾t+=⋂s>tF‾s\overline{\mathscr {F}}_{t+} = \bigcap_{s > t} \overline{\mathscr {F}}_{s}, t⩾0t \geqslant 0.

Let (Xt,Ft)t⩾0\left(X_{t}, \mathscr {F}_{t}\right)_{t \geqslant 0} be a submartingale (martingale) with a.s. right-continuous sample paths. Suppose that for each t∈R+t \in \mathbb {R}_{+} there exists δ=δ(t)>0\delta = \delta (t) > 0 such that

E[sup⁡s∈[t,t+δ]∣Xs∣]<∞(76) \mathbb {E}\left[\sup _{s \in [t, t+\delta ]}\left|X_{s}\right|\right] < \infty \tag {76}

Show that then Xˉ=(Xt,F‾t+)\bar{X} = \left(X_{t}, \overline{\mathscr {F}}_{t+}\right) is a submartingale (martingale).

Verify that condition (76) holds for the Wiener and Poisson processes (recall that the sample paths of the Wiener process are a.s. continuous, and the sample paths of the Poisson process are a.s. right-continuous).

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Problem IV.36

Suppose that on a financial market the evolution of a riskless asset BB (bank account) and of a risky asset SS (stock) is described by the system of difference equations (n⩾1n \geqslant 1)

ΔBn=rnBn−1(77) \Delta B_{n} = r_{n} B_{n-1} \tag {77} ΔSn=ρnSn−1(78) \Delta S_{n} = \rho _{n} S_{n-1} \tag {78}

where, as usual, we set ΔXn=Xn−Xn−1\Delta X_{n} = X_{n}-X_{n-1}. In addition, assume that all the random variables Bn,Sn,rnB_{n}, S_{n}, r_{n} and ρn\rho_{n} are defined on a filtered probability space (Ω,F,(Fn)1⩽n⩽N,P)\left(\Omega , \mathscr {F},\left(\mathscr {F}_{n}\right)_{1 \leqslant n \leqslant N}, \mathbb {P}\right), where FN=F,N<∞\mathscr {F}_{N} = \mathscr {F}, N < \infty. For n=1,…,Nn = 1, \ldots , N set

Un=∑k=0nrk,Vn=∑k=0nρk U_{n} = \sum _{k = 0}^{n} r_{k}, \quad V_{n} = \sum _{k = 0}^{n} \rho _{k}

where {rn}\left\{ r_{n}\right\} and {ρn}\left\{ \rho_{n}\right\} are stochastic sequences, i.e. they are Fn∣B(R)\mathscr {F}_{n} \mid \mathscr {B}(\mathbb {R})-measurable for each (relevant) nn. We shall also assume that F0={∅,Ω}\mathscr {F}_{0} = \left\{ \varnothing , \Omega \right\} and Fn=σ{S0,…,Sn}\mathscr {F}_{n} = \sigma \left\{ S_{0}, \ldots , S_{n}\right\} for n⩾1n \geqslant 1.

Prove that the solutions of equations (77) and (78) can be written in the form of (discrete) stochastic exponentials, i.e.

Bn=B0En(U),Sn=S0En(V) B_{n} = B_{0} \mathscr {E}_{n}(U), \quad S_{n} = S_{0} \mathscr {E}_{n}(V)

where

En(X)=∏k=1n(1+ΔXk),n⩾1,E0(X)=1 \mathscr {E}_{n}(X) = \prod _{k = 1}^{n}\left(1+\Delta X_{k}\right), \quad n \geqslant 1, \quad \mathscr {E}_{0}(X) = 1
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Problem IV.37

Suppose that on a financial market the evolution of a riskless asset BB (bank account) and of a risky asset SS (stock) is described by the system of difference equations (n⩾1n \geqslant 1)

ΔBn=rnBn−1ΔSn=ρnSn−1 \begin{aligned} \Delta B_{n} & = r_{n} B_{n-1} \\ \Delta S_{n} & = \rho _{n} S_{n-1} \end{aligned}

where, as usual, we set ΔXn=Xn−Xn−1\Delta X_{n} = X_{n}-X_{n-1}. In addition, assume that all the random variables Bn,Sn,rnB_{n}, S_{n}, r_{n} and ρn\rho_{n} are defined on a filtered probability space (Ω,F,(Fn)1⩽n⩽N,P)\left(\Omega , \mathscr {F},\left(\mathscr {F}_{n}\right)_{1 \leqslant n \leqslant N}, \mathbb {P}\right), where FN=F,N<∞\mathscr {F}_{N} = \mathscr {F}, N < \infty.

Definition 14. An investment strategy, or portfolio, is a two-dimensional stochastic sequence π=(βn,γn)0⩽n⩽N\pi = \left(\beta_{n}, \gamma_{n}\right)_{0 \leqslant n \leqslant N} such that for all nn

βn∈Fn∣B(R),γn∈Fn−1∣B(R) \beta _{n} \in \mathscr {F}_{n} \mid \mathscr {B}(\mathbb {R}), \quad \gamma _{n} \in \mathscr {F}_{n-1} \mid \mathscr {B}(\mathbb {R})

The capital of the portfolio π\pi is the stochastic sequence

Xπ={Xnπ}, where Xnπ=βnBn+γnSn,0⩽n⩽N. X^{\pi } = \left\{ X_{n}^{\pi }\right\} , \quad \text{ where } X_{n}^{\pi } = \beta _{n} B_{n}+\gamma _{n} S_{n}, \quad 0 \leqslant n \leqslant N .

Definition 15. The class of portfolios π\pi possessing, for all n=1,…,Nn = 1, \ldots , N, the property

Bn−1Δβn+Sn−1Δγn=0 B_{n-1} \Delta \beta _{n}+S_{n-1} \Delta \gamma _{n} = 0

is called self-financing and is denoted by SFS F.

Verify that the capital of a self-financing portfolio can be represented in the form

Xnπ=X0π+∑k=0n(βkΔBk+γkΔSk) X_{n}^{\pi } = X_{0}^{\pi }+\sum _{k = 0}^{n}\left(\beta _{k} \Delta B_{k}+\gamma _{k} \Delta S_{k}\right)

where ΔB0=ΔS0=0\Delta B_{0} = \Delta S_{0} = 0.

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