Chapter 4

The Integral of a Random Variable and its Basic Properties

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Problem 4.1

Construct an example where r.v.s XnX_{n} and XX, defined on a probability space (Ω,A,P)(\Omega , \mathcal{A}, P), are such that Xn→n→∞→XX_{n} \xrightarrow [n \rightarrow \infty ]{\rightarrow } X pointwise (hence Xn a.s. n→∞XX_{n} \underset {n \rightarrow \infty }{\text{ a.s. }} X and Xn→n→∞PXX_{n} \xrightarrow [n \rightarrow \infty ]{P} X), but E[Xn→]→n→∞E[X]\mathbb {E}\left[X_{n}^{\rightarrow }\right] \underset {n \rightarrow \infty }{\rightarrow } \mathbb {E}\left[X\right].

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Problem 4.2

If the r.v. X≥0X \geq 0 and ∫Xdμ=0\int X d \mu = 0, then show that μ(X≠0)=0\mu (X \neq 0) = 0.

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Problem 4.3

If the r.v. XX takes on the values n,n=1,2,…n, n = 1,2, \ldots, with probability 1 , so that ∑n=1∞P(X=n)=1\sum_{n = 1}^{\infty } \mathbb {P}\left(X = n\right) = 1, show that E[X]=∑n=1∞P(X≥n)\mathbb {E}\left[X\right] = \sum_{n = 1}^{\infty } \mathbb {P}\left(X \geq n\right).

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Problem 4.4
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(i)

For a r.v. XX and any two disjoint events AA and BB, show that

(XIA+B)+=(XIA)++(XIB)+,(XIA+B)−=(XIA)−+(XIB)−. \begin{aligned} & \left(X I_{A+B}\right)^{+} = \left(X I_{A}\right)^{+}+\left(X I_{B}\right)^{+}, \\ & \left(X I_{A+B}\right)^{-} = \left(X I_{A}\right)^{-}+\left(X I_{B}\right)^{-} . \end{aligned}
(ii)

More generally, for any finite collection of pairwise disjoint events AiA_{i}, i=1,…,ni = 1, \ldots , n, show that

(XI∑i=1nAi)+=∑i=1n(XIAi)+,(XI∑i=1nAi)−=∑i=1n(XIAi)−. \left(X I_{\sum _{i = 1}^{n} A_{i}}\right)^{+} = \sum _{i = 1}^{n}\left(X I_{A_{i}}\right)^{+},\left(X I_{\sum _{i = 1}^{n} A_{i}}\right)^{-} = \sum _{i = 1}^{n}\left(X I_{A_{i}}\right)^{-} .
Problem 4.5

If for the pairwise disjoint events Ai,i=1,…,nA_{i}, i = 1, \ldots , n, the integral ∫∑i=1nAiX\int_{\sum_{i = 1}^{n} A_{i}} X exists, then the integrals ∫AiX,i=1,…,n\int_{A_{i}} X, i = 1, \ldots , n, exist, and ∫∑i=1nAiX=∑i=1n∫AiX\int_{\sum_{i = 1}^{n} A_{i}} X = \sum_{i = 1}^{n} \int_{A_{i}} X.

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Problem 4.6

Let XX and YY be two simple r.v.s such that ∫X=∫Y=∞\int X = \int Y = \infty or ∫X=∫Y=−∞\int X = \int Y = -\infty. Then show that ∫(X+Y)\int (X+Y) exists and ∫(X+Y)=∫X+∫Y(=∞\int (X+Y) = \int X+\int Y( = \infty or −∞)-\infty ).

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Problem 4.7

Let μ1\mu_{1} and μ2\mu_{2} be two measures on (Ω,A)(\Omega , \mathcal{A}) and suppose that ∫Xd(μ1+μ2)\int X d\left(\mu_{1}+\mu_{2}\right) exists. Then ∫Xdμi\int X d \mu_{i} exist, i=1,2i = 1,2, and ∫Xd(μ1+μ2)=∫Xdμ1+∫Xdμ2\int X d\left(\mu_{1}+\mu_{2}\right) = \int X d \mu_{1}+\int X d \mu_{2}.

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Problem 4.8

If the r.v.s Xi,i=1,…,nX_{i}, i = 1, \ldots , n, are integrable, then so is the r.v. ∑i=1nXi\sum_{i = 1}^{n} X_{i} and ∫∑i=1nXi=∑i=1n∫Xi\int \sum_{i = 1}^{n} X_{i} = \sum_{i = 1}^{n} \int X_{i}.

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Problem 4.9

Let XX and YY be r.v.s with finite second moments, and set E[X]=μ1,E[Y]=μ2,0<Var⁡[X]=σ12,0<Var⁡[Y]=σ22\mathbb {E}\left[X\right] = \mu_{1}, \mathbb {E}\left[Y\right] = \mu_{2}, 0 < \operatorname {Var}\left[X\right] = \sigma_{1}^{2}, 0 < \operatorname {Var}\left[Y\right] = \sigma_{2}^{2}. Then the covariance and the correlation coefficient of XX and YY, denoted respectively, by Cov⁡[X,Y]\operatorname {Cov}\left[ X, Y \right] and ρ(X,Y)\rho (X, Y), are defined by: Cov⁡[X,Y]=E[(X−μ1)(Y−μ2)]=E[XY]−μ1μ2\operatorname {Cov}\left[ X, Y \right] = \mathbb {E}\left[\left(X-\mu_{1}\right)\left(Y-\mu_{2}\right)\right] = \mathbb {E}\left[X Y\right]-\mu_{1} \mu_{2} and ρ(X,Y)=Cov⁡[X,Y]/σ1σ2\rho (X, Y) = \operatorname {Cov}\left[ X, Y \right] / \sigma_{1} \sigma_{2}.

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(i)

Then show that −σ1σ2≤Cov⁡[X,Y]≤σ1σ2-\sigma_{1} \sigma_{2} \leq \operatorname {Cov}\left[ X, Y \right] \leq \sigma_{1} \sigma_{2}, and Cov⁡[X,Y]=σ1σ2\operatorname {Cov}\left[ X, Y \right] = \sigma_{1} \sigma_{2} if and only if P(Y=μ2+σ2σ1(X−μ1))=1\mathbb {P}\left(Y = \mu_{2}+\frac{\sigma_{2}}{\sigma_{1}}\left(X-\mu_{1}\right)\right) = 1, and Cov⁡[X,Y]=−σ1σ2\operatorname {Cov}\left[ X, Y \right] = -\sigma_{1} \sigma_{2} if and only if P(Y=μ2−σ2σ1(X−μ1))=1\mathbb {P}\left(Y = \mu_{2}-\frac{\sigma_{2}}{\sigma_{1}}\left(X-\mu_{1}\right)\right) = 1.

(ii)

Also, −1≤ρ(X,Y)≤1-1 \leq \rho (X, Y) \leq 1, and ρ(X,Y)=1\rho (X, Y) = 1 if and only if

P(Y=μ2+σ2σ1(X−μ1))=1, \mathbb {P}\left(Y = \mu _{2}+\frac{\sigma _{2}}{\sigma _{1}}\left(X-\mu _{1}\right)\right) = 1,

and ρ(X,Y)=−1\rho (X, Y) = -1 if and only if P(Y=μ2−σ2σ1(X−μ1))=1\mathbb {P}\left(Y = \mu_{2}-\frac{\sigma_{2}}{\sigma_{1}}\left(X-\mu_{1}\right)\right) = 1.

Problem 4.10

Let Ω={−2,−1,3,7}\Omega = \left\{ -2,-1,3,7\right\}, and let the measure μ\mu be defined by: μ({−2})=2\mu (\left\{ -2\right\} ) = 2, μ({−1})=1,μ({3})=3,μ({7})=7\mu (\left\{ -1\right\} ) = 1, \mu (\left\{ 3\right\} ) = 3, \mu (\left\{ 7\right\} ) = 7. Let XX be a r.v. defined by: X(−2)=X(−1)=−1X(-2) = X(-1) = -1, and X(3)=X(7)=1X(3) = X(7) = 1. Then, for A={−2,3,7}A = \left\{ -2,3,7\right\}, compute the integral ∫AXdμ\int_{A} X d \mu.

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Problem 4.11

Let (Ω,A,μ)=((−5,5),B(−5,5),λ)(\Omega , \mathcal{A}, \mu ) = \left((-5,5), \mathcal{B}_{(-5,5)}, \lambda \right) where λ\lambda is the Lebesgue measure, and let the r.v. XX be defined as follows:

X(ω)={12,ω∈(−5,2)13,ω=21,ω∈(2,3]0,ω∈(3,5) X(\omega ) = \left\{ \begin{array}{l} \frac{1}{2}, \omega \in (-5,2) \\ \frac{1}{3}, \omega = 2 \\ 1, \omega \in (2,3] \\ 0, \omega \in (3,5) \end{array}\right.

Then, for A=[−1,4]A = [-1,4], compute the integral ∫AXdμ\int_{A} X d \mu.

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Problem 4.12

Let Ω={0,1,2,…}\Omega = \left\{ 0,1,2, \ldots \right\}, let A\mathcal{A} be the discrete σ\sigma-field of subsets of Ω\Omega, and on A\mathcal{A}, define the function μ\mu by: μ(A)=\mu (A) = number of nonnegative integers in AA. Then show that μ\mu is a measure, and indeed, a σ\sigma-finite measure.

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Problem 4.13

Let g:R→(0,∞)g: \mathbb {R} \rightarrow (0, \infty ) be nondecreasing in (0,∞)(0, \infty ) and symmetric about 0 (g(−x)=g(x),x∈R)(g(-x) = g(x), \quad x \in \mathbb {R}), and let XX be a r.v. such that E[g(X)]<∞\mathbb {E}\left[g(X)\right] < \infty. Then show that: P(∣X∣≥c)≤E[g(X)]/g(c)\mathbb {P}\left(\left|X\right| \geq c\right) \leq \mathbb {E}\left[g(X)\right] / g(c) for every c>0c > 0.

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Problem 4.14

For a r.v. XX, show that

P(∣X∣≥c)≤(1+1c)E[∣X∣1+∣X∣] for every c>0 \mathbb {P}\left(\left|X\right| \geq c\right) \leq \left(1+\frac{1}{c}\right) \mathbb {E}\left[\frac{\left|X\right|}{1+\left|X\right|}\right] \text{ for every } c > 0
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Problem 4.15

Let XX be a simple r.v. defined on (Ω,A)(\Omega , \mathcal{A}), so that (see Definition 13 in Chapter 1) X=∑j=1nαjIAjX = \sum_{j = 1}^{n} \alpha_{j} I_{A_{j}}, where {Aj,j=1,…,n}\left\{ A_{j}, j = 1, \ldots , n\right\} is a (measurable) partition of Ω\Omega, and the αj,j=1,…,n\alpha_{j}, j = 1, \ldots , n, are assumed to be distinct. Next, let X=∑i=1mβiIBiX = \sum_{i = 1}^{m} \beta_{i} I_{B_{i}} be any other representation of XX. Then show that each Aj,j=1,…,nA_{j}, j = 1, \ldots , n, is the sum of some BiB_{i} 's.

Remark 6. Under the assumption that the αj,j=1,…,n\alpha_{j}, j = 1, \ldots , n, are distinct, the partition {Aj,j=1,…,n}\left\{ A_{j}, j = 1, \ldots , n\right\} may be called irreducible. For such a partition, there is a unique representation of the respective simple r.v. and consequently, Theorem 1 is rendered superfluous.

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Problem 4.16

If XX and YY are two identically distributed integrable r.v.s, then

E[XI(∣X∣≤c)]=E[YI(∣Y∣≤c)] \mathbb {E}\left[X I_{(\left|X\right| \leq c)}\right] = \mathbb {E}\left[Y I_{(\left|Y\right| \leq c)}\right]

for any constant cc.

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