Math Problems Vault

Problem 9.1

If $X$ is an integrable r.v. and $\left\{ A_{n}, n = 1,2, \ldots \right\}$ are pairwise disjoint events, show that

\int _{\sum _{i = 1}^{\infty } A_{i}} X d P = \sum _{i = 1}^{\infty } \int _{A_{i}} X d P

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Problem 9.2

Let $X$ be an integrable r.v. and, for $B \in \mathcal{A}$ with $\mathbb {P}\left(B\right) > 0$ , consider the conditional probability on $\mathcal{A}, \mathbb {P}\left( \cdot \mid B \right)$ . Then show that

\int _{\Omega } X d \mathbb {P}\left( \cdot \mid B \right) = \frac{1}{\mathbb {P}\left(B\right)} \int _{B} X d P

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Problem 9.3

Consider the probability space $(\Omega , \mathcal{A}, P) = \left([0,1], \mathcal{B}_{[0,1]}, \lambda \right)$ , where $\lambda$ is the Lebesgue measure, and let $\mathcal{F}$ be the $\sigma$ -field generated by the class $\left\{ \left[0, \frac{1}{4}\right],\left(\frac{1}{4}, \frac{2}{3}\right]\right.$ , $\left.\left(\frac{2}{3}, 1\right]\right\}$ . Also, let $X$ be the r.v. defined by: $X(\omega ) = \omega^{2}, \omega \in \Omega$ . Then show that

\mathbb {E}\left[ X \mid \mathcal{F} \right] = \alpha _{1} I_{\left[0, \frac{1}{4}\right]}+\alpha _{2} I_{\left(\frac{1}{4}, \frac{2}{3}\right]}+\alpha _{3} I_{\left(\frac{2}{3}, 1\right]},

and compute $\alpha_{1}, \alpha_{2}, \alpha_{3}$ .

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Problem 9.4

Let $X$ and $Y$ be $\mathcal{B}$ -measurable and integrable r.v.s. We further assume that $\int_{B} X d P \geq \int_{B} Y d P$ for every $B \in \mathcal{B}(\subseteq \mathcal{A})$ . Then show that $X \geq Y$ a.s.

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Problem 9.5

(i)(ii)(iii)

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(i)

Let $X \geq 0$ a.s. on a set $A$ with $\mathbb {P}\left(A\right) > 0$ , and suppose that $\int_{A} X d P = 0$ . Then show that $X = 0$ a.s. on $A$ .

(ii)

Let $X \geq 0$ , integrable, and $X = 0$ a.s. on a set $A \in \mathcal{B}(\subseteq A)$ , with $\mathbb {P}\left(A\right) > 0$ . Then show that $\mathbb {E}_{\mathcal{B}}\left[X\right] = 0$ a.s. on $A$ .

(iii)

Problem 9.6

If $\mathbb {E}\left[\left|X\right|^{r}\right] < \infty$ , then show that $\left|\mathbb {E}_{\mathcal{B}}\left[X\right]\right|^{r} \leq \mathbb {E}_{\mathcal{B}}\left[\left|X\right|^{r}\right]$ a.s., $r \geq 1$ .

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Problem 9.7

If the r.v. $X$ is integrable, then $\mathbb {E}_{\mathcal{B}}\left[X\right]$ is finite a.s.

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Problem 9.8

Recall that a function $g: I$ (open interval) $\subseteq \mathbb {R} \rightarrow \mathbb {R}$ is said to be convex if $g\left(\alpha_{1} x_{1}+\alpha_{2} x_{2}\right) \leq \alpha_{1} g\left(x_{1}\right)+\alpha_{2} g\left(x_{2}\right)$ for all $\alpha_{1}, \alpha_{2} \geq 0$ with $\alpha_{1}+\alpha_{2} = 1$ , and all $x_{1}, x_{2} \in I$ . Prove the following generalization: if $g$ is as above, then:

g\left(\alpha _{1} x_{1}+\cdots +\alpha _{n} x_{n}\right) \leq \alpha _{1} g\left(x_{1}\right)+\cdots +\alpha _{n} g\left(x_{n}\right)

for any $n \geq 2$ , any $\alpha_{1}, \ldots , \alpha_{n} \geq 0$ with $\alpha_{1}+\cdots +\alpha_{n} = 1$ , and all $x_{1}, \ldots , x_{n} \in I$ .

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Problem 9.9

Recall the following lemma (Lemma 2). Let $\left(\Omega , \mathcal{B}_{Y}, P_{Y}\right)$ , $\left(\Omega^{\prime }, \mathcal{B}_{Y}^{\prime }, P_{Y}^{\prime }\right)$ be as before (i.e., $\mathcal{B}_{Y}^{\prime } = \left\{ A^{\prime } \subseteq \Omega^{\prime } ; Y^{-1}\left(A^{\prime }\right) = A \text{ for some } A \in \mathcal{A}\right\}$ , $\mathcal{B}_{Y} = Y^{-1}\left(\mathcal{B}_{Y}^{\prime }\right)$ , and $P_{Y}, P_{Y}^{\prime }$ are the probability measures induced on $\mathcal{B}_{Y}, \mathcal{B}_{Y}^{\prime }$ , respectively, by $P$ ), and let $g : \left(\Omega^{\prime }, \mathcal{B}_{Y}^{\prime }\right) \rightarrow \left(\mathbb {R}, \mathcal{B}\right)$ be measurable, so that $g(Y) : (\Omega , \mathcal{B}_{Y}) \rightarrow (\mathbb {R}, \mathcal{B})$ is measurable (a r.v.). Then

\int _{B^{\prime }} g(y) \, dP_{Y}^{\prime } = \int _{B} g(Y) \, dP_{Y}, \quad B^{\prime } \in \mathcal{B}_{Y}^{\prime }, \quad B = Y^{-1}\left(B^{\prime }\right),

in the sense that, if one of these integrals exists, then so does the other, and both are equal.

Fill in the details in proving Lemma 2.

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Problem 9.10

Let $X$ and $Y$ be discrete r.v.s and recall (from the application following Theorem 11) that $\mathcal{B}_{Y}^{\prime }$ is the $\sigma$ -field of subsets of $\mathbb {R}$ defined by $\mathcal{B}_{Y}^{\prime } = \left\{ B^{\prime } \subseteq \mathbb {R}\right.$ ; $Y^{-1}\left(B^{\prime }\right) = A$ for some $\left.A \in \mathcal{A}\right\}$ . For $x, y \in \mathbb {R}$ with $\mathbb {P}\left(Y = y\right) > 0$ , consider the conditional probability

\mathbb {P}\left( X = x \mid Y = y \right) = \mathbb {P}\left(X = x, Y = y\right) / \mathbb {P}\left(Y = y\right)

and show that, for each fixed $x$ , the function $\mathbb {P}\left( X = x \mid Y = \cdot \right)$ is $\mathcal{B}_{Y}^{\prime }$ -measurable.

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Problem 9.11

The Dominated Convergence Theorem in its classical form states: If $\left|X_{n}\right| \leq Y$ a.s., $n \geq 1, \mathbb {E}\left[Y\right] < \infty$ , and either $X_{n} \xrightarrow [n \rightarrow \infty ]{\text{ a.s. }} X$ or $X_{n} \xrightarrow [n \rightarrow \infty ]{P} X$ , then $\mathbb {E}\left[X_{n_{n}}^{ \rightarrow \infty }\right] \mathbb {E}\left[X\right]$ finite.

In the framework of conditional expectations, we have shown that: If $\left|X_{n}\right| \leq Y$ a.s., $n \geq 1, \mathbb {E}\left[Y\right] < \infty$ , and $X_{n} \xrightarrow [n \rightarrow \infty ]{\text{ a.s. }} X$ with $\mathbb {E}\left[X\right]$ finite, then $\mathbb {E}_{\mathcal{B}}\left[X_{n}\right] \xrightarrow [n \rightarrow \infty ]{\text{ a.s. }} \mathbb {E}_{\mathcal{B}}\left[X\right]$ for any $\sigma$ -field $\mathcal{B} \subseteq \mathcal{A}$ (see Theorem 7 (iii')).

By means of an example, show that the convergence $X_{n} \xrightarrow {\text{ a.s. }} X$ cannot be replaced by $X_{n} \xrightarrow [n \rightarrow \infty ]{P} X$ and still conclude that $\mathbb {E}_{\mathcal{B}}\left[X_{n}\right] \xrightarrow {\xrightarrow {\text{ a.s. }}} \mathbb {E}_{\mathcal{B}}\left[X\right]$ .

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Problem 9.12

(i)(ii)

Let the r.v.s $X$ and $Y$ have the Bivariate Normal distribution with parameters $\mu_{1}, \mu_{2}$ in $\mathbb {R}, 0 < \sigma_{1}, \sigma_{2} < \infty$ , and $\rho \in [-1,1]$ , so that their joint probability density function (p.d.f.) is given by

p_{X, Y}(x, y) = \frac{1}{2 \pi \sigma _{1} \sigma _{2} \sqrt{1-\rho ^{2}}} e^{-q / 2}

where

\begin{aligned} q = \frac{1}{1-\rho ^{2}} \left[\left(\frac{x-\mu _{1}}{\sigma _{1}}\right)^{2}-2 \rho \left(\frac{x-\mu _{1}}{\sigma _{1}}\right)\left(\frac{y-\mu _{2}}{\sigma _{2}}\right)+\left(\frac{y-\mu _{2}}{\sigma _{2}}\right)^{2}\right], \quad x, y \in \mathbb {R} \end{aligned}

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(i)

Show that the exponent may be written thus:

\begin{aligned} \left(1-\rho ^{2}\right) q = & {\left[\left(\frac{y-\mu _{2}}{\sigma _{2}}\right)-\rho \left(\frac{x-\mu _{1}}{\sigma _{1}}\right)\right]^{2}+\left(1-\rho ^{2}\right)\left(\frac{x-\mu _{1}}{\sigma _{1}}\right)^{2} } \\ = & \left(\frac{y-b}{\sigma _{2}}\right)^{2}+\left(1-\rho ^{2}\right)\left(\frac{x-\mu _{1}}{\sigma _{1}}\right)^{2}, \\ & \text{ where } b = \mu _{2}+\frac{\rho \sigma _{2}}{\sigma _{1}}\left(x-\mu _{1}\right) . \end{aligned}

(ii)

From part (i), it follows that:

\begin{aligned} p_{X, Y}(x, y) = & \frac{1}{\sqrt{2 \pi } \sigma _{1}} \exp \left[-\frac{\left(x-\mu _{1}\right)^{2}}{2 \sigma _{1}^{2}}\right] \\ & \times \frac{1}{\sqrt{2 \pi }\left(\sigma _{2} \sqrt{1-\rho ^{2}}\right)} \exp \left[-\frac{(y-b)^{2}}{2\left(\sigma _{2} \sqrt{1-\rho ^{2}}\right)^{2}}\right] . \end{aligned}

From this expression, and without any actual integration, conclude that the r.v. $X$ is distributed as $N\left(\mu_{1}, \sigma_{1}^{2}\right)$ ; i.e., $X \sim N\left(\mu_{1}, \sigma_{1}^{2}\right)$ , and by symmetry, $Y \sim N\left(\mu_{2}, \sigma_{2}^{2}\right)$ .

Problem 9.13

In reference to Exercise 12 (ii), and without any actual operations, conclude that the conditional distribution of the r.v. $Y$ , given $X = x$ , is $N\left(b, \sigma_{2}^{2}\left(1-\rho^{2}\right)\right)$ , where $b = \mu_{2}+\frac{\rho \sigma_{2}}{\sigma_{1}}\left(x-\mu_{1}\right)$ ; and by symmetry, the conditional distribution of $X$ , given $Y = y$ , is $N\left(c, \sigma_{1}^{2}\left(1-\rho^{2}\right)\right)$ , where $c = \mu_{1}+\frac{\rho \sigma_{1}}{\sigma_{2}}\left(y-\mu_{2}\right)$ .

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Problem 9.14

(i)(ii)(iii)

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(i)

In reference to Exercises 12 and 13 , and by writing $\mathbb {E}\left[X Y\right] = \mathbb {E}\left[ \mathbb {E}\left[X Y\right] \mid X \right]$ , show that $\mathbb {E}\left[X Y\right] = \mu_{1} \mu_{2}+\rho \sigma_{1} \sigma_{2}$ .

(ii)

Use the definition of the covariance of two r.v.s $X$ and $Y$ (with finite second moments), $\operatorname {Cov}\left[ X, Y \right] = \mathbb {E}\left[(X-\mathbb {E}\left[X\right])(Y-\mathbb {E}\left[Y\right])\right] = \mathbb {E}\left[X Y\right]-(\mathbb {E}\left[X\right])(\mathbb {E}\left[Y\right])$ , in order to conclude that, in the present case, $\operatorname {Cov}\left[ X, Y \right] = \rho \sigma_{1} \sigma_{2}$ .

(iii)

From part (ii), conclude that, in the present case, the correlation coefficient of the r.v.s $X$ and $Y, \rho (X, Y)$ , is equal to $\rho$ .

Problem 9.15

If the r.v.s $X$ and $Y$ have the Bivariate Normal distribution with parameters $\mu_{1}, \mu_{2}$ in $\mathbb {R}, 0 < \sigma_{1}, \sigma_{2} < \infty$ and $\rho \in [-1,1]$ , set

U = \frac{X-\mu _{1}}{\sigma _{1}} \text{ and } V = \frac{Y-\mu _{2}}{\sigma _{2}}

and show that the r.v.s $U$ and $V$ have the Bivariate Normal distribution with parameters $0,0,1,1$ and $\rho$ , by transforming the joint p.d.f. of $X$ and $Y$ into the joint p.d.f. of $U$ and $V$ .

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Problem 9.16

(i)(ii)

Let $X$ and $Y$ be r.v.s defined on the probability space $(\Omega , \mathcal{A}, P)$ , and suppose that $\mathbb {E}\left[X^{2}\right] < \infty$ . Then show that

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(i)

The conditional variance of $X$ , given $Y$ , is given by the formula

\begin{aligned} \operatorname {Var}\left[ X \mid Y \right] & = \mathbb {E}\left[\left\{ [X-\mathbb {E}\left[ X \mid Y \right]]^{2} \mid Y\right\} \right] \\ & = \mathbb {E}\left[ X^{2} \mid Y \right]-[\mathbb {E}\left[ X \mid Y \right]]^{2} \text{ a.s. } \end{aligned}

(ii)

$\operatorname {Var}\left[X\right] = \mathbb {E}\left[\operatorname {Var}\left[ X \mid Y \right]\right]+\operatorname {Var}\left[\mathbb {E}\left[ X \mid Y \right]\right]$ .

Problem 9.17

(i)(ii)(iii)(iv)

(Wald)

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(i)

Let $X_{1}, X_{2}, \ldots$ be r.v.s and let $N$ be a r.v. taking the values $1,2, \ldots$ , all defined on the probability space $(\Omega , \mathcal{A}, P)$ . Define the function $X$ as

X(\omega ) = X_{1}(\omega )+\cdots +X_{N(\omega )}(\omega )

and show that $X$ is a r.v.

(ii)

Now suppose that the $X_{i}$ s are independent and identically distributed with $\mathbb {E}\left[X_{1}\right] = \mu \in \mathbb {R}$ , that $N$ is independent of the $X_{i} \mathrm{~ s}$ , and that $\mathbb {E}\left[N\right] < \infty$ . Then show that $\mathbb {E}\left[ X \mid N \right] = \mu N$ , and therefore $\mathbb {E}\left[X\right] = \mu (\mathbb {E}\left[N\right])$ .

(iii)

If in addition to the assumptions made in part (ii), it also holds that $\operatorname {Var}\left[X_{1}\right] = \sigma^{2} < \infty$ and $\operatorname {Var}\left[N\right] < \infty$ , then show that $\operatorname {Var}\left[ X \mid N \right] = \sigma^{2} N$ .

(iv)

Use parts (ii) and (iii) here and part (ii) of Exercise 16 in order to conclude that $\operatorname {Var}\left[X\right] = \sigma^{2}(\mathbb {E}\left[N\right])+\mu^{2} \operatorname {Var}\left[N\right]$ .

Problem 9.18

Let $\mathcal{B}$ be a sub- $\sigma$ -field in $(\Omega , \mathcal{A}, P)$ which is equivalent to the trivial $\sigma$ -field $\left\{ \varnothing , \Omega \right\}$ (in the sense that, for every $B \in \mathcal{B}$ , either $\mathbb {P}\left(B\right) = 0$ or $\mathbb {P}\left(B\right) = 1$ ), and let $X$ be a $\mathcal{B}$ -measurable (integrable) r.v. Then show that $X = \mathbb {E}\left[X\right]$ a.s.

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Problem 9.19

In reference to Example 2, show that, for each $t = 0,1, \ldots$ , there are, indeed, $\binom {n+t-1}{t} \mathbf{x}_{i} = \left(x_{i 1}, \ldots , x_{i n}\right)$ where each $x_{i 1}, \ldots , x_{i n}$ ranges from 0 to $t$ , and $x_{i 1}+\cdots +x_{i n} = t, i = 1, \ldots ,\binom {n+t-1}{t}$ .

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