Solutions to the Selected Exercises in R. Durrett's Probability: Theory and Examples, 5th edition.

The 2rd part of those solutions can be find in this pdf file.

4.1.2. Prove Chebyshev‘s inequality. If $a>0$$a>0$ then

Proof: Notice that

Therefore, form Theorem 4.1.9. (b) we have

4.1.4. Use regular conditional probability to get the conditional Holder inequality from the unconditional one, i.e., show that if $p,q\in (1,\mathrm{\infty })$$p,q\in (1,\infty)$ with $1/p+1/q=1$$1/p+1/q = 1$ then

Proof: Note that $({\mathbb{R}}^2,\mathcal{B}({\mathbb{R}}^2))$$(\mathbb R^2, \mathcal B(\mathbb R^2))$ is a nice space. Therefore, according to Theorem 4.1.17. there exists a $\mu$$\mu$ which is the regular conditional distribution for $(X,Y)$$(X,Y)$ given $\mathcal{G}$$\mathcal G$. In another word,

• For each $A\in \mathcal{B}({\mathbb{R}}^2)$$A \in \mathcal B(\mathbb R^2)$, $\omega \mapsto \mu (\omega ,A)$$\omega \mapsto \mu(\omega, A)$ is a version of $P((X,Y)\in A|\mathcal{G})$$P((X,Y)\in A|\mathcal G)$.

• For a.e. $w$$w$, $A\mapsto \mu (\omega ,A)$$A\mapsto \mu(\omega, A)$ is a probability measure on $({\mathbb{R}}^2,\mathcal{B}({\mathbb{R}}^2))$$(\mathbb R^2, \mathcal B(\mathbb R^2))$.

Now for a.e. $w$$w$, it follows from the unconditional Holder inequality that

Using Theorem 4.1.16., the above inequality implies that

as desired. $◻$$\Box$

4.1.6. Show that if $\mathcal{G}\subset \mathcal{F}$$\mathcal G\subset \mathcal F$ and $E{X}^2<\mathrm{\infty }$$EX^2 < \infty$ then

Proof: Note that

Therefore we can verify the following Pythagorean law:

Also note that $E[E(X|\mathcal{F})|\mathcal{G}]=E(X|\mathcal{G})$$E[E(X|\mathcal F)|\mathcal G] = E(X|\mathcal G)$. So using this Pythagorean law three times, we get that the desired equality is equivalent to

This is trival. $◻$$\Box$

4.1.9. Show that if $X$$X$ and $Y$$Y$ are random variables with $E(Y|\mathcal{G})=X$$E(Y|\mathcal G) = X$ and $E{Y}^2=E{X}^2<\mathrm{\infty }$$EY^2 = EX^2< \infty$, then $X=Y$$X = Y$ a.s.

Proof: Using the Pythagorean law (see the proof of exercise 4.1.6.), we have

So we have $X=Y$$X = Y$ a.s. $◻$$\Box$

4.1.10. If $E|Y|<\mathrm{\infty }$$E|Y|<\infty$ and $E(Y|\mathcal{G})$$E(Y|\mathcal G)$ has the same distribution as $Y$$Y$, then $E(Y|\mathcal{G})=Y$$E(Y|\mathcal G) = Y$ a.s.

Proof: First we proof that for each random variable $X$$X$ satisfies the condition of this exercise, we have $\{E(X|\mathcal{G})\ge 0\}\stackrel{a.s.}{=}\{X\ge 0\}$$\{E(X|\mathcal G)\geq 0\} \overset{a.s.}{=} \{X\geq 0\}$. In fact, on one hand, Jensen's inequality implies that

On the other hand, the condition that $E(X|\mathcal{G})\stackrel{d}{=}X$$E(X|\mathcal G)\overset{d}{=} X$ says that

So we must have $|E(X|\mathcal{G})|=E(|X||\mathcal{G})$$|E(X|\mathcal G)|= E(|X||\mathcal G)$ as surely.

This leads us to

which forces that

So we must have $\{E(X|\mathcal{G})\ge 0\}\subset \{X=|X|\}=\{X\ge 0\}$$\{E(X|\mathcal G)\geq 0\} \subset \{X=|X|\}= \{X\geq 0\}$. Noticing again we have

so we must have $\{E(X|\mathcal{G})\ge 0\}\stackrel{a.s.}{=}\{X\ge 0\}$$\{E(X|\mathcal G)\geq 0\} \overset{a.s.}= \{X\geq 0\}$ as required. Now take $X=Y-c$$X = Y-c$, we get

This complete the proof. $◻$$\Box$

4.2.2. Give an example of a submartingale $X_n$$X_n$ so that $X_n^2$$X_n^2$ is a supermartingale.

Proof: $X_n=0$$X_n = 0$. $◻$$\Box$

4.2.3. Show that if $X_n$$X_n$ and $Y_n$$Y_n$ are submartingales w.r.t. ${\mathcal{F}}_n$$\mathcal F_n$ then $X_n\vee Y_n$$X_n \vee Y_n$ is also.

Proof: Obviously $X_n\vee Y_n$$X_n\vee Y_n$ is adapted to ${\mathcal{F}}_n$$\mathcal F_n$. From $X_n\vee Y_n\le |X_n|+|Y_n|$$X_n \vee Y_n \leq |X_n|+|Y_n|$, we know $X_n\vee Y_n$$X_n\vee Y_n$ is integrable. Finally, we have

4.2.4. Let $X_n,n\ge 0,$$X_n,n\geq 0,$ be a submartingale with $sup{X}_n<\mathrm{\infty }$$\sup X_n < \infty$. Let ${\xi }_n=X_n-X_{n-1}$$\xi_n = X_n - X_{n-1}$ and suppose $E(sup{\xi }_n^{+})<\mathrm{\infty }$$E(\sup \xi_n^+)<\infty$. Show that $X_n$$X_n$ converges a.s..

Proof: For each $m\ge 0,$$m \geq 0,$ define stopping time ${\tau }_m:=inf\{k:X_k>m\}$$\tau_m:= \inf\{k: X_k > m\}$. From $\underset{n}{sup}{X}_{n\wedge {\tau }_m}^{+}\le X_{{\tau }_m}^{+}=(X_{{\tau }_m-1}+{\xi }_{{\tau }_m}{)}^{+}\le X_{{\tau }_m-1}^{+}+{\xi }_{{\tau }_m}^{+}\le m+\underset{n}{sup}{\xi }_n^{+}$$\sup_n X^+_{n\wedge \tau_m} \leq X^+_{\tau_m} = (X_{\tau_m - 1}+\xi_{\tau_m})^+\leq X_{\tau_m-1}^+ +\xi_{\tau_m}^+ \leq m+\sup_n \xi_n^+$, we know $E\underset{n}{sup}{X}_{n\wedge {\tau }_m}^{+}<\mathrm{\infty }$$E\sup_n X^+_{n\wedge \tau_m} < \infty$. According to Theorem 4.2.11, this says that $X_{n\wedge {\tau }_m}$$X_{n\wedge \tau_m}$ convergence a.s.. (Note that $X_{n\wedge {\tau }_m}$$X_{n\wedge \tau_m}$ is also a submartingale due to Theorem 4.2.9.). Therefore $X_n$$X_n$ convergence on the event $\{{\tau }_m=\mathrm{\infty }\}$$\{\tau_m = \infty\}$. Note that the condition $sup{X}_n<\mathrm{\infty }$$\sup X_n < \infty$ implies that $\bigcup _{m=1}^{\mathrm{\infty }}\{{\tau }_m=\mathrm{\infty }\}\stackrel{a.s.}{=}\mathrm{\Omega }$$\bigcup_{m=1}^\infty \{\tau_m = \infty\} \overset{a.s.}{=} \Omega$. Therefore, $X_n$$X_n$ converges a.s.. $◻$$\Box$

4.2.6. Let $(Y_k{)}_{k\in \mathbb{N}}$$(Y_k)_{k \in \mathbb N}$ be nonnegative i.i.d. random variables with $E{Y}_m=1$$E Y_m = 1$ and $P(Y_m=1)<1$$P(Y_m = 1)< 1$. By example 4.2.3 that $X_n=\prod _{m\le n}{Y}_m$$X_n = \prod_{m\leq n}Y_m$ defines a martingale. (i) Show that $X_n\stackrel{}{\to }0$$X_n \xrightarrow[]{} 0$ a.s.. (ii) Use the strong law of large numbers to conclude $(1/n)\log X_n\stackrel{}{\to }c<0$$(1/n) \log X_n \xrightarrow[]{} c < 0$.

Proof. (i) Since $X_n$$X_n$ is a non-negative martingale, it convergence a.s.ly to a limit, say $X$$X$. Fix a $u>0$$u > 0$ such that $P(|Y_{\cdot }-1|\ge u)>0$$P(|Y_\cdot - 1| \geq u)> 0$. Then for each $ϵ>0$$\epsilon > 0$, since $X_n$$X_n$ is independent of $Y_{n+1}$$Y_{n+1}$, we have

The left hand side converges to $0$$0$ as $n\to \mathrm{\infty }$$n\to \infty$, so we must have $P(|X_n|>ϵ)$$P(|X_n|>\epsilon)$ converges to $0$$0$ as well. Therfore, $X=0$$X = 0$ a.s..

(ii) We can assume that $P(Y_m=0)=0$$P(Y_m = 0) = 0$, since if $P(Y_m=0)>0$$P(Y_m = 0 )>0$, it is easy to see that

which implies that $(1/n)\log X_n\to -\mathrm{\infty }$$(1/n ) \log X_n \to -\infty$.

Now, assuming $P(Y_m=0)=0$$P(Y_m = 0) = 0$, we can write

According to strong law of large numbers (Theorem 2.4.1. and 2.4.5.), we only have to show that $E\log Y_1\in [-\mathrm{\infty },0)$$E \log Y_1 \in [-\infty, 0)$.

Define $Y_1^{(n)}=Y{\mathbf{1}}_{n^{-1}<Y<n}$$Y_1^{(n)} = Y\mathbf 1_{n^{-1}< Y < n}$, then both $Y_1^{(n)}$$Y_1^{(n)}$ and $\log Y_1^{(n)}$$\log Y_1^{(n)}$ are integrable. From Jensen's inequality, we have

By monotonicity, taking $n\to \mathrm{\infty }$$n\to \infty$, we have

Now, we only have to show that $E\log Y_1\ne 0$$E\log Y_1 \neq 0$. In fact, if $E\log Y_1=0$$E \log Y_1 = 0$, we have $\log Y_1$$\log Y_1$ is integrable. So from $E\log Y_1=\log E{Y}_1$$E \log Y_1 = \log E Y_1$ and Exercise 1.6.1. we have $Y_1=1$$Y_1 = 1$ a.s., which contradicts to the condition $P(Y_1=1)<1$$P(Y_1 = 1)< 1$. $◻$$\Box$

4.2.8. Let $X_n$$X_n$ and $Y_n$$Y_n$ be positive integrable and adapted to ${\mathcal{F}}_n$$\mathcal F_n$. Suppose

with $\sum Y_n<\mathrm{\infty }$$\sum Y_n < \infty$ a.s.. Prove that $X_n$$X_n$ converges a.s. to a finite limit.

Proof: Let

which is positive, integrable and adapted to ${\mathcal{F}}_n$$\mathcal F_n$. From

we know $(W_n)$$(W_n)$ is a supersomartingale. Theorem 4.1.12. says that $(W_n)$$(W_n)$ converges a.s. to a finite limit, say $W$$W$. Since $Y_m$$Y_m$ are positive, we have

From the condition $\sum Y_n<\mathrm{\infty }$$\sum Y_n < \infty$ a.s. and the fact that the left hand side of above is non-decreasing, we have $\prod _{m=1}^n(1+Y_m)$$\prod_{m=1}^n (1+Y_m)$ converges a.s. to a finite limit. Therefore $X_n$$X_n$ also converges a.s. to a finite limit. $◻$$\Box$

4.2.9. Suppose $X_n^1$$X_n^1$ and $X_n^2$$X_n^2$ are supermartingales w.r.t. ${\mathcal{F}}_n$$\mathcal F_n$, and $N$$N$ is a stopping time so that $X_N^1\ge X_N^2$$X_N^1 \geq X_N^2$. Then

and

are supermartinales.

Proof: Clearely, $Y_n$$Y_n$ and $Z_n$$Z_n$ are integrable and adapted to ${\mathcal{F}}_n$$\mathcal F_n$. Note that

Therefore,

$◻$$\Box$ 4.3.1. Give an example of a martingale $X_n$$X_n$ with $\underset{n}{sup}|X_n|<\mathrm{\infty }$$\sup_n |X_n| < \infty$ and $P(X_n=ai.o.)=1$$P(X_n = a \ i.o.) = 1$ for $a=-1,0,1$$a = {-1,0,1}$.

Proof: Suppose that $(U_k{)}_{k\in \mathbb{N}}$$(U_k)_{k\in \mathbb N}$ are i.i.d. r.v. with uniform distribution in $(0,1)$$(0,1)$. Let $X_0=0$$X_0 = 0$. For each $n\ge 1$$n\geq 1$, if $X_n=0$$X_n = 0$, let $X_{n+1}={\mathbf{1}}_{U_{n+1}\ge 1/2}-{\mathbf{1}}_{U_{n+1}<1/2}$$X_{n+1} = \mathbf 1_{U_{n+1}\geq 1/2}-\mathbf 1_{U_{n+1}< 1/2}$; if $X_n\ne 0$$X_n \neq 0$, let $X_{n+1}=n^2{X}_n{\mathbf{1}}_{U_{n+1}\le n^{-2}}$$X_{n+1} = n^2 X_{n} \mathbf 1_{U_{n+1} \leq n^{-2}}$. Then $(X_n)$$(X_n)$ is a martingale since

Note that

So B.C. lemma says that $P(X_n\le 1\text{for }n\text{ large engough})=1$$P( X_n \leq 1~\text{for $n$ large engough}) = 1$, which says that $\underset{n}{sup}|X_n|<\mathrm{\infty }$$\sup_n |X_n|< \infty$ a.s..

It is elementary to see that

Notice that we always have $\sum n^{-2}{\mathbf{1}}_{X_n\ne 0}<\mathrm{\infty }$$\sum n^{-2} \mathbf 1_{X_n \neq 0} < \infty$, so the above identity says that

Now, using Theorem 4.3.4. and above we have

in the sense of a.s.. This can only happen if $P(X_n=0i.o.)=1$$P(X_n = 0~i.o.) = 1$.

We can also verify that

So from what we have proved, we know that a.s.ly

Using Theorem 4.3.4., we have that $P(X_n=1i.o.)=1$$P(X_{n} = 1~i.o.) = 1$. Similarly, we have $P(X_n=-1,i.o.)=1$$P(X_n = -1,~i.o.) = 1$. $◻$$\Box$

4.3.3. Let $X_n$$X_n$ and $Y_n$$Y_n$ be positive integrable and adpted to ${\mathcal{F}}_n$$\mathcal F_n$. Suppose $E(X_{n+1}|{\mathcal{F}}_n)\le X_n+Y_n$$E(X_{n+1}|\mathcal F_n) \leq X_n+Y_n$, with $\sum Y_n<\mathrm{\infty }$$\sum Y_n <\infty$ a.s.. Prove that $X_n$$X_n$ converges a.s. to a finite limit.

Proof: Define $M_n=X_n-\sum _{k=1}^{n-1}{Y}_k$$M_n = X_n - \sum_{k=1}^{n-1} Y_k$. Then $(M_n)$$(M_n)$ is a supermartingale, since

Define stopping times

Then it is easy to see that, for each $m\in \mathbb{N}$$m \in \mathbb N$,

is a non-negative supermartingale. Therefore, $M_n$$M_n$ convergences a.s.ly on event $\{N_m=\mathrm{\infty }\}$$\{N_m = \infty\}$. Finally, notice that event

is with probability $1$$1$. $◻$$\Box$

4.3.5. Show $\sum _{n=2}^{\mathrm{\infty }}P(A_n|{\cap }_{m=1}^{n-1}{A}_m^c)=\mathrm{\infty }$$\sum_{n=2}^\infty P(A_n | \cap _{m=1}^{n-1} A_m^c) = \infty$ implies $P({\cap }_{m=1}^{\mathrm{\infty }}{A}_m^c)=0$$P(\cap_{m=1}^\infty A_m^c) = 0$.

Proof: Note that, there is a partition $\{{\tilde{A}}_n:n\in \mathbb{N}\}$$\{\tilde A_n: n\in \mathbb N\}$ for the event $\bigcup _{m=1}^{\mathrm{\infty }}{A}_m$$\bigcup_{m=1}^\infty A_m$ satisfying that

Define a filtration $({\mathcal{F}}_n)$$(\mathcal F_n)$ such that

Notice also that $\{{\tilde{A}}_1,\dots ,{\tilde{A}}_n,\bigcap _{m=1}^n{A}_m^c\}$$\{\tilde A_1,\dots, \tilde A_n, \bigcap_{m=1}^n A_m^c\}$ is a partition for the underlying probablity space $\mathrm{\Omega }$$\Omega$. Therefore, according to Example 4.1.5., we have

From the condition of this exercise, we have

Now, using Theorem 4.3.4. we get that

in the sense of almost sure. This can only happen if $P({\cap }_{m=1}^{\mathrm{\infty }}{A}_m^c)=0$$P(\cap_{m=1}^\infty A_m^c) = 0$. $◻$$\Box$

For the next two exercises, in the context of Kakutani dichotomy for infinite product measures on page 235, suppose $F_n$$F_n$, $G_n$$G_n$ are concentrated on $\{0,1\}$$\{0,1\}$ and have $F_n(0)=1-{\alpha }_n$$F_n(0) = 1-\alpha_n$, $G_n(0)=1-{\beta }_n$$G_n(0) = 1-\beta_n$.

4.3.9. Show that if $\sum {\alpha }_n<\mathrm{\infty }$$\sum \alpha_n < \infty$ and $\sum {\beta }_n=\mathrm{\infty }$$\sum\beta_n = \infty$ then $\mu \perp \nu$$\mu \perp \nu$.

Proof: Let $A:=\{{\xi }_n\ne 0i.o.\}$$A:= \{ \xi_n \neq 0~i.o.\}$. According to B.C. lemma, condition $\sum {\alpha }_n<\mathrm{\infty }$$\sum \alpha _n < \infty$ says that $\mu (A)=0$$\mu(A) = 0$; condition $\sum {\beta }_n=\mathrm{\infty }$$\sum \beta_n = \infty$ says that $\nu (A)=1$$\nu(A) = 1$. So we must have $\mu \perp \nu$$\mu \perp \nu$. $◻$$\Box$

4.3.10. Suppose $0<{\alpha }_n,{\beta }_n<1$$0 < \alpha_n, \beta_n < 1$. Show that $\sum |{\alpha }_n-{\beta }_n|<\mathrm{\infty }$$\sum |\alpha_n - \beta_n|< \infty$ is sufficient for $\mu \ll \nu$$\mu \ll \nu$ in general.

Proof: Let $(U_k{)}_{k\in \mathbb{N}}$$(U_k)_{k\in \mathbb N}$ be i.i.d. r.v. uniform distribution on $[0,1]$$[0,1]$ w.r.t. probability space $(\mathrm{\Omega },\mathcal{F},P)$$(\Omega, \mathcal F, P)$. Define $\{0,1{\}}^{\mathbb{N}}\times \{0,1{\}}^{\mathbb{N}}$$\{0,1\}^\mathbb N\times \{0,1\}^\mathbb N$ random element $({\xi }^1,{\xi }^2)$$(\xi^1, \xi^2)$ by

Then we have ${\xi }^1$$\xi^1$ has distribution $\mu$$\mu$ and ${\xi }^2$$\xi^2$ has distribution $\nu$$\nu$. Note that

therefore, according to B.C. lemma, we have

This says that there exists $K\in \mathbb{N}$$K \in \mathbb N$ such that $P(\bigcap _{k>K}\{{\xi }_k^1={\xi }_k^2\})>0$$P\big(\bigcap_{k> K} \{\xi_k^1 = \xi_k^2\}\big)>0$. On the other hand, it is obvious that

so from the independency, we have

Now, suppose that $\mu \ll \nu$$\mu \ll \nu$ is not true, then according to Kakutani dichotomy, we have $\mu \perp \nu$$\mu \perp \nu$. This says that, there exists a subset $A\subset {\mathbb{R}}^{\mathbb{N}}$$A \subset \mathbb R^\mathbb N$, wuch that $\mu (A)=\nu (A^c)=1$$\mu(A) = \nu(A^c) = 1$. In this case, we have

which says that

This is a contradiction. $◻$$\Box$

4.3.13. Galton and Watson who invented the process that bears their names were interested in the survival of family names. Suppose each family has exactly 3 children but coin flips determine their sex. In the 1800s, only male children kept the family name so following the male offspring leads to a branching process with $p_0=1/8$$p_0 = 1/8$, $p_1=3/8$$p_1 = 3/8$, $p_2=3/8$$p_2 = 3/8$, $p_3=1/8$$p_3 = 1/8$. Compute the probability $\rho$$\rho$ that the family name will die out when $Z_0=1$$Z_0 = 1$.

Proof: According to Theorem 4.3.12. we know that $\rho$$\rho$ is the only solution of

in $[0,1)$$[0,1)$, where

Solving this gives that $\rho =\sqrt{5}-2$$\rho = \sqrt 5 - 2$. $◻$$\Box$ 4.4.3. Suppose $M\le N$$M \leq N$ are stopping times. If $A\in {\mathcal{F}}_M$$A \in \mathcal F_M$ then $L=M{\mathbf{1}}_A+N{\mathbf{1}}_{A^c}$$L = M \mathbf 1_A+N\mathbf 1_{A^c}$ is a stopping time.

Proof: According to Theorem 7.3.6. we have $A^c\in {\mathcal{F}}_M\subset {\mathcal{F}}_N$$A^c \in \mathcal F_M \subset \mathcal F_N$. Therefore, for each $t\ge 0$$t\geq 0$, we have

From the above, we have $\{L\le t\}=(A\cap \{M\le t\})\cup (A^c\cap \{N\le t\})\in {\mathcal{F}}_t$$\{L \leq t\} = (A \cap \{M \leq t\} ) \cup (A^c \cap \{N \leq t\}) \in \mathcal F_t$. $◻$$\Box$

4.4.5. Prove the following variant of the conditional variance formula. If $\mathcal{F}\subset \mathcal{G}$$\mathcal F \subset \mathcal G$ then

Proof: Note that $E(E[Y|\mathcal{G}]|\mathcal{F})=E[Y|\mathcal{F}]$$E\big(E[Y|\mathcal G] \big|\mathcal F\big) = E[Y|\mathcal F]$. So according to the Pythagorean law (see the Slution to Excise 4.1.6.) we get the desired result. $◻$$\Box$

4.4.7. Let $X_n$$X_n$ be a martingale with $X_0=0$$X_0 = 0$ and $E{X}_n^2<\mathrm{\infty }$$EX_n^2 < \infty$. Show that

Proof: According to Theorem 4.2.6. we have $(X_n+c{)}^2$$(X_n +c)^2$ is a submartingale where $c$$c$ is an arbitrary real number. Therefore, for each $c\in \mathbb{R}$$c\in \mathbb R$, according to Doob's inequality

Now, taking $c=\frac{E{X}_n^2}{\lambda }$$c = \frac{EX_n^2}{ \lambda}$ we have

$◻$$\Box$

4.4.8. Let $X_n$$X_n$ be a submartingale and ${\log }^{+}x=max(\log x,0)$$\log ^+ x = \max (\log x, 0)$. Prove

where ${\overline{X}}_n=\underset{k=1}{\overset{n}{max}}{X}_k$$\bar X_n = \max_{k=1}^n X_k$.

Proof: Fix an $M>1$$M > 1$. Note that

Doob's inequality then says that

Now, use the calculus fact that $a\log b\le a{\log }^{+}a+b/e$$a \log b \leq a\log^+a + b/e$, we have

This says that

Finally, taking $M\to \mathrm{\infty }$$M \to \infty$, using monotone convergence theorem, we get the desired result. $◻$$\Box$

4.4.9. Let $X_n$$X_n$ and $Y_n$$Y_n$ be martingales with $E{X}_n^2<\mathrm{\infty }$$EX_n^2 < \infty$ and $E{Y}_n^2<\mathrm{\infty }$$EY_n^2< \infty$. Show that

Proof: Since $E[X_{n+1}-X_n|{\mathcal{F}}_n]=0$$E[X_{n+1} - X_n|\mathcal F_n] = 0$ and $Y_n\in {\mathcal{F}}_n$$Y_n\in\mathcal F_n$ we have by Theorem 4.4.7. that $E[(X_{n+1}-X_n)Y_n]=0$$E[(X_{n+1}- X_n) Y_n] = 0$. Similarly we have $E[(Y_{n+1}-Y_n)X_n]=0$$E[(Y_{n+1} - Y_n)X_n] = 0$. Now it is easy to calculate that

From this to the desired result is trival. $◻$$\Box$

4.4.10. Let $X_n,n\ge 0$$X_n, n\geq 0$, be a martingale and let ${\xi }_n=X_n-X_{n-1}$$\xi_n = X_n - X_{n-1}$ for $n\ge 1$$n \geq 1$. If $E{X}_0^2,\sum _{m=1}^{\mathrm{\infty }}E{\xi }_m^2<\mathrm{\infty }$$E X_0^2, \sum_{m=1}^\infty E\xi_m^2 < \infty$ then $X_n\to X_{\mathrm{\infty }}$$X_n \to X_\infty$ a.s. and in $L^2$$L^2$.

Proof: Using the result in Excise 4.4.9, we have

Therefore $\underset{n}{sup}E{X}_n^2=E{X}_0^2+\sum _{m=1}^{\mathrm{\infty }}E{\xi }_m^2<\mathrm{\infty }$$\sup_n EX_n^2 = EX_0^2 + \sum_{m=1}^\infty E\xi_m^2 < \infty$. According to Theorem 4.4.6. we get the desired result. $◻$$\Box$

4.6.4. Let $X_n$$X_n$ be r.v.'s taking values in $[0,\mathrm{\infty })$$[0,\infty)$. Let $D=\{X_n=0\text{for some}n\ge 1\}$$D = \{X_n = 0~\text{for some}~n\geq 1\}$ and assume

Use Theorem 4.6.9 to conclude that $P(D\cup \{\underset{n}{lim}{X}_n=\mathrm{\infty }\})=1$$P(D \cup \{\lim_n X_n = \infty\}) = 1$.

Proof: Let ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n),n\ge 1$$\mathcal F_n = \sigma(X_1,\dots,X_n), n\geq 1$ and ${\mathcal{F}}_{\mathrm{\infty }}=\sigma ({\cup }_n{\mathcal{F}}_n)$$\mathcal F_\infty = \sigma (\cup_{n} \mathcal F_n)$. According to $D\in {\mathcal{F}}_{\mathrm{\infty }}$$D \in \mathcal F_\infty$, we have by Levy's 0-1 law that $E[D|{\mathcal{F}}_n]\to 1_D$$E[D|\mathcal F_n] \to 1_D$ a.s.. For each $x>0$$x>0$, and each element $\omega \in \{X_n\le xi.o.\}$$\omega\in \{X_n \leq x~i.o.\}$, there exists a sequence of integers $(n_i{)}_{i\in \mathbb{N}}$$(n_i)_{i\in \mathbb N}$ such that for each $i\in \mathbb{N}$$i \in \mathbb N$, we have $X_{n_i}(\omega )\le x$$X_{n_i}(\omega) \leq x$. Therefore, for this $\omega$$\omega$,

So we must have $1_D=1$$1_D = 1$ on this event $\{X_n\le xi.o.\}$$\{X_n \leq x~i.o.\}$. This says that $\{X_n\le xi.o.\}\subset D$$\{X_n \leq x~i.o.\} \subset D$ for each $x>0$$x>0$. Therefore, ${\cup }_{x\in \mathbb{N}}\{X_m\le xi.o.\}\subset D$$\cup_{x \in \mathbb N}\{ X_m \leq x~i.o.\} \subset D$. Finally, noticing that ${\cup }_{x\in \mathbb{N}}\{X_m\le xi.o.\}=\{\underset{n}{lim}{X}_n=\mathrm{\infty }{\}}^c$$\cup_{x \in \mathbb N}\{ X_m \leq x~i.o.\} =\{\lim_n X_n = \infty\}^c$, we must have the desired result. $◻$$\Box$

4.6.5. Let $Z_n$$Z_n$ be a branching process with offspring distribution $p_k$$p_k$. Use the last result to show that if $p_0>0$$p_0 > 0$ then $P(\underset{n}{lim}{Z}_n=0\text{or}\mathrm{\infty })=1$$P(\lim_n Z_n = 0~\text{or}~ \infty) = 1$.

Proof: Let $D:=\{\underset{n}{lim}{Z}_n=0\}=\{Z_n=0,\mathrm{\exists }n\in \mathbb{N}\}$$D:= \{\lim_n Z_n = 0\} = \{Z_n = 0,\exists n\in \mathbb N\}$ be the event of extinction. Let $({\xi }_i^n{)}_{i,n\in \mathbb{N}}$$(\xi_i^n)_{i,n\in \mathbb N}$ be i.i.d. r.v. used in (4.3.4.). Let ${\mathcal{F}}_n=\sigma (Z_1,\dots ,Z_n)$$\mathcal F_n= \sigma(Z_1,\dots, Z_n)$. Now for each $x>0$$x>0$, on event $\{0<Z_n\le x\}$$\{0<Z_n \leq x\}$ we have

On event $\{Z_n=0\}$$\{Z_n = 0\}$, we have $P(D|{\mathcal{F}}_n)=1\ge p_0^x>0$$P(D|\mathcal F_n) = 1 \geq p_0^x>0$. Now, using Exercise 4.6.4. we have