Ex. 14.4.7 from A First Look at Rigorous Probability Theory

Question

I have found myself stumped by the following question from Rosenthal's A First Look at Rigorous Probability Theory and would appreciate hints.

Exercise $\mathbf{14.4.7}$: Let $C \in \bf{R}$ and let $\{Z_i\}$ be an i.i.d. collection of random variables with $\textbf{P}[Z_i = -1] = 3/4$ and $\textbf{P}[Z_i = C] = 1/4$. Let $X_0 = 5$ and $X_n = 5 + Z_1 + Z_2 + \ldots + Z_n$, for $n \geq 1$.
(a) Find a value of $C$ such that $\{X_n\}$ is a martingale
(b) For this value of $C$ prove or disprove that there is a random variable $X$ such that as $n \to \infty$, $X_n \to X$ with probability $1$
(c) For this value of $C$ prove or disprove that $\textbf{P}[X_n = 0 \text{ for some } n \in \textbf{N}] = 1$

I believe that for part (a) $C$ must be $23$ and for part (b) it is sufficient to note that for such a sequence to converge, as it is integer-valued, it must eventually be constant which is impossible almost everywhere in this case. Part (c) however, eludes me.

Answer 1

I guess it can be solved using martingale theory, but I failed to find it out.

Wald's equation. Let $X_1,,X_2,\cdots$ be $i.i.d.$ with $E|X_i|<\infty$ and $S_{n}=X_{1}+\cdots+X_n$. If $N$ is a stopping time with $EN<\infty$, then $ES_N=EX_1EN$.

Back to this problem. Let $S_n=Z_{1}+\cdots+Z_n$. Let $a<0<b$ be integers and let $N=\inf\{n:S_n\notin(a,b)\}$. We observe that if $x\in(a,b)$, then $$P(x+S_{b-a}\notin(a,b))\geq\left(\frac14\right)^{\frac{b-a}{3}}$$ since $\frac{b-a}{3}$ steps of size $+3$ in a row will take us out of the interval. Iterating the last inequality, it follows that $$P(N>n(b-a))\leq \left(1-\left(\frac14\right)^{\frac{b-a}{3}}\right)^n$$ so $EN=\sum_{k=1}^\infty P(N\geq k)\leq\sum_{n=0}^\infty (b-a+1)P(N>n(b-a))<\infty$. Hence $N<\infty$ a.s. and $S_N\in\{a,b,b+1,b+2\}$. Applying Wald's equation now gives $ES_N=0$ or $$aP(S_N=a)+bP(S_N=b)+(b+1)P(S_N=b+1)+(b+2)P(S_N=b+2)=0.$$ Also we have $$P(S_N=a)+P(S_N=b)+P(S_N=b+1)+P(S_N=b+2)=1.$$ So $$P(S_N=a)=\frac{b+P(S_N=b+1)+2P(S_N=b+2)}{b-a}.$$ Letting $T_a=\inf\{n:S_n=a\}$, we can get $$\frac{b}{b-a}\leq P(T_a<T_b\wedge T_{b+1}\wedge T_{b+2})\leq\frac{b+3}{b-a}.$$ Letting $b\to\infty$ gives $$P(T_a<\infty)=1$$ for all $a<0$. Taking $a=-5$ concludes the proof.

Answer 2

We can rewrite event $$A = \{ X_n = 0 \ \text{ for some }\ n \in \mathbb{N} \}$$ as union of simple events: $$A = \bigcup_k \{ X_k = 0 \}$$.

They are exact same events. Then from union bound we have: $$\mathbb{P}(A) \leq \sum_k \mathbb{P}(X_k = 0)$$

Let us not that these summands can be non-zero iff $k$ is divisible by four, as every move by 3 upward has to be offset by 3 moves by one downward.

Therefore it can be written as: $$\mathbb{P}(X_k = 0) = \begin{cases} 0 & \ \text{if k not divisible by four} \\ \binom{k}{k/4} ( (3/4)^3 1/4)^{k/4} = \binom{k}{k/4} (27/256)^{k/4} & \ \text{otherwise} \end{cases}$$

Therefore right hand-side of the union bound is convergent by the ratio test, and as such, from BC lemma, event $A$ happens almost never.