1
$\begingroup$

Let $(X_1,X_2,...)$ be i.i.d random variables, with $P(X_t=1)=P(X_t=-1)=1/2$. Then

$S_t= \frac{1}{t}\sum_{i=1}^{t}X_i $ is a zero mean random walk. Let $\tau$ be the stopping time corresponding to the first time that $S_t$ hits $M \in (0,1/2)$, \begin{equation} \tau=\inf\{t \geq 1 \mid S_t \geq M \} \end{equation} I am trying to show that $P(\tau = \infty)>0$.

For starters, clearly $P(\tau < \infty)>0$ since $P(\tau=1)=1/2$. Also, the probability that $S_t$ exceeds $M$ infinitely often is zero, since $S_t$ converges to $0$. But I would like to show that with positive probability, $S_t$ can converge to zero without ever exceeding $M$.

$\endgroup$
1
$\begingroup$

I'll recast the problem in terms of the random walk $W_n=W_0+\sum_{i=1}^n (X_i-M).$

For $j\geq 0$, define the (identically distributed) stopping times $$\tau_j=\inf\left(n> j: \sum_{i=j+1}^n (X_i-M)>0\right)=\inf(n> j: W_n>W_{j}). $$

If $\mathbb{P}(\tau_0<\infty)=1,$ then $\mathbb{P}(\tau_j<\infty)=1$ for all $j\geq 0$, and hence $\mathbb{P}\left(\cap_{j=0}^\infty [\tau_j <\infty]\right)=1.$ It follows that, with probability 1, the sequence $(W_j(\omega))_{j\geq 0}$ does not achieve its maximum.

But this contradicts the fact, due to the strong law of large numbers, that $W_j\to-\infty$ with probability one.

Therefore $\mathbb{P}(\tau_0<\infty)=1$ is false, that is, $\mathbb{P}(\tau_0=\infty)>0.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.