I've encountered a problem when using the random walk to approximate a Brownian motion:

Consider a sequence of iid random variable ${X_{i}}\ $ where $i=1,2,...\ $ with binary outcomes such that: $X_{i}=1$ with probability $p$, and $X_{i}=-1$ with $1-p$.

Define the partial sums as $Y_{n}=X_{1}+X_{2}+...+X_{n}$.

And the maximum as $Z_{n}=\max_{1\leq i \leq n} {Y_{i}}$

Let $a\ $ be a positive number. The problem is to find the value of

$$ \lim_{n\rightarrow\infty}\Pr\lbrace Z_{n}\geq a\rbrace $$

I tried to consider an example where $a=0.5$, then $$ \Pr\{Z_{1}\geq0.5\}=p $$ $$ \Pr\{Z_{2}\geq0.5\}=\Pr\{\max_{i=1,2}Y_{i}\geq0.5\}=\Pr\{\max\{X_{1},X_{1}+X_2\}\geq0.5\} $$ $$ =1-\Pr\{X_1<0.5, X_1+X_2<0.5\}=1-\Pr\{X_1+X_2<0.5\mid X_1<0.5\}\Pr\{X_1<0.5\} $$ $$ =1-1*(1-p)=p $$ and $$ \Pr\{Z_3\geq0.5\}=1-\Pr\{X_1<0.5, X_1+X_2<0.5, X_1+X_2+X_3<0.5\} $$ but it seems complicated, according to the answer for a similar question: Distribution of maximum of partial sums of independent random variables, Markov property doesn't hold here. And the result is: $$ \Pr\{Z_3\geq0.5\}=p^2+p-p^3 $$

So is there a closed-form expression for this binary case? Many thanks!

  • $\begingroup$ For the closed-form, I don't know but I'd doubt. In any case, you don't need it to solve your problem. $\endgroup$ – user52227 Oct 3 '17 at 8:32

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