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Consider the following basic random walk : $S_n=\sum_{k=1}^n X_k$ where the $(X_k)$ are i.i.d. with $P(X_k=(-1))=P(X_k=1)=\frac{1}{2}$. Let $M_n=\max(0,S_1,S_2, \ldots ,S_n)$ and $\mu(n,t)=P(M_n \leq t)$.

Using the reflection principle , (as explained here), and putting $q=\lfloor \frac{n-1}{2} \rfloor$ one can see that

$$ \mu(n,t)=P\bigg( -2\lfloor \frac{t}{2} \rfloor-1\leq S_{2q+1} \leq 2\lceil \frac{t}{2} \rceil-1\bigg)=\frac{1}{2^{2q+1}} \sum_{j=-\lfloor \frac{t}{2} \rfloor}^{\lceil \frac{t}{2} \rceil} \binom{2q+1}{q+j} \tag{1} $$

The Stirling formula yields $\binom{2q+1}{q+j} \sim_{q \to \infty} \frac{2^{2q+1}}{\sqrt{\pi q}}$ for any $j$, and hence for a fixed $t$,

$$ \mu(n,t) \sim_{n\to \infty} (t+1) \sqrt{\frac{2}{\pi n}} \tag{2} $$

I ask if this limit is also an uniform upper bound, in the sense that

$$ \mu(n,t) \leq (t+1) \sqrt{\frac{2}{\pi n}} \text{ whenever } \ \mu(n,t) \ \text{is not trivial (i.e. equal to } 0 \ \text{ or } \ 1). \tag{3} $$

UPDATE 02/14/2013 I’ve just realized that the two-variables inequality (3) follows from the much simpler one-variable inequality

$$ \frac{\binom{2q+1}{q}}{2^{2q+1}} \leq \sqrt{\frac{1}{\pi (q+1)}} \tag{4} $$

Indeed, the numerator in the RHS of (1) is a sum of $t+1$ binomials, and the largest such binomial is $\binom{2q+1}{q}$. So (1) implies that

$$ \mu(n,t) \leq (t+1) \frac{\binom{2q+1}{q}}{2^{2q+1}} \tag{5} $$

Inequality (4) looks much easier to prove.

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I changed {\sf max} to \max. That is standard usage and has the effect of providing proper spacing before and after $\max$ in things like $a\max b$, and also, in "displayed" contexts, of suitably positioning subscripts, thus: $\displaystyle\max_{x\in S} f(x)$. –  Michael Hardy Feb 13 '13 at 15:56

1 Answer 1

up vote 1 down vote accepted

The answer is YES, and indeed inequality (4) is true. To see this, put

$$ u_q=\frac{\binom{2q+1}{q}}{2^{2q+1}}\sqrt{q+1} \tag{5} $$

Then the usual Stirling formula stuff shows that $u_q \to \frac{1}{\sqrt{\pi}}$. To show (4), it will therefore suffice to show that $(u_q)$ is increasing. So let us compute $\frac{u_{q+1}}{u_q}$. Using

$$ \binom{2q+3}{2q+1}=\frac{2(2q+3)}{q+2} \binom{2q+1}{q}, $$

we obtain

$$ \frac{u_{q+1}}{u_q} =\frac{q+\frac{3}{2}}{\sqrt{(q+1)(q+2)}}, $$

and this is $\gt 1$ by the AM-GM inequality, qed.

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