Using martingale inequalities to control the distribution of the supremum of a biased simple random walk

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I have another question about an exercise on martingales. I completely solved it, up to the las point, I know how the method in order to prove the last point, but I am not capable of doing it... perhaps somebody can help me?

Let $(S_n)_{n\geq 0}$ be a simple random walk on $\mathbb{Z}$: $$S_0=0, \quad S_n=X_1+...+X_n \quad n\geq 1,$$ where the random variables are i.i.d. such that $\mathbb{P}\{X_i=1\}=p$ and $\mathbb{P}\{X_i=-1\}=q$ where $p+q=1$. Suppose that $0<p<q$.

  • Let $(Z_n)=(\frac{q}{p})^{S_n}$. Show that $(Z_n)_{n\geq 0}$ is a positive martingale.
  • Derive by a maximal inequality (applied to $(Z_n)_{n\geq 0}$) that $$\mathbb{P}\big\{\sup_{n \geq 0} S_n \geq k\big\}\leq\big(\frac{p}{q}\big)^k\quad \text{ and } \quad \mathbb{E}[\sup_{n \geq 0} S_n]\leq \frac{p}{q-p}.$$

The only thing I do not know to show is $\mathbb{E}[\sup_{n \geq 0} S_n]\leq \frac{p}{q-p}$. Of course one has to write out the definition of $\mathbb{E}[\sup_{n \geq 0} S_n]$ and then use the fact $\mathbb{P}\big\{\sup_{n \geq 0} S_n \geq k\big\}\leq\big(\frac{p}{q}\big)^k$, but I am not able to get a series which converges to the limit $\frac{p}{q-p}$.

Any help is welcome! Thank you!

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For every nonnegative integer valued random variable $R$, $$ \mathbb E(R)=\sum_{k\geqslant1}\mathbb P(R\geqslant k). $$ Hence, $$ \mathbb E\left(\sup\limits_{n\geqslant0}S_n\right)=\sum_{k\geqslant1}\mathbb P\left(\sup\limits_{n\geqslant0}S_n\geqslant k\right)\leqslant\sum_{k\geqslant1}\left(\frac{p}q\right)^k=\ ... $$