2
$\begingroup$

I'm working on the following problem for my first Stochastic Processes course.

Let $(S_n)_{n \in \mathbb{N}}$ be a random walk starting at $0$ with independent increments $X_n$ that satisfy $\mathbb{E}\big(e^{\delta X_n}\big) = 1, \forall n$ and $\delta > 0$. Show that $\mathbb{P}(S_n > a \text{ for some $n$}) \leq e^{-\delta a}$, where $a > 0$.

I'm having a lot of trouble with the "for some $n$" part. For example, I am able to prove the following: pick any $n$. Now, notice that,

$\{S_n > a\} \iff \{e^{\delta S_n} >e^{\delta a} \}$

From Markov's inequality, it follows that

$\mathbb{P}(e^{\delta S_n} >e^{\delta a}) \leq \frac{1}{e^{\delta a}}$

because $\mathbb{E}(e^{\delta S_n}) = 1$. Hence, $\mathbb{P}(S_n \geq a) \leq e^{-\delta a}$

However, this is not the required result. This was proved for a given $n$. Is there a way to adapt the argument and prove the correct statement. Also, it can be the case in which the professor messed up things with a typo or so on, I don't know. Any ideas on how to follow from here? Thanks in advance!!

$\endgroup$
  • $\begingroup$ Have you thought about using Borel-Cantelli? I haven’t thought about it much, but this problem has the feel of one where B-C might be useful. $\endgroup$ – Theoretical Economist Aug 28 '18 at 21:08
2
$\begingroup$

I think your professor is correct; I think the proof involves:

a) Proving that $e^{\delta S_n}$ is a martingale.

b) Using Doob's martingale inequality, which states that $P(X_k \geq a$ for some k$\in [n]) \leq \frac{\mathbb{E}(X_n)^+}{a}$ for a submartingale. (where the $(f(x))^+ = \max(0,f(x))$, but in your case since your martingale is non-negative it does not matter.)

c) Now, the right hand side does not depend on $n$, so you should be able to take the limit with dominated convergence theorem.

(You can prove Doob's martingale inequality by using a stopping time where you check when your process exceeds $a$ or reaches time $n$.)

$\endgroup$
  • 1
    $\begingroup$ It worked!! However, dominated convergence was not needed. I could do it using the continuity of probability measures. $\endgroup$ – Raul Guarini Aug 29 '18 at 16:32
  • 1
    $\begingroup$ Thanks a lot @E-A $\endgroup$ – Raul Guarini Aug 29 '18 at 16:32

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.