0
$\begingroup$

Let $X_1, X_2, X_3,...$ be a sequence of nonnegative random variables such that $\lim\limits_{n\to\infty} E[X_n]=0$.

I want to know if the sequence $Y_n=1-e^{-X_n}$ converges in probability and if so, what is the limit.

My thought is to approach this problem using the Markov inequality:

$$P[X>\epsilon]\leq\frac{E[X]}{\epsilon}$$

Then, by plugging in $Y_n$, I get,

$$P[Y_n>\epsilon]\leq\frac{E[1-e^{-X_n}]}{\epsilon}$$

and since $\lim\limits_{n\to\infty} E[X_n]=0$, then $E[1-e^{-X_n}]$ must also go to $0$ as $n\to\infty$. Therefore, $Y_n$ converges in probability to $0$.

Is this a correct proof?

$\endgroup$
  • $\begingroup$ What is the meaning of $X_n$? $\endgroup$ – callculus Jan 8 at 19:48
  • $\begingroup$ Of course, you might be asked to include a proof of the step that $E(X_n)\to0$ implies $E(1-e^{-X_n})\to0$. Can you do that? $\endgroup$ – Did Jan 8 at 20:58
  • 1
    $\begingroup$ @callculus There is no specific meaning other than what is stated. X_i are a sequence of rvs and the limit of that sequence has a mean of zer0. $\endgroup$ – Avedis Jan 8 at 21:02
  • 1
    $\begingroup$ @callculus It doesnt mean what you implied. It could, but the problem is specifically not commenting. You are given the facts that you have a sequence, the rv's are non-negative, and the expectation of the limit is 0. $\endgroup$ – Avedis Jan 8 at 21:08
  • 1
    $\begingroup$ You are again asserting that $E(X_n)\to0$ implies $E(e^{-X_n})\to1$, not proving it. Can you prove this? $\endgroup$ – Did Jan 8 at 21:31
1
$\begingroup$

It is well known that $$0\leq 1-e^{-x}\leq x$$ for all $x\geq 0$ (this fact can be easily proven using Mean Value Theorem for instance). In the particular problem you are considering the inequality translates to $$0\leq Y_n \leq X_n$$ That gives the result $$\lim_{n\to\infty} \mathbb E|Y_n|=0$$ Hence $Y_n$ converges to $0$ in $L^1$ so it converges in probability as well (by Markov's inequality). You indeed said the result; here is a way to prove it.

$\endgroup$

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.