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:


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


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?

  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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
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    $\begingroup$ 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

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.


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