Does sequence converge in probability?

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?

• What is the meaning of $X_n$? – callculus Jan 8 at 19:48
• 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? – Did Jan 8 at 20:58
• @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. – Avedis Jan 8 at 21:02
• @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. – Avedis Jan 8 at 21:08
• You are again asserting that $E(X_n)\to0$ implies $E(e^{-X_n})\to1$, not proving it. Can you prove this? – 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.