Consider a sequence of non-negative random variables $\{X_n\}$ such that $X_n\overset{a.s}\longrightarrow X$, such that $\mathbb{P}(X>0)=1$.
I have to show that the below infinite series is convergent:
$$\sum_{n=1}^\infty\mathbb{E}(e^{-e^nX_n})$$
Intuitively, I can see how this might happen, since the exponent is negative, $e^n\uparrow\infty$ and $X_n$ converges almost surely to a positive random variable.
But I am having trouble figuring out how to show this mathematically.
The only way I could think of proceeding was to show that for large $n$, we can have $X_n$ to be arbitrarily close to $X$ and if $X$ was almost surely greater than some small positive constant, then $X_n$ is also positive, for all large n. And this ensures the convergence of the series.
But this is incorrect because there need not exist a particular $N$ after which $X_n$ is arbitrarily close to $X$ over the entire sample space (the convergence need not be uniform).
I do not know how else to proceed now, it would be great if someone could help me with an idea or give me a hint.
Thank you so much!
Best Answer
You cannot prove a false statement.
To construct a counter-example, let $U$ be uniformly distributed over $[0, 1]$ and consider the random variables $X_n$ and $X$ defined by
$$ X_n = \mathbf{1}_{\{U > \frac{1}{n}\}} + e^{-n}\mathbf{1}_{\{U \leq \frac{1}{n}\}} \qquad\text{and}\qquad X = 1. $$
Then $X_n \to X$ a.s. and $\mathbf{P}(X = 1) = 1$, but
$$ \mathbf{E}[e^{-e^n X_n}] \geq \mathbf{E}[e^{-1}\mathbf{1}_{\{U \leq \frac{1}{n}\}}] = \frac{1}{en}. $$
Hence it follows that
$$ \sum_{n=1}^{\infty} \mathbf{E}[e^{-e^n X_n}] = \infty. $$