$\lim_{n \to \infty} E[e^{-X*\frac{t}{n}}]^n \approx \lim_{n \to \infty} E(1-\frac{E[X] t}{n})^n = e^{-E[X]*t}$

Question

X is a random variable.

I don't understand this passage. Please someone can explain it to me each equality at a time? Thank you!

$\lim_{n \to \infty} E[e^{-X\frac{t}{n}}]^n \approx \lim_{n \to \infty} E(1-\frac{E[X] t}{n})^n = e^{-E[X]t}$

Answer 1

You need some assumption on $X$. I will assume that it is a positive random variable and that $t \geq 0$. It is not clear what $\approx$ means so I will give a proof in two steps. The inequality $e^{-s} \geq 1-s$ is true for all $s \geq 0$ so $[Ee^{-tX/n}]^{n} \geq [1-tEX/n)]^{n}$ and the limit of this is $e^{-tEX}$. On the other hand $(EY)^{n} \leq EY^{n}$ for any positive random variable $Y$ so $[Ee^{-tX/n}]^{n} \leq E[e^{-tX}]$. You can combine these two fact to get the required result.