# Convergence in distribution with exponential limit distribution

Let $X_1,X_2, \ldots$ be independent, identically distributed, positive random variables with probability density function $f$, which is continuous in $(0, \infty)$ and $\lambda :=\lim_{x \searrow 0} f(x) > 0$.

I want to show $Z_n = n \cdot\min(X_1, \ldots, X_n) \rightarrow Z \sim \mathrm{Exp}(\lambda)$.

I calculated $F_n(z) = 1 - \left(1-F\left( \frac{z}{n}\right)\right)^n$, but now I don't know how to go on. Can anybody help?

## 1 Answer

You can show and use the following deterministic result:

If $(a_n)_{n\geqslant 1}$ is a sequence of real numbers which converges to $a$, then $$\lim_{n\to\infty }\left(1+\frac{a_n}n \right)^n=e^a .$$