Assume that $X_1$, $X_2$,… are independent random variables uniformly distributed on $[0,1]$. Let $Y^{(n)}=n\inf\{X_i,1\leq i\leq n\}$. I am asked to prove that it converges weakly to an exponential random variable, i.e. for any continuous bounded function $f:\mathbb R^{+}\rightarrow\mathbb R$,
$\displaystyle \mathbb E\left(f(Y^{(n)})\right)\xrightarrow[n\rightarrow\infty]{}\int_{\mathbb R^{+}}f(u)e^{-u}du$
Definition: A sequence of distribution functions is said to converge weakly to a limit $F$ (written $F_n\Rightarrow F)$ if $F_n(y)\rightarrow F(y)$ for all $y$ that are continuity points of $F$. A sequence of random variables $X_{n}$ is said to converge weakly or converge in distribution to a limit $X_{\infty}$ (written $X_n\Rightarrow X_{\infty}$) if their distribution functions $F_n(x)=\mathbb P(X_n\leq x)$ converge weakly.
Best Answer
Weak convergence is equivalent to cdf convergence at all continuity points of the limiting cdf. In this case it is easier to look at the complementary problem: \begin{align*} P(Y^{(n)} > y) &= P\left( \inf_{1 \leq i \leq n}{X_i} > \frac{y}{n} \right)\\ &= P\left( \text{Every $X_i$ is greater than } \frac{y}{n} \right) \\ &= P\left(X_1 > \frac{y}{n} \right)^n \quad \text{by iid-ness} \\ &= \left( 1 - y/n\right)^{n} \quad \text{for $n$ large enough} \end{align*} Which converges to $e^{-y}$ which is exactly what we want.