# Weak Convergence to Exponential Random Variable

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.

-

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.
Why is $P(X_{1}>\frac{y}{n})^{n}$ equal to $(1-y/n)^{n}$ and not $1-(y/n)^{n}$? If instead it were $P(M_{1}>1-\frac{t}{n})^{n}$, would it equal $(1-1-\frac{t}{n})^{n}$? –  Jessy Cat Dec 12 '13 at 18:18
$X_1$ is uniformly distributed in $[0,1]$. So $P(X > x) = 1 - x$ for every $0 \leq x \leq 1$. The probability $1 -(y/n)^n$ can be interpreted as $$1 - P(X_1 < y/n)^n = 1 - P(\text{All } X_i < y/n) = P(\text{At least one } X_i \geq y/n) = P(\sup X_i > y/n)$$ which is not what we look for. –  Bunder Dec 17 '13 at 10:43