Exponential Distribution in terms of Beta Distribution

Theorem
Let $\sequence {X_n}$ be a sequence of independent random variables with:


 * $X_n \sim \BetaDist 1 n$

for each natural number $n$, where $\BetaDist 1 n$ denotes the beta distribution with parameters $1$ and $n$.

Then:


 * $n X_n \xrightarrow d X$

with:


 * $X \sim \Exponential 1$

where:
 * $\Exponential 1$ denotes the exponential distribution with parameter $1$,
 * $\xrightarrow d$ denotes convergence in distribution.

Proof
We aim to show that for each real $x > 0$, we have:


 * $\displaystyle \lim_{n \mathop \to \infty} \map \Pr {X_n \le x} = \map \Pr {X \le x}$

From the definition of the exponential distribution, we have:


 * $\map \Pr {X \le x} = 1 - e^{-x}$

Note that, from the definition of the beta distribution:


 * $0 \le X_n \le 1$

So, if $n \le x$, we have:


 * $\map \Pr {X_n \le x} = 1$

Take $n \ge x$.

Then, we have:

From the definition of the exponential function as a sequence, we have:


 * $\displaystyle \lim_{n \mathop \to \infty} \paren {1 - \frac x n}^n = e^{-x}$

so:


 * $\displaystyle \lim_{n \mathop \to \infty} \paren {1 - \paren {1 - \frac x n}^n} = 1 - e^{-x} = \map \Pr {X \le x}$

That is:


 * $\displaystyle \lim_{n \mathop \to \infty} \map \Pr {X_n \le x} = \map \Pr {X \le x}$

as required.