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:

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

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:

\(\ds \map \Pr {n X_n \le x}\)

\(=\)

\(\ds \map \Pr {X_n \le \frac x n}\)

\(\ds \)

\(=\)

\(\ds \int_0^{\frac x n} \frac {t^{1 - 1} \paren {1 - t}^{n - 1} } {\map \Beta {1, n} } \rd t\)

Definition of Beta Distribution

\(\ds \)

\(=\)

\(\ds \int_0^{\frac x n} \frac {\paren {1 - t}^{n - 1} } {\frac {\map \Gamma 1 \map \Gamma n} {\map \Gamma {n + 1} } } \rd t\)

Definition of Beta Function

\(\ds \)

\(=\)

\(\ds \int_0^{\frac x n} \frac {\paren {1 - t}^{n - 1} } {\frac {\map \Gamma n} {n \map \Gamma n} } \rd t\)

Gamma Difference Equation

\(\ds \)

\(=\)

\(\ds \int_0^{\frac x n} n \paren {1 - t}^{n - 1} \rd t\)

\(\ds \)

\(=\)

\(\ds \intlimits {-\paren {1 - t}^n} 0 {\frac x n}\)

Primitive of Power

\(\ds \)

\(=\)

\(\ds 1 - \paren {1 - \frac x n}^n\)

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

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

so:

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

That is:

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

as required.

$\blacksquare$