Standard Continuous Uniform Distribution in terms of Exponential Distribution

Theorem
Let $X$ and $Y$ be independent random variables.

Let $\beta$ be a strictly positive real number.

Let $X$ and $Y$ be random samples from the exponential distribution with parameter $\beta$.

Then:


 * $\dfrac X {X + Y} \sim \operatorname U \openint 0 1$

where $\operatorname U \openint 0 1$ is the uniform distribution on $\openint 0 1$.

Proof
Note that the support of $\operatorname U \openint 0 1$ is $\openint 0 1$.

It is therefore sufficient to show that for $0 < z < 1$:


 * $\map \Pr {\dfrac X {X + Y} \le z} = z$

Note that if $x, y > 0$ then:


 * $0 < \dfrac x {x + y} < 1$

Note also that:


 * $\dfrac x {x + y} \le z$

with $0 < z < 1$ is equivalent to:


 * $x \le z x + z y$

which is in turn equivalent to:


 * $\paren {1 - z} x \le z y$

That is:


 * $x \le \dfrac z {1 - z} y$

We therefore have:


 * $\map \Pr {\dfrac X {X + Y} \le z} = \map \Pr {X \le \dfrac z {1 - z} Y}$

Let $f_{X, Y}$ be the joint probability density function of $X$ and $Y$.

Let $f_X$ and $f_Y$ be the probability density function of $X$ and $Y$ respectively.

From Condition for Independence from Joint Probability Density Function, we have for each $x, y \in \R_{> 0}$:


 * $\map {f_{X, Y} } {x, y} = \map {f_X} x \map {f_Y} y$

We therefore have: