Standard Continuous Uniform Distribution in terms of Exponential Distribution

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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:

\(\ds \map \Pr {\dfrac X {X + Y} \le z}\) \(=\) \(\ds \map \Pr {X \le \frac z {1 - z} Y}\)
\(\ds \) \(=\) \(\ds \int_0^\infty \int_0^{\frac z {1 - z} y} \map {f_{X, Y} } {x, y} \rd x \rd y\)
\(\ds \) \(=\) \(\ds \int_0^\infty \int_0^{\frac z {1 - z} y} \map {f_X} x \map {f_Y} y \rd x \rd y\)
\(\ds \) \(=\) \(\ds \int_0^\infty \frac 1 \beta \map \exp {-\frac y \beta} \paren {\int_0^{\frac z {1 - z} y} \frac 1 \beta \map \exp {-\frac x \beta} \rd x} \rd y\) Definition of Exponential Distribution
\(\ds \) \(=\) \(\ds \int_0^\infty \frac 1 \beta \map \exp {-\frac y \beta} \intlimits {-\map \exp {-\frac x \beta} } 0 {\frac z {1 - z} y} \rd y\) Primitive of Exponential of $a x$
\(\ds \) \(=\) \(\ds \int_0^\infty \frac 1 \beta \map \exp {-\frac y \beta} \paren {1 - \map \exp {-\frac z {\beta \paren {1 - z} } y} } \rd y\) Exponential of Zero
\(\ds \) \(=\) \(\ds 1 - \frac 1 \beta \int_0^\infty \map \exp {-y \paren {\frac z {\beta \paren {1 - z} } + \frac 1 \beta} } \rd y\)
\(\ds \) \(=\) \(\ds 1 - \frac 1 \beta \int_0^\infty \map \exp {-\frac y {\beta \paren {1 - z} } } \rd y\)
\(\ds \) \(=\) \(\ds 1 + \frac 1 \beta \intlimits {\beta \paren {1 - z} \map \exp {-\frac y {\beta \paren {1 - z} } } } 0 \infty\) Primitive of Exponential of $a x$
\(\ds \) \(=\) \(\ds 1 + \frac 1 \beta \paren {0 - \beta \paren {1 - z} }\) Exponential Tends to Zero and Infinity, Exponential of Zero
\(\ds \) \(=\) \(\ds 1 - \paren {1 - z}\)
\(\ds \) \(=\) \(\ds z\)

$\blacksquare$

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