# Signal X and Y are independent and identically distributed exponential random variable with mean...

## Question:

Signal X and Y are independent and identically distributed exponential random variable with mean 1.

(a) Find the cdf of Z = |X - Y|

(b) Find E(Z).

## Exponential distribution

The exponential distribution is a continuous distribution used to model the waiting times or inter-event times. If there are two or more exponential variables with the same parameter value, we can model then using gamma distribution.

Given Information

{eq}*X \sim Exp\left( 1 \right) {/eq}

{eq}*Y \sim Exp\left( 1 \right) {/eq}

{eq}Z = \left| {X - Y} \right| {/eq}

(a)

Finding the distribution of Z:

{eq}\begin{align*} p\left( {Z \le z} \right) &= p\left( {Y \le y} \right).p\left( {X \le z + y} \right)\\ &= \int\limits_{y = 0}^z {\left( {{\lambda _y}{e^{ - {\lambda _y}y}}} \right)\left( {{\lambda _x}{e^{ - {\lambda _x}\left( {z + y} \right)}}} \right)dy} \\ &= {e^{ - z}}\int\limits_0^z {{e^{ - 2y}}dy} \\ &= {e^{ - z}}\left| {\dfrac{{{e^{ - 2y}}}}{{ - 2}}} \right|_0^z\\ &= {e^{ - z}}\left( {\dfrac{{\left( {{e^{ - 2z}} - 1} \right)}}{{ - 2}}} \right)\\ &= \dfrac{{\left( {{e^{ - z}} - {e^{ - 3z}}} \right)}}{2} \end{align*} {/eq}

Hence, the CDF of Z is {eq}\dfrac{{\left( {{e^{ - z}} - {e^{ - 3z}}} \right)}}{2}{/eq}.

(b)

The PDF of Z is calculated while differentiating the CDF:

{eq}\begin{align*} f\left( Z \right) &= \dfrac{1}{2}\left[ {{e^{ - z}}\left( { - 1} \right) - {e^{ - 3z}}\left( { - 3} \right)} \right]\\ &= \dfrac{{3{e^{ - 3z}} - {e^{ - z}}}}{2} \end{align*} {/eq}

Now,

{eq}\begin{align*} E\left( Z \right) &= \int\limits_0^\infty {z.f\left( z \right)dz} \\ &= \dfrac{1}{2}\left\{ {\int\limits_0^\infty {3z{e^{ - 3z}}dz} - \int\limits_0^\infty {z{e^{ - z}}dz} } \right\} \end{align*} {/eq}

The expression seems like the PDF of gamma. So applying the transformation:

{eq}\begin{align*} E\left( Z \right) &= \dfrac{1}{2}\left\{ {\dfrac{{3 \times \left| \!{\overline {\, 2 \,}} \right. }}{{{3^2}}}\int\limits_0^\infty {\dfrac{{{3^2}}}{{\left| \!{\overline {\, 2 \,}} \right. }}z{e^{ - 3z}}dz} - \dfrac{{\left| \!{\overline {\, 2 \,}} \right. }}{{{1^2}}}\int\limits_0^\infty {\dfrac{{{1^2}}}{{\left| \!{\overline {\, 2 \,}} \right. }}z{e^{ - z}}dz} } \right\}\\ &= \dfrac{1}{2}\left\{ {\dfrac{3}{9} \times 1 - 1 \times 1} \right\}\\ &= - \dfrac{1}{3} \end{align*} {/eq}

Hence, the expectation of Z is 0.3333 as{eq}Z = \left| {X - Y} \right|{/eq}. 