Question

Suppose that \({{\bf{X}}_{\bf{1}}}\;{\bf{and}}\;{{\bf{X}}_{\bf{2}}}\)are i.i.d. random variables andthat the p.d.f. of each of them is as follows:

\({\bf{f}}\left( {\bf{x}} \right){\bf{ = }}\left\{ \begin{array}{l}{{\bf{e}}^{{\bf{ - x}}}}\;\;\;\;\;\;{\bf{for}}\;{\bf{x > 0}}\\{\bf{0}}\;\;\;\;\;\;\;\;{\bf{otherwise}}\end{array} \right.\)

Find the p.d.f. of \({\bf{Y = }}{{\bf{X}}_{\bf{1}}} - {{\bf{X}}_{\bf{2}}}\)

Short answer

The p.d.f of Y is

\[\]\(g\left( y \right) = \left\{ \begin{array}{l}\frac{1}{2}{e^{ - \left| y \right|}}\;\;\;\;\;{\rm{for}}\; - \infty < y < \infty \\0\;\;\;\;\;\;\;\;\;\;\;{\rm{otherwise}}\end{array} \right.\)

Step-by-step solution

Given information

The p.d.f of two i.i.d variables\({X_1},{X_2}\)are,

\(f\left( {{x_i}} \right) = \left\{ \begin{array}{l}{e^{ - {x_i}}}\;\;\;\;\;\;{\rm{for}}\;{x_i} > 0\\0\;\;\;\;\;\;\;\;\;{\rm{otherwise}}\end{array} \right.\)

Obtainjoint p.d.f. for two variables by cdf approach

Since the random variables are independent, the joint pdf of both variables is a product of the marginalpdf of both variables.

Hence,

\(\begin{aligned}f\left( {{x_1},{x_2}} \right) &= f\left( {{x_1}} \right)f\left( {{x_1}} \right)\\ &= {e^{ - {x_1}}}{e^{ - {x_2}}}\\f\left( {{x_1},{x_2}} \right) &= \left\{ \begin{aligned}{e^{ - \left( {{x_1} + {x_2}} \right)}}\;\;\;\;\;\;for\;{x_1} > 0,{x_2} > 0\\0\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{otherwise}}\end{aligned} \right.\end{aligned}\)

Let, \(Y = {X_1} - {X_2} \Rightarrow {X_1} = Y + {X_2}\)

Assuming that \({X_2} = U \Rightarrow {X_1} = U + Y\)

Substitute the values in the pdf obtained above,

\(\begin{array}{l}f\left( {{x_1},{x_2}} \right) = \left\{ \begin{array}{l}{e^{ - \left( {z + y + z} \right)}}\;\;\;\;\;\;\\0\end{array} \right.\\f\left( {{x_1},{x_2}} \right) = \left\{ \begin{array}{l}{e^{ - \left( {2z + y} \right)}}\\0\;\;\;\;\;\;\;\end{array} \right.\end{array}\)

Obtain the CDF of the variable

Consider two cases.

Case 1:

\(\begin{aligned}For\;y \le 0,\\{f_Y}\left( y \right) &= \int\limits_{ - \infty }^\infty {{e^{ - 2u - y}}du} \\ &= {e^{ - y}}\left( {\frac{{{e^{ - 2u}}}}{2}} \right)_0^\infty \;\;\;\;\;\;\;\;\;\;\;\;\;\left( {{\rm{The}}\;{\rm{limits}}\;{\rm{for}}\;{\rm{the}}\;{\rm{exponential}}\;{\rm{distribution}}} \right)\\ &= {e^{ - y}}\left( {\frac{{0 - 1}}{{ - 2}}} \right)\\ &= \frac{1}{2}\left( {{e^{ - y}}} \right)\end{aligned}\)

Case 2:

\(For\;y > 0,\)

\({f_Y}\left( y \right) = \frac{1}{2}{e^{ - \left| y \right|}}\)

The above result holds true as the difference of two exponential distributions would have positive values.

Conclude the results

The PDF of the variable Y is obtained as follows,

\({f_Y}\left( y \right) = \left\{ \begin{array}{l}\frac{{{e^{ - \left| y \right|}}}}{2}\;\;\;\;\;\;\;\;\;\;\;{\rm{for}} - \infty < y < \infty \\0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{otherwise}}\end{array} \right.\)