Difference of Exponential Random Variables / Linear Transformations of RVs

Question

Suppose X and Y are both distributed exponentially with parameter $\lambda$ and $\mu$ respectively. I am trying to find the distribution of X - Y via this method and it does not seem to be working, could you show me what I am doing wrong?

Define Z1 = X - Y, Z2 = Y so $\begin{bmatrix} Z1 \\ Z2 \end{bmatrix}$ = $\begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}$ $\begin{bmatrix} X \\ Y \end{bmatrix}$ and so $\begin{bmatrix} X \\ Y \end{bmatrix}$ = $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ $\begin{bmatrix} Z1 \\ Z2 \end{bmatrix}$

So the joint pdf of Z1 and Z2 is $ g(z_1, z_2) = 1/det A * f_X(z_1 + z_2)f_Y(z2) = \lambda e^{-\lambda(z_1+z_2)} \mu e^{-\mu z_2} =\lambda \mu e^{-(\lambda + \mu)z_2 -\lambda z_1} $.

Hence the marginal density of Z1 is $g(z_1) = \int_0^\infty \lambda \mu e^{-(\lambda + \mu)z_2 -\lambda z_1} dz_2 = [ \frac{-\lambda \mu}{\lambda + \mu} e^{-(\lambda + \mu)z_2 - \lambda z_1}]^{\infty}_0 = \frac{\lambda \mu}{\lambda + \mu} e^{-\lambda z_1}$ Which is wrong?

Answer 1

You have the correct joint density function for $Z_1,Z_2$:

$$f_{Z_1,Z_2}(z_1,z_2) = \lambda\mu e^{-(\lambda + \mu)z_2-\lambda z_1}.$$

However, the ranges $X\gt 0,\;Y\gt 0$ imply the ranges for $Z_1,Z_2$ are:

$$Z_2 \gt 0,\quad Z_2 \gt -Z_1.$$

So to calculate the marginal density for $Z_1$ we must work with two ranges:

If $z_1 \geq 0$ then we proceed exactly as you have done, giving

$$f_{Z_1}(z_1) = \dfrac{\lambda\mu}{\lambda + \mu}e^{-\lambda z_1}.$$

If $z_1 \lt 0$ then:

\begin{eqnarray*} f_{Z_1}(z_1) &=& \int_{-z_1}^\infty{\lambda\mu e^{-(\lambda + \mu)z_2-\lambda z_1}\;dz_2} \\ &=& \left[\dfrac{-\lambda\mu}{\lambda + \mu} e^{-(\lambda + \mu)z_2-\lambda z_1}\right]_{-z_1}^\infty \\ &=& \dfrac{\lambda\mu}{\lambda + \mu}e^{\mu z_1}. \end{eqnarray*}