\[f_Z (z) = (1/2)\lambda e^{\lambda |z|}\] To avoid confusion, let’s rewrite Z = X1 - X2 as Z = X - Y = X + (-Y)
Using the convolution formula \(f_Z(z) = \int_{-\infty}^{\infty}{f_X(x)f_{Y}(z-x) dx}\) and the property \(f_{-Y}(z-x) = f_{Y}(x-z)\) we can rewrite the convolution as:
\[f_Z(z) = \int_{-\infty}^{\infty}{f_X(x)f_{-Y}(z-x) dx} = \int_{-\infty}^{\infty}{f_X(x)f_{Y}(x-z) dx}\]
Recall the exponential distribution:
\[ f(x) = \lambda e ^ {- \lambda x} \qquad x > 0, \qquad f(x) = 0 \qquad x \le 0 \] In our problem we must consider both cases when z < 0 and when z \(\ge\) 0.
When z < 0:
\[f_Z(z) = \int_{0}^{\infty} {\lambda e^{-\lambda x} \lambda e^{-\lambda (x-z)} dx}\] \[f_Z(z) = \lambda^2 e^{\lambda z}\int_{0}^{\infty} {e^{-2\lambda x} dx}\] \[f_Z(z) = \lambda^2 e^{\lambda z} (-\frac{1}{2\lambda}) [e^{-2\lambda x}]_0^\infty\] \[f_Z(z) = \frac{1}{2} \lambda e^{\lambda z} [e^{-2\lambda x}]_0^\infty\] \[f_Z(z) = \frac{1}{2} \lambda e^{\lambda z}\] Now for Z \(\ge\) 0, we can use the fact that X and Y are independent and randomly distributed so Z = X-Y has the same distribution as -Z = Y - X which makes the two distribution symmetric about the y-axis. For symmetric functions \(f_Z(z) = f_Z(-z)\) therefore:
\[f_Z(z) = \frac{1}{2} \lambda e^{\lambda z} \qquad \ when \ z < 0 \] \[f_Z(z) = \frac{1}{2} \lambda e^{-\lambda z} \qquad \ when \ z \ge 0 \] Combining the two, we get the result:
\[f_Z (z) = \frac{1}{2}\lambda e^{-\lambda |z|}\]