[math] \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}[/math]

Let [math]X[/math] and [math]Y[/math] be independent real-valued random variables

with density functions [math]f_X(x)[/math] and [math]f_Y(y)[/math], respectively. Show that the density function of the sum [math]X + Y[/math] is the convolution of the functions [math]f_X(x)[/math] and [math]f_Y(y)[/math]. Hint: Let [math]\bar X[/math] be the joint random variable [math](X, Y)[/math]. Then the joint density function of [math]\bar X[/math] is [math]f_X(x)f_Y(y)[/math], since [math]X[/math] and [math]Y[/math] are independent. Now compute the probability that [math]X+Y \le z[/math], by integrating the joint density function over the appropriate region in the plane. This gives the cumulative distribution function of [math]Z[/math]. Now differentiate this function with respect to [math]z[/math] to obtain the density function of [math]z[/math].