Show that if [math]X[/math] is a random variable with mean [math]\mu[/math] and variance [math]\sigma^2[/math], and if [math]X^* = (X - \mu)/\sigma[/math] is the standardized version of [math]X[/math], then

[[math]] g_{X^*}(t) = e^{-\mu t/\sigma} g_X\left( \frac t\sigma \right)\ . [[/math]]