Show that $\frac{X-\mu}{\sigma}\sim N(0,1)$ using moment functions

moment-generating-functionsnormal distributionprobability distributions

Let $X\sim N(\mu,\sigma^2)$. Show that $Z=\frac{X-\mu}{\sigma}\sim N(0,1)$ using moment generating functions.

\begin{align*}
M_Z(t)&=M_{\frac{X-\mu}{\sigma}}(t)\\
&=M_{X-\mu}\left(\frac t\sigma\right)\\
&=e^{-\mu t}M_X\left(\frac t\sigma\right)\\
&=e^{-\mu t}\cdot e^{\frac{t\mu}{\sigma}+\frac{t^2}{2}}\\
&=e^{-\mu t+\frac{t\mu}{\sigma}+\frac{t^2}{2}}
\end{align*}

I'm not sure how to conclude that $Z\sim N(0,1)$.

Best Answer

As mentioned in the comments, your error is that $$M_{X-\mu}(t/\sigma) = e^{-\mu t/\sigma}M_X(t/\sigma)$$

Fixing that gives a final expression of $M_Z(t) = e^{t^2/2},$ as expected.

Related Question