Exercise 3.3 (Normal kurtosis). The kurtosis of a random variable is defined to be the ratio of its fourth central moment to the square of its variance. For a normal random variable, the kurtosis is 3 . This fact was used to obtain (3.4.7). This exercise verifies this fact.
Let $X$ be a normal random variable with mean $\mu$, so that $X-\mu$ has mean zero. Let the variance of $X$, which is also the variance of $X-\mu$, be $\sigma^2$. In (3.2.13), we computed the moment-generating function of $X-\mu$ to be $\varphi(u)=\mathbb{E} e^{u(X-\mu)}=e^{\frac{1}{2} u^2 \sigma^2}$, where $u$ is a real variable. Differentiating this function with respect to $u$, we obtain
$$
\varphi^{\prime}(u)=\mathbb{E}\left[(X-\mu) e^{u(X-\mu)}\right]=\sigma^2 u e^{\frac{1}{2} \sigma^2 u^2}
$$
and, in particular, $\varphi^{\prime}(0)=\mathbb{E}(X-\mu)=0$. Differentiating again, we obtain
$$
\varphi^{\prime \prime}(u)=\mathbf{E}\left[(X-\mu)^2 e^{u(X-\mu)}\right]=\left(\sigma^2+\sigma^4 u^2\right) e^{\frac{1}{2} \sigma^2 u^2}
$$
and, in particular, $\varphi^{\prime \prime}(0)=\mathbf{E}\left[(X-\mu)^2\right]=\sigma^2$. Differentiate two more times and obtain the normal kurtosis formula $\mathrm{E}\left[(X-\mu)^4\right]=3 \sigma^4$.