Let $U$ be a uniform random variable on $[−1,1]$. Given $U$, let $Z$ be a normal random variable with mean $U$ and variance $1$. Prove that $Z$ is not a normal random variable by showing that its kurtosis differs from that of a normal random variable.
I didn’t quite understand how to use given $U$, let $Z$… I guess that mean $E[Z｜U]=U$，$Var[Z|U]=1$ and $Z|U$ is normal. But I want to calculate $\text{kurt}[Z]$. I need $E[Z^4]$ and so on. How could I use these conditions to solve it?
On
You are correct -- $Z$ is conditionally normal, but its unconditional distribution will be more dispersed.
First, note that $\text{Kurt}\left({Z|U=u}\right) = 3 \;\;\forall u$ since $Z|U$ is normally distributed for $U=u$
Given the symmetry of $U$ and $Z|U$, we know that $E[Z]=0$. We can also get this from the properties of conditional expectation $E[Z] = E[E[Z|U]] = E[U] = 0 = \mu_Z$
By the definition of kurtosis, we get:
$$\text{Kurt}\left({Z}\right) :=\frac{E\left[(Z-\mu_Z)^4\right]}{\sigma^4} = \frac{E\left[Z^4\right]}{\sigma^4}=\frac{E[Z^4]}{1}=E[E[Z^4|U]]= $$ $$E\left[{\displaystyle U ^{4}+6U^{2}+3}\right] = \frac{1}{5} + \frac{6}{3} +3 = 5.2 \neq 3$$
To compute the expectations necessary to calculate the kurtosis, we can use the tower property of conditional expectations, $\mathbb{E}\left[\mathbb{E}\left[Y\left|U\right.\right]\right] = \mathbb{E}\left[Y\right]$ for any integrable random variable.
Using the tower property, we have for the mean, $\mathbb{E}\left[Z\right] = \mathbb{E}\left[\mathbb{E}\left[Z\left|U\right.\right]\right] = \mathbb{E}\left[U\right] = 0$ and for the variance, $$\text{Var}\left(Z\right) = \mathbb{E}\left[Z^2\right] = \mathbb{E}\left[\mathbb{E}\left[Z^2\left|U\right.\right]\right] = \mathbb{E}\left[\text{Var}\left(Z\left|U\right.\right) + \mathbb{E}\left[Z\left|U\right.\right]^2\right] = 1 + \mathbb{E}\left[U^2\right] = \frac{4}{3}.$$ With these computations, we can calculate the expectation defining kurtosis. We write $\mu$ for the mean and $\sigma^2$ the variance, $$\text{Kurt}\left[Z\right] = \mathbb{E}\left[\left(\frac{Z - \mu}{\sigma}\right)^4\right] = \frac{9}{16}\mathbb{E}\left[Z^4\right]=\frac{9}{16}\mathbb{E}\left[\mathbb{E}\left[Z^4\left|U\right.\right]\right] = \frac{9}{16}\mathbb{E}\left[U^4 + 6U^2 + 3\right] = 2.925, $$ the second to last equality is the fourth moment of a normal distribution.