Consider $Z_1$and $Z_2$, two independent random variables that both follow the normal distribution $N(\mu, \sigma ^2)$:
$X = Z_2$
$Y = Z_1 + Z_1(Z_2 – \mu)^p$ , where $p$ is a positive integer.
Recall that if $X ∼ N(\mu, \sigma ^2)$, then:
- $\mu_{2k} = \mathbb{E}[(X-\mu)^{2k}] = \frac{(2k)!}{2^kk!}\sigma^{2k} $
- $\mu_{2k+1} = \mathbb{E}[(X-\mu)^{2k+1}] = 0 $
Find $Var(\mathbb{E}[Y|X])$ and $Var[Y|X]$.
I found:
$\mathbb{E}[Y|X] = \mathbb{E}[Z_1 + Z_1(Z_2 – \mu)^p | Z_2]$
$ = \mathbb{E}[Z_1|Z_2] + \mathbb{E}[Z_1(Z_2 – \mu)^p | Z_2]$ ; $Z_1$ and $Z_2$ are independent, and $Z_2 = z_2$ so $(Z_2 – \mu)^p$ becomes a constant
$ = \mathbb{E}[Z_1] + (z_2 – \mu)^p\mathbb{E}[Z_1|Z_2]$
$ = \mathbb{E}[Z_1] + (z_2 – \mu)^p\mathbb{E}[Z_1] $
$ = \mu + (z_2 – \mu)^p \mu$
and the variance of a constant is $0$ so $Var(\mathbb{E}[Y|X]) = 0$ but I feel like I may be wrong somewhere…
Best Answer
There shouldn’t be a lowercase $z_2$ in your result – that’s a notation for fixed values of random variables, and no fixed value of $Z_2$ is in play here. You’re right in treating $Z_2$ as constant in evaluating $\mathbb E[Z_1\mid Z_2]$, but that doesn’t mean it’s actually constant – you should keep it as $Z_2$, not $z_2$, in the result, which is then not constant and has non-zero variation. This is somewhat analogous to forming a partial derivative by keeping the remaining variables constant, which doesn’t mean that they’re actually constant and that the result doesn’t depend on them.