Why does $-E \left(\frac{1}{2 \sigma^4}- \frac{(X-\mu)^2}{\sigma^6} \right) = \frac{1}{2 \sigma^4}$
Shouldn't it be $-\frac{1}{2 \sigma^4}$? Note that $X \sim N(\mu, \sigma^2)$.
Why does $-E \left(\frac{1}{2 \sigma^4}- \frac{(X-\mu)^2}{\sigma^6} \right) = \frac{1}{2 \sigma^4}$
Shouldn't it be $-\frac{1}{2 \sigma^4}$? Note that $X \sim N(\mu, \sigma^2)$.
Your expression evaluates to $-\frac{1}{2\sigma^4} +\operatorname {Var} X \cdot \frac{1}{\sigma^6} =-\frac{1}{2\sigma^4} + \frac{\sigma^2}{\sigma^6} = \ldots $
Note. I have used the fact that $\operatorname{Var} X = \operatorname{E}(X-\mu)^2$.
Taking advantage of the fact that $X \sim N(\mu, \sigma^2)$, we immediately get $\operatorname{Var} X = \sigma^2$ without much work. ;)
No, it is not.
Here is a full answer:
By definition: $\begin{align}\mathbb{E}((X-\mu)^2)=\sigma^2 \tag{$\ast$}\end{align}$
Now, it is easy to see what you have:
$\begin{align} -\mathbb{E}\left(\frac 1 {2\sigma^4}-\frac{(X-\mu)^2}{\sigma^6}\right)&=-\frac 1 {2\sigma^4}+\frac 1 {\sigma^6}\mathbb{E}((X-\mu)^2)\\&\overset{( \ast)}{=}\frac{\sigma^2}{\sigma^6}-\frac{1}{2\sigma^4}\\&=\frac{1}{2\sigma^4}\end{align}$