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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 $X \sim N(\mu, \sigma^2)$.

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The minus sign in front turns the thing positive. – André Nicolas Apr 24 '12 at 15:22
Or, as has been mis-attributed as a poem by W. H. Auden, "Minus times minus equals plus, The reason for this we need not discuss. " – Dilip Sarwate Apr 24 '12 at 15:42

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. ;)

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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:

$$\begin{align} -\mathbb{E}\left(\frac 1 {2\sigma^4}-\frac{(X-\mu)^2}{\sigma^6}\right)&=-\frac 1 {2\sigma^4}+\frac 1 {\sigma^6}\mathbb{E}((X-\mu)^2)\\&\overset{( \ast)}{=}\frac{\sigma^2}{\sigma^6}-\frac{1}{2\sigma^4}\\&=\frac{1}{2\sigma^4}\end{align}$$

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Oh, please! Applying $$\mathbb{E}(X^2)-(\mathbb{E}(X))^2=\operatorname{Var}(X)=\sigma^2 \tag{$\ast$}$$ is overkill. Can't you simply say that the definition of the variance is $\sigma^2 = {\mathbb E}((X-\mu)^2)$ and jump directly to the conclusion, as kuku did, (which incidentally holds even when $X$ is not a normal random variable, as longs as it has finite variance). – Dilip Sarwate Apr 24 '12 at 15:24
Right!! Will edit to reflect that. Thank you @DilipSarwate . Whew What world am I in? – user21436 Apr 24 '12 at 15:25

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