Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
The minus sign in front turns the thing positive. –  André Nicolas Apr 24 '12 at 15:22
1  
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
add comment

2 Answers

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. ;)

share|improve this answer
add comment

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:

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.