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How do I show that the second derivative is always negative?

I've computed the second derivative to be:

$\displaystyle\frac{n}{2\sigma^4}-\frac{1}{\sigma^6}\sum\limits_{i=1}^n(x_i-\mu)^2$

Then I don't know what to do next, mainly because I don't know how to deal with the summation in the second term.

Also, if $\mu$ is unknown, then $\mu= \bar{x}$. How will that change the answer?

Note:

$n>0$

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

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There is an equation in the material you provided that tells us that the summation term is equal to $n\sigma^2$. I expect you can take it from there.

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  • $\begingroup$ since $x_1, x_2, \dots , x_n$ is a sample, I thought the summation term would be equal to $(n-1)s^2$ where $s^2$ is the sample variance. By the way, can you please tell me where in the material I can find the explanation of the summation term? $\endgroup$ – mauna Jan 22 '14 at 23:00
  • $\begingroup$ The fourth displayed equation says $\sigma^2= \tfrac1n \times \sum\ldots$, so $\sum\ldots = n \sigma^2$. Use this to eliminate the sum from the expression for the second derivative. $\endgroup$ – bubba Jan 23 '14 at 2:43
  • $\begingroup$ thanks! I can't believe I miss that. One last thing, what about the case when $\mu$ is not known? Then $\mu=\bar{x}$. Is it correct to say $\sum \limits_{i=1}^n (x_i-\bar{x})^2 = (n-1)s^2$, where $s^2$ is the sample variance? $\endgroup$ – mauna Jan 23 '14 at 17:01
  • $\begingroup$ > Is it correct ... Yes; that's the definition of sample variance. $\endgroup$ – bubba Jan 24 '14 at 9:03

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