enter image description here

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

I've computed the second derivative to be:


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?



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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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