2
$\begingroup$

Show that, if $\sigma$ is unknown, the likelihood ratio statistic for testing a value of $\alpha$ is given by $$D = n \log\left(1 + \frac{1}{n-1}T^2\right)\;,$$ where $$T = \frac{\hat{α} -\alpha}{\sqrt{s^2/n}}$$

So far, I have the following: $\hat\alpha=\overline{y}$ and $\hat\sigma=\sqrt{\frac{\sum \left ( y_i-\bar{y} \right )^2}{n-1}}$.

Now, when I plug this in to my ratio, I have: $$D=2\left [ l(\hat{\mu}, \hat{\sigma})-l(\mu_0,\hat{\sigma}) \right ]=\frac{n(\bar{y}-\mu_0)^2}{\hat{\sigma}}=\frac{(\bar{y}-\mu_0)^2}{c\hat{\sigma}}\text{ where }c=\frac{1}{n}$$

So, just to clarify the following things:

  1. I accidentally typed $n-1$. I meant $n$ in the denominator for $\sigma$.

  2. My log-likelihood function is: $l(\mu, \sigma)=-n\log(\sigma)-\frac{\sum{(y_i-\bar{y})^2}}{2\sigma^2}$

  3. When I expand my ratio statistic and simplify, the logs cancel because they are identical, and I am left with the equivalent expression of $T^2$. I don't understand how I am supposed to get the expression that I am asked for. Any ideas of what I am doing wrong?

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ This is not an exact duplicate of a question posted twice recently, since, although it seems to be the same homework problem from the same instructor, it asks different questions about it. $\endgroup$ – Michael Hardy Mar 18 '12 at 23:43
0
$\begingroup$

I'll use $\alpha_0$ for the value of $\alpha$ specified by the null hypothesis.

Under the null hypothesis $\alpha=\alpha_0$, you should get the MLE $$\hat\sigma_\text{NH}^2 = \frac 1 n \sum_{i=1}^n (y_i - \alpha_0)^2. $$

Under the alternative hypothesis $\alpha\ne\alpha_0$, you should get two MLEs: $$ \hat\alpha = \overline{y}=\frac{y_1+\cdots+y_n}{n} $$ $$ \hat\sigma_\text{AH}^2 = \frac 1 n \sum_{i=1}^n (y_i-\overline{y})^2. $$ (I don't know why you have $n-1$ instead of $n$. We're looking for MLEs, not unbiased estimators.)

The likelihood function is $$ L(\alpha,\sigma) = \text{constant}\cdot\frac{1}{\sigma^n} \exp\left( \frac{-1}{2} \sum_{i=1}^n \left(\frac{y_i-\alpha}{\sigma} \right)^2\right), $$ so $$ \ell = \log L = -n\log\sigma - \frac 1 2 \sum_{i=1}^n \left(\frac{y_i-\alpha}{\sigma} \right)^2. $$ When you look at $\ell(\alpha_0,\hat\sigma)$, you might want to write $$ \sum_{i=1}^n (y_i-\alpha_0)^2 = n(\alpha_0-\overline{y})^2 + \sum_{i=1}^n (y_i-\overline{y})^2. $$

Could it be that you omitted the $-n\log\sigma$ term?

| cite | improve this answer | | | | |
$\endgroup$

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.