1
$\begingroup$

I'm having a little trouble with the following problem:

"Let $X \sim N(0,\sigma^2)$ and consider testing $H_0: \sigma = \sigma_0$ versus $H_A: \sigma = \sigma_1$, where $\sigma_1 > \sigma_0$. The values $\sigma_0$ and $\sigma_1$ are fixed.

For a sample $X_1, \ldots, X_n$:

(1) What is the likelihood ratio?

(2) What is the rejection region of a level $\alpha$ test?"

So far, I've gotten that the likelihood ratio is: \begin{equation*} \Lambda = \bigg( \frac{ \sigma_1 } { \sigma_0 } \bigg)^n \exp\bigg\{ \frac{1}{2}\Big(\frac{1}{\sigma_1^2}-\frac{1}{\sigma_0^2} \Big) \sum\limits_{i = 1}^n X_i^2 \bigg\} \end{equation*} However, the answer that I was given for the likelihood ratio doesn't have $\Big( \frac{ \sigma_1 } { \sigma_0 } \Big)^n$, it has $\frac{ \sigma_1 } { \sigma_0 }$. Is this a mistake in the answer?

For the rejection region, I had done the following: \begin{equation*} - 2 \log(\Lambda) = - 2 n \log\bigg( \frac{ \sigma_1 } { \sigma_0 } \bigg) + \Big(\frac{1}{\sigma_1^2}-\frac{1}{\sigma_0^2} \Big) \sum\limits_{i = 1}^n X_i^2 \end{equation*}

Since $-2\log(\lambda) \sim \chi^2_1$, we see that \begin{equation*} - 2 n \log\bigg( \frac{ \sigma_1 } { \sigma_0 } \bigg) + \Big(\frac{1}{\sigma_1^2}-\frac{1}{\sigma_0^2} \Big) \sum\limits_{i = 1}^n X_i^2 > \chi^2_1(\alpha) \end{equation*}

\begin{equation*} \bigg| \frac{1}{\sigma_1^2}-\frac{1}{\sigma_0^2} \bigg| \geq \frac{2z(\alpha/2) + 4 n \log(\sigma_1/\sigma_0) } {x^2} \end{equation*}

This rejection region doesn't seem right to me though. Could someone explain to me where I went wrong?

$\endgroup$
5
  • $\begingroup$ You've got an extra factor of $\dfrac 1 2$ inside the exponential. $\qquad$ $\endgroup$ Mar 5, 2016 at 0:24
  • $\begingroup$ Right, thank you for pointing that out $\endgroup$
    – CLL
    Mar 5, 2016 at 0:24
  • $\begingroup$ When you write \text{exp} or \text{log}, instead of \exp and \log, you don't automatically get proper spacing. Consider what you get from A\exp B and A\exp(B) and A\text{exp} B and $A\text{exp}(B)$: $$A\exp B$$ $$A\exp(B)$$ $$A\text{exp} B$$ $$A\text{exp}(B)$$ The second one has less space to the right of $\exp$ than the first one; that's built in to the software; the last two lack proper spacing. $\qquad$ $\endgroup$ Mar 5, 2016 at 0:29
  • $\begingroup$ I wasn't aware of the spacing difference, but that is good to know. Thank you for writing that out $\endgroup$
    – CLL
    Mar 5, 2016 at 0:35
  • $\begingroup$ Typo: That fourth one was A\text{exp} B. $\qquad$ $\endgroup$ Mar 5, 2016 at 0:38

1 Answer 1

1
$\begingroup$

You are right that there should be an $n$th power.

You reject the null hypothesis if $\Lambda$ is too small (since the density under the null hypothesis is in the numerator. That happens if $$ \left( \frac 1 {\sigma_1^2} - \frac 1 {\sigma_0^2} \right) \sum_{i=1}^n X_i^2 $$ is too small. If $\sigma_1>\sigma_0$, that happens if $\displaystyle \sum_{i=1}^n X_i^2$ is too big. If $\sigma_1<\sigma_0$, it happens if that sum is too small.

Under the null hypothesis you have $\displaystyle \frac 1 {\sigma_0^2} \sum_{i=1}^n X_i^2 \sim \chi^2_n$.

Suppose we let $\chi_n^2(\alpha)$ the the positive number satisfying $\Pr(\chi_n^2 > \chi_n^2(\alpha)) = \alpha$. If we reject the null hypothesis when our sum of squares is too big, then we reject the null hypothesis if that sum is bigger than $\sigma_0^2 \chi_n^2(\alpha)$. The rejection region is then the interval $(\sigma^2 \chi_n^2(\alpha),\infty)$.

If the opposite inequality on the two hypothetical variances holds, then mutatis mutandis.

$\endgroup$
3
  • $\begingroup$ Is the rejection region supposed to be $\sigma_0^2 \chi^2_n(\alpha)$? I'm confused by what $\sigma^2_n(\alpha)$ means $\endgroup$
    – CLL
    Mar 6, 2016 at 7:51
  • $\begingroup$ Also, would you mind elaborating on how you got the null distribution? I'm confused about where the $1/\sigma_1^2$ term went $\endgroup$
    – CLL
    Mar 6, 2016 at 7:54
  • $\begingroup$ @cooperl1 : Sorry -- I meant $\sigma^2 \chi_n^2(\alpha)$, with $\chi_n^2(\alpha)$ defined as above. $\qquad$ $\endgroup$ Mar 6, 2016 at 16:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .