This question is a follow-up to this discussion: https://math.stackexchange.com/questions/634914/distribution-of-likelihood-ratio-in-a-test-on-the-unknown-variance-of-a-normal-s/635371?noredirect=1#comment1339933_635371

Could someone audit the reasoning below?

I am trying to derive the distribution of the likelihood ratio statistic for the hypothesis test below.

Let $X_1 ... X_{n}$ be a random sample from a $N(\mu,\sigma^2)$ distribution, where $\mu$ is known and $\sigma^2$ is unknown. I want to test the hypothesis $H_0 : \sigma^2 = \sigma_{0}^{2} $ vs. $H_1 : \sigma^2 \neq \sigma_{0}^{2}$ (and, trivially, $\sigma^2 >0$).

The generic joint pdf for the n independent random variables (ie. the likelihood function for the random sample) is:

$L=\prod_{i=1}^{n} \large\left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)\cdot e^{-\frac{(X_i - \mu)^2}{2\sigma^2}}= \large\left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)^{n}\cdot e^{-\frac{\sum_{i=1}^{n} (X_i - \mu)^2}{2\sigma^2}}$

Under the null hypothesis, the maximum value taken by $L$ is: $\large\left(\frac{1}{\sqrt{2\pi\sigma_{0}^{2}}}\right)^{n}\cdot e^{-\frac{\sum_{i=1}^{n} (X_i - \mu)^2}{2\sigma_{0}^{2}}}$

If we do not constrain $\sigma^{2}$ to be equal to $\sigma_{0}^{2}$, then $L$ is maximised by the maximum likelihood estimator $\hat{\sigma}^{2}=\frac{\sum_{i=1}^{n} (X_i - \mu)^2}{n}$

In this case, the maximum likelihood becomes: $\large\left(\frac{1}{\sqrt{2\pi\hat{\sigma_{0}}^{2}}}\right)^{n}\cdot e^{-\frac{\sum_{i=1}^{n} (X_i - \mu)^2}{2\hat{\sigma_{0}}^{2}}}$

Setting these as numerator and denominator, respectively, I get the following likelihood ratio statistic

$\Lambda = \LARGE\frac{\large\left(\frac{1}{\sqrt{2\pi\sigma_{0}^{2}}}\right)^{n}\cdot e^{-\frac{\sum_{i=1}^{n} (X_i - \mu)^2}{2\sigma_{0}^{2}}}}{\large(\frac{1}{\sqrt{2\pi\hat{\sigma_{0}}^{2}}})^{n}\cdot e^{-\frac{\sum_{i=1}^{n} (X_i - \mu)^2}{2\hat{\sigma_{0}}^{2}}}} \\ = \frac{ \left(\frac{1}{\sqrt{2\pi\sigma_0^2}}\right)^{n/2}\cdot e^{-\frac{\sum_{i=1}^n (X_i - \mu)^2}{2\sigma_0^2}}} {\left(\frac{1}{\sqrt{2\pi\hat{\sigma}^{2}}}\right)^{n}\cdot e^{-(n/2)}} = \left(\frac {\hat{\sigma}^2}{\sigma_0^2}\right)^{n/2} \cdot \exp\left \{-\frac 12\left(\frac{\sum_{i=1}^n (X_i - \mu)^2}{\sigma_0^2}-n\right)\right\}$

Using the expression for $\hat \sigma^2$ we get

$$\Lambda = \left(\frac {\hat{\sigma}^2}{\sigma_0^2}\right)^{n/2} \cdot \exp\left \{-\frac n2\left(\frac {\hat{\sigma}^2}{\sigma_0^2}-1\right)\right\}$$

Now, under the null, the random variable denoted

$$z_i^2 = \left(\frac {x_i - \mu}{\sigma_0}\right)^2 \sim \chi^2(1)$$

We have

$$ \frac {\hat{\sigma}^2}{\sigma_0^2} = \frac 1n\sum_{i=1}^{n} \left(\frac {x_i - \mu}{\sigma_0}\right)^2 = \frac 1n \sum_{i=1}^{n}z_i^2$$

So we can write $$\Lambda = \left(\frac 1n \sum_{i=1}^{n}z_i^2 \right)^{n/2} \cdot \exp\left \{-\frac n2\left( \frac 1n \sum_{i=1}^{n}z_i^2-1\right)\right\}$$

Taking minus log we have

$$-\ln \Lambda = -\frac n2 \ln \left(\frac 1n \sum_{i=1}^{n}z_i^2 \right) +\frac n2\left( \frac 1n \sum_{i=1}^{n}z_i^2-1\right)$$

Manipulating the second term in the RHS,

$$= -\frac n2 \ln \left(\frac 1n \sum_{i=1}^{n}z_i^2 \right) + \sqrt {\frac 1 2} \left( \frac {\sum_{i=1}^{n}(z_i^2-1)}{\sqrt {2}}\right)$$ and multiplying throughout by $\sqrt {2}$ we obtain

$$-\sqrt {2} \ln \Lambda = -\frac{n}{\sqrt{2}} \ln \left(\frac 1n \sum_{i=1}^{n}z_i^2 \right) + \left( \frac {\sum_{i=1}^{n}(z_i^2-1)}{\sqrt {2}}\right)$$ $$= -\frac{1}{\sqrt{2}} \ln \left[\left(\frac 1n \sum_{i=1}^{n}z_i^2\right)^n \right] + \left( \frac {\sum_{i=1}^{n}(z_i^2-1)}{\sqrt {2}}\right)$$

The second term in the RHS is a standardized sum of i.i.d $\chi^2(1)$ random variables, and this quantity will converge to a $N(0,1)$. Then (abusing notation a bit),

$$\operatorname{plim}\left(-\sqrt {2} \ln \Lambda\right) = \operatorname{plim}\left(-\frac{1}{\sqrt{2}} \right)\cdot \operatorname{plim}\left\{\ln\left[ \left(\frac 1n \sum_{i=1}^{n}z_i^2 \right)\right]^n \right\} + \operatorname{plim}\left( \frac {\sum_{i=1}^{n}(z_i^2-1)}{\sqrt {2}}\right) $$

$$= -\frac{1}{\sqrt{2}} \cdot \left\{\operatorname{plim}\ln\left[\left(1 \right)^n\right] \right\}+ N(0,1)=-\frac{1}{\sqrt{2}} \cdot\infty + N(0,1) = -\infty$$

At this point, I am completely stuck as I can find neither the test statistic nor its distribution. Could someone flag any mistakes I have made and possibly help me find the test statistic and its distribution?

  • 1
    $\begingroup$ I haven't looked the entire post over--it's too long and this site is not for exhaustive review of derivations (or code or analogous work)--but I can't help wondering where the "$\infty$" in the last line came from. $\endgroup$
    – whuber
    Commented Jan 12, 2014 at 17:47
  • $\begingroup$ Well, thanks for commenting anyways. It's better than complete silence, I suppose :) $\endgroup$
    – user114618
    Commented Jan 12, 2014 at 21:11

1 Answer 1


the last line is wrong. it's not infinite; instead, it's zero. therefore it $ -\sqrt {\frac 2n} \ln \Lambda \rightarrow_d N(0,1)$ and all are done


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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