# Likelihood ratio test.

Consider a random sample from a distribution with density function $$f(y)=\frac{1}{2 \lambda} e^{-\frac{y}{2 \lambda}}$$ when $$\lambda,y>0$$. I want to construct a likelihood ratio test to determine when we can discard $$H_0 :\lambda = 1$$ in favor of $$H_1 :\lambda \ne 1$$, with significance level $$\alpha=0.01$$. I obtain the following likelihood: $$L(\theta)=\prod_i^n \frac{1}{2 \lambda} e^{-\frac{1}{2 \lambda} y_i}=\frac{1}{2^n \lambda^n} e^{-\frac{1}{2 \lambda} \sum_{i=1}^n y_i}=\frac{1}{2^n \lambda^n} e^{-\frac{n \bar{y}}{2 \lambda}}$$ and the MLE for $$\lambda$$: $$\hat{\lambda}=\frac{1}{\bar{y}}$$ The likelihood ratio is then: $$\Lambda=\frac{L\left(\lambda_0=1\right)}{L\left(\hat{\lambda}=\bar{y}^{-1}\right)}=\frac{\bar{y}^n}{2^n} e^{-\frac{n \bar{y}}{2}+n}$$ We discard $$H_0$$ if $$\Lambda\le c$$ for some $$c$$ determined by the signifinance level. My book states that we can use that $$-2 \log \Lambda \stackrel{D}{\rightarrow} \chi^2(1)$$ under the null hypothesis $$H_0 :\lambda=\lambda_0$$. Hence, I assume that we can discard $$H_0$$ if $$-2 \log\Lambda\ge \chi^2_{.01}$$ where $$\chi^2_{.01}$$ is the $$.01$$-quantile of the chi-squared distribution with 1 degree of freedom: $$\chi^2_{.01}=1.57\times10^{-4}$$.

Is this the correct procedure, or have I misunderstood something?

• I am afraid I cannot help you however your computation is good so if you followed your book which I assume is reliable it should be ok. But I have a question : why do we not take a threshold at $c=1$ ? If I understand, the idea is to compare what is "more likely to be true" in the ratio so having something smaller than $1$ tells us that $H0$ is more likely to be false ? Commented Dec 26, 2022 at 18:26
• I am not taking $c=1$. I am not even computing the value of $c$ explicitly, because I use the fact that $-2 \log \Lambda \stackrel{D}{\rightarrow} \chi^2(1)$. Generally, $c$ should be such that $P(\Lambda \leq c)=\alpha$, but I do not need to pay attention to that since I can use the condition $-2 \log \Lambda \geq \chi_{.01}^2$ (at least, that is the idea).
– Manó
Commented Dec 26, 2022 at 18:43
• @coboy, you write "having something smaller than ... tells us that $H_0$ is more likely to be false". That is correct, we discard $H_0$ if $\Lambda \le c$, but the negative sign switches the equality $\Lambda \leq c \Leftrightarrow-\Lambda \geq -c$.
– Manó
Commented Dec 26, 2022 at 19:00
• Thank you a lot Manó for your answer, it is something I don't know very well (but I should) and you make it clear, I hope you will find an answer to your question ! Commented Dec 26, 2022 at 20:15
• Thank you, @coboy, and happy holidays. :)
– Manó
Commented Dec 26, 2022 at 20:49

I think your critical value is off.

Under the null hypothesis, the test statistic $$-2\log(\Lambda_n)$$ is asymptotically $$\chi^2_1$$ distributed. To carry out a LR test at a given significance level $$\alpha\in(0,1)$$, your critical value should therefore be the $$(1-\alpha)$$-quantile of $$\chi^2_1$$. Based on your description, $$\alpha=.01$$, and hence the right critical value is the $$.99$$-quantile, $$\chi^2_.99(=6.63)$$ in your notation.

Note that the $$\chi^2_1$$ distribution lives on the positive reals. Hence, it is only a surprise if we see something far into the right tail, something "large". Your current critical value, $$\chi^2_.01=.000157$$, is very close to zero. Seeing something larger than that number isn't surprising at all.

• I am using the book Introduction to Mathematical Statistics by Hogg, McKean and Craig. In the examples they give (none of which are exponential distributions) they use the condition "Reject $H_0$ in favor of $H_1$ if test statistic $\ge \chi^2_\alpha(1)$". You write "Seeing something larger than that number isn't surprising at all". However, recall that we test if $\lambda$ is given by one specific value, so it should not be a surprise to see a larger value, unless $\lambda$ is indeed $\lambda_0$.
– Manó
Commented Dec 26, 2022 at 20:27
• Take a closer look at Hogg, McKean and Craig for the definition of $\chi^2_{\alpha}(1)$. I taught from the very same book for years. These authors take it to mean the $(1-\alpha)$-quantile of a $\chi^2$ with one degree of freedom (i.e. the number leaving $\alpha$ probability to the right), and not the $\alpha$-quantile as you write (leaving $\alpha$ probability to the left). Beware that these authors use some unconventional language such as "upper" and "lower" quantiles, which may be the source of your confusion. Commented Dec 26, 2022 at 21:20
• Oh, I had not noticed that; thanks a lot for pointing it out! I am learning from two books simultaneously, and in the other book $\chi_p^2$ means the $p$th quantile.
– Manó
Commented Dec 26, 2022 at 22:45