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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?

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  • $\begingroup$ 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 ? $\endgroup$
    – G2MWF
    Commented Dec 26, 2022 at 18:26
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    $\begingroup$ 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). $\endgroup$
    – Manó
    Commented Dec 26, 2022 at 18:43
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    $\begingroup$ @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 $\Lambda \leq c \Leftrightarrow-\Lambda \geq -c$. $\endgroup$
    – Manó
    Commented Dec 26, 2022 at 19:00
  • $\begingroup$ 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 ! $\endgroup$
    – G2MWF
    Commented Dec 26, 2022 at 20:15
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    $\begingroup$ Thank you, @coboy, and happy holidays. :) $\endgroup$
    – Manó
    Commented Dec 26, 2022 at 20:49

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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.

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  • $\begingroup$ 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 $\lambda_0$. $\endgroup$
    – Manó
    Commented Dec 26, 2022 at 20:27
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    $\begingroup$ 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. $\endgroup$
    – jerkajerka
    Commented Dec 26, 2022 at 21:20
  • $\begingroup$ 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. $\endgroup$
    – Manó
    Commented Dec 26, 2022 at 22:45

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