Let's say I have a sample $X_1, X_2, \cdots ,X_n$ from a Bernoulli distribution, and I want to test the following: $$ H_0 \colon p=p_{0} \quad\text{vs}\quad H_1 \colon p= p_{1} $$ where $p$ is the parameter of a Bernoulli distribution. The likelihood ratio test is then: $$\Lambda(x) = \left(\frac{1 - p_0}{1 - p_1}\right)^n \left(\frac{p_0 (1 - p_1)}{p_1 (1 - p_0)}\right)^{\sum x_{i}} $$ Suppose we want to determine the exact distribution of $\Lambda$. In order to do that, do we suppose the $x_{i}$ considered a Bernoulli with parameter $p_0$ or $p_1$?.

EDIT: In other words, is the distribution of $\Lambda$ determined under $H_0$?

  • 1
    $\begingroup$ Can't answer specifically about the likelihood ratio test, but in all cases hypothesis testing use the distribution of the statistics under the bull hypotheaia, that is p = p0. Wait for somebody to confirm this, though. $\endgroup$
    – mugen
    Commented Mar 26, 2017 at 12:18
  • $\begingroup$ The exact distribution of lambda can be computed for any value of p. The answer depends on why you want to determine the exact distribution of Lambda $\endgroup$
    – WNG
    Commented Mar 26, 2017 at 16:49
  • $\begingroup$ Note that $\log(\Lambda)=a+b\sum x_i$ for constants $a$ and $b$. $\endgroup$
    – Glen_b
    Commented Jun 3, 2020 at 10:02

1 Answer 1


As the likelihood ratio is a function of the data $x$, it is a statistic. You calculate the statistic and test a hypothesis under $H_0$, so here we assume $p_0$ is the true value of the parameter.

Regarding the exact distribution of $\Lambda(x)$:

Say $H_0$ and $H_1$ have parameter spaces $\Theta_0$ (of dimension $d_0$) and $\Theta_1$(of dimension $d_1$) respectively. If $\Theta_0$ is a subspace of $\Theta_1$ (i.e. testing against a general alternative $H_1$), then the distribution of $\Lambda(x)$ is given by the Wilks' Theorem. Here $H_0$ is either simple or composite, but the general alternative $H_1$ is composite.

The theorem states that, as the sample size $n\rightarrow \infty$, then the statistic $-2 \operatorname{log}\Lambda(x)$ converges in distribution to the chi-squared distribution $\chi_{d_1-d_0}^2$ i.e. with degrees of freedom the difference in dimensions in parameter spaces.

Here the null and alternative hypothesis are both simple, so we can't use Wilks' theorem to find the distribution. Instead, after finding the likelihood ratio $\Lambda(x)$ for the observed data, we can use it to apply the Neyman-Pearson lemma in order to find the most powerful test among all tests with a fixed significance level $\alpha$. Let me know if the theorem doesn't make sense.

Hope this helps.

  • $\begingroup$ This doesn't answer my question. my question was: is the distribution of the likelihood ratio test determined under $H_0$ or not? $\endgroup$
    – noob
    Commented Mar 26, 2017 at 12:36
  • $\begingroup$ Yes, you compute the statistic and tests hypotheses under $H_0$. I gave the explanation for the distribution as you wrote you'd also like to determine the exact distribution of $\Lambda(x)$. I will edit it in a moment. $\endgroup$ Commented Mar 26, 2017 at 12:44
  • $\begingroup$ So in my example, we determine the distribution of $\Lambda$ under he assumption that the $x_i$s are Bernoullis with parameter $p_0$? $\endgroup$
    – noob
    Commented Mar 26, 2017 at 12:50
  • 2
    $\begingroup$ Yep, if $H_0$ is true, then you do a likelihood ratio test (LRT) or use Neyman-Pearson lemma for $H_0$ against $H_1$. The chi-squared distribution above also assumes $H_0$ is true. $\endgroup$ Commented Mar 26, 2017 at 12:59

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