Suppose $X_1,X_2,...,X_n$ are i.i.d. $\mathrm{Bernoulli}(\theta)$. We are interested in testing the hypotheses $$H_0:\theta\leq\theta_0$$vs. $$H_1:\theta>\theta_0$$ Show that if we use the Likelihood Ratio Test, then we will end up with the test $$\text{Reject}\space H_0 \space\text{if}\space\sum_{i=1}^nX_i>b$$for some positive constant $b$.

We know that for a likelihood ratio test, if $T(X)$ is a sufficient statistic then it is equivalent to consider the rejection region w.r.t. $T$. Now $T(X)=\sum_{i=1}^nX_i$ is a sufficient statistic which follows a $\mathrm{Binomial}(n,\theta)$ distribution.

I observe that my LRT statistic has the following form:

$$\lambda(X)=\dfrac{\sup_{\theta\leq \theta_0}g(T(X)|\theta)}{\sup_{\theta\in[0,1]}g(T(X)|\theta)}$$ where $g$ is the pmf of $T(X)$;$g(y)=$$n\choose y$$\theta^y(1-\theta)^{n-y}$.

Also I observe, after some calculations, that $$\lambda(X)=1,\space\text{if}\space T(X)\leq n\theta_0 $$ and $$\lambda(X)=\left(\dfrac{n\theta_0}{T(X)}\right)^{T(X)}\left(\dfrac{n-n\theta_0}{n-T(X)}\right)^{n-T(X)}\space\text{if}\space T(X)>n\theta_0$$

I do not know what my conclusion will be once I write $\lambda(X)<c$ for some $c\in[0,1]$. Clearly $T(X)>n\theta_0$ is the case to be considered but the expression of $\lambda(X)$ is not really yielding anything.

  • 3
    $\begingroup$ Well you already knew that $T(x)$ partitions the sample space in the same way as the likelihood ratio; now you should be able to say something about how the likelihood ratio changes as $T(x)$ increases, allowing you to propose a test (you also know the distribution of $T(X)$) & to state an important property it has. $\endgroup$ Commented Jun 5, 2015 at 16:49
  • $\begingroup$ Yes I have precisely got that, even before I saw your comment. It seems that LRT is not "routine" in nature: you need to consider the behavior of the LRT statistic as your sufficient statistic increases or decreases, so LRT is considerably harder than Neyman-Pearson Lemma. $\endgroup$ Commented Jun 6, 2015 at 13:41
  • $\begingroup$ Well, the Neyman-Pearson lemma concerns LRTs where both hypotheses are simple; you'll typically use some sufficient statistic other than the likelihood ratio itself as a test-statistic, & still therefore need to examine the relationship between it & the likelihood ratio. Here you have composite hypotheses (it's sometimes called a generalized LRT) but the approach is the same, & I was hinting that there's a result similar to the N-P lemma that applies. $\endgroup$ Commented Jun 6, 2015 at 16:14
  • $\begingroup$ That this is still the most powerful test at a given significance level? $\endgroup$ Commented Jun 7, 2015 at 2:36
  • $\begingroup$ Yes: if for a test of any simple hypotheses where $\theta_1>\theta_0$, the likelihood ratio's a monotonic function of $T(X)$, the generalized LRT is uniformly most powerful for composite hypotheses where $\theta_1>\theta_0$. (Though you may not have covered that yet, & now I look back at the question it's not asked for.) Can I ask what you're still stuck on, if anything? $\endgroup$ Commented Jun 7, 2015 at 11:22

2 Answers 2


You can take the derivative of $\log(\lambda)$ in order to get a rejection region. This is problem 8.3 in (1).

(1) Casella, G., and Berger, R. L. (2002). Statistical inference. Duxbury Press.

  • $\begingroup$ We are looking for high-quality and self-contained answers - could you expand your answer a little bit so to (at least) summarize what does Casella and Berger say about it? Also, this is obvious case, but nonetheless, could you provide proper reference rather then names of the authors? Thanks $\endgroup$
    – Tim
    Commented Jul 5, 2016 at 7:22

Let $k=\sum_ix_i$. The likelihood $L(\theta|X)=\theta^k(1-\theta)^{n-k}$ is unimodal and maximized at $\theta=\frac{k}{n}$. Hence, the likelihood ratio $\lambda(X)$ is given by $$\lambda(X)=\frac{\theta_0^k(1-\theta_0)^{n-k}}{\left(\frac{k}{n}\right)^k\left(1-\frac{k}{n}\right)^{n-k}}$$ when $\theta_0<\frac{k}{n}$ and by $\lambda(X)=1$ otherwise.

To deal with the case where $\theta_0<\frac{k}{n}$, it is sufficient to show that $\log(\lambda(X))<c$ is equivalent to $k>b$ for some $b$. Note that $\log(\lambda(X))$ is actually a function of $k$, so we just need to show that $\log(\lambda(X))$ is decreasing as a function of $k$, i.e. that the derivative with respect to $k$ is negative.

One may compute (via a fairly messy but straightforward computation) $$\frac{d}{dk}\log(\lambda(X))=\log\theta_0-\log(1-\theta_0)+\log(n-k)-\log k.$$ Then using $\theta_0<\frac{k}{n}$, and substituting, one obtains that the above expression is always less than zero, as desired.


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.