# LRT for one-sided Bernoulli parameter

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.

• 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. – Scortchi - Reinstate Monica Jun 5 '15 at 16:49
• 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. – Landon Carter Jun 6 '15 at 13:41
• 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. – Scortchi - Reinstate Monica Jun 6 '15 at 16:14
• That this is still the most powerful test at a given significance level? – Landon Carter Jun 7 '15 at 2:36
• 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? – Scortchi - Reinstate Monica Jun 7 '15 at 11:22

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