LRT in hypothesis testing $X_1,\ldots,X_n$ is a random sample from $\operatorname{Poisson}(\lambda)$
$$P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}, x=0,1,2,3,\ldots$$
Consider the test:
$H_0: \lambda \le \lambda_0$ vs $H_1: \lambda > \lambda_0$
Show that the LRT gives a rejection region on the form $T>c$
The likelihood function is
$$L=\lambda^t e^{-n\lambda}\frac{1}{x_1! \cdots x_n!}$$
where $t=x_1+\cdots+x_n$
$$\frac{\partial}{\partial \lambda} \log L=\frac t \lambda - n = 0$$
$\lambda=\frac{t}{n} $ maximizes $L$
it follows that if $\lambda_0 \le \frac{t}{n}$ LRT is:
$$\frac{\sup_{\lambda \le \lambda_0 L(\lambda\mid x_1,\ldots,x_n)}}{\sup_\lambda L(\lambda\mid x_1,\ldots,x_n)}= \left( \frac{\lambda_0}{t/n}\right)^te^{-n(\lambda_0-t/n)} = e^{t\log(\lambda_0)-t\log(t/n)-n \lambda_0+t}$$
How do I proceed from here to find the rejection region and to show that it is on the form $T>c$?
 A: You probably already know everything below that comes before the line below that is labeled $(1),$ so you can consider the part above that to be an appendix for others who want to understand your question, and start reading after that.
You have
$$
\frac d {d\lambda} \log L(\lambda) = \frac t \lambda - n \quad \begin{cases} > 0 & \text{if } \lambda > t/n, \\ <0 & \text{if } \lambda > t/n. \end{cases}
$$
Therefore the value of $\lambda$ that maximizes $L(\lambda)$ on the interval $0\le\lambda\le\lambda_0$ is
$$
\begin{cases} \lambda_0 & \text{if } t/n \ge \lambda_0, \\ t/n & \text{if } t/n \le \lambda_0, \end{cases}
$$
(these two pieces agree when $t/n=\lambda_0$) and the value of $\lambda$ that maximizes $L(\lambda)$ on the interval $\lambda\ge 0$ is $t/n.$
The likelihood ratio is therefore
$$
\begin{cases} 1 & \text{if } t/n\le\lambda_0, \\[12pt] \dfrac{\lambda_0^t e^{-n\lambda_0}}{(t/n)^t e^{-t}} = \left( \dfrac{\lambda_0}{t/n} \right)^t e^{t-n\lambda_0} & \text{if } t/n\ge\lambda_0. \end{cases}
$$
The logarithm of that last expression is
$$
t\log\lambda_0 - t\log \frac t n + t - n\lambda_0. \tag 1
$$
The variable $t$ takes nonnegative integer values, so it may seem strange to differentiate with respect to it. But it is the restriction to integers, of a function of a real variable $t,$ and if we can show that is a decreasing function of $t$ on the interval $[\lambda_0,\infty),$ then we're done. We have
$$
\frac d {dt} \left( t\log\lambda_0 - t\log \frac t n + t - n\lambda_0 \right) = \log\left(\frac{\lambda_0}{t/n}\right).
$$
This is negative if $t/n > \lambda_0.$
