# Likelihood ratio test of $n$ iid $\operatorname{Poisson}(\theta)$ random variables.

Let $X_1, X_2, ..., X_n$ be iid random variables, each with a Poisson distribution with parameter $\theta$. Find the form of the likelihood ratio test of $H_0: \theta = 1$ against $H_1:\theta=1.21$. By using the Central Limit Theorem to approximate the distribution of $\sum_i X_i$, show that the smallest value of $n$ required to make $\alpha=0.05$ and $\beta \leq0.1$ is approximately $212$.

• Where are you stuck? – Stefan Hansen Mar 11 '13 at 16:07
• I can show that the likelihood ratio is $e^{-0.21n}1.21^{\sum{x_i}}$, and can use the condition on $\alpha$ to show that we reject $H_0$ if $\bar{X}>\frac{1.645}{\sqrt{n}} +1$. However, I can't see where the $212$ is coming from. – user55225 Mar 11 '13 at 16:29
• thank you, I am now doing so – user55225 Mar 12 '13 at 1:33

We have the likelihood $\Lambda_{{x}}(H_0;H_1) = \frac{\prod_{i=1}^n [e^{-1.21} 1.21^{x_i} / x_i !]}{\prod_{i=1}^n [e^{-1} 1^{x_i} / x_i !]}=e^{-0.21n}1.21^{\sum{x_i}}$.

This is an increasing function of $\sum{x_i}$ (n fixed) and hence is an increasing function of $\bar{x}$. We can use the central limit theorem to approximate the distribution of $\bar{X}$ as:

$\sqrt{n}(\bar{X} - 1) \to N(0,1)$ under $H_0$.

Now if we want $\alpha = 0.05$, then $\alpha = P(\bar{X}>k | H_0) = P(\sqrt{n}(\bar{X} - 1) > \kappa |H_0) = 0.05$.

This gives $\kappa = z_{0.05} = 1.645$, where $z_{\alpha}$ is the upper $100\alpha$% point of the $N(0,1)$ distribution.

So we reject $H_0$ is $\bar{X}>\frac{1.645}{\sqrt{n}}+1$

Now under $H_1$, again by the Central Limit Theorem we have $\frac{\sqrt{n}(\bar{X} - 1.21)}{1.1} \to N(0,1)$, so we want: $P($do not reject $H_0$ | $H_0$ false$)=\beta\leq0.1$

We do not reject $H_0$ if $\sqrt{n}(\bar{X}-1)<1.645$, so we want $P(\sqrt{n}(\bar{X}-1)<1.645 | \theta=1.21)\leq0.1$

$P(\frac{\sqrt{n}(\bar{X}-1)}{1.1}<\frac{1.645}{1.1} | \theta=1.21)=P(\frac{\sqrt{n}(\bar{X}-1.21)}{1.1}<\frac{1.645}{1.1} -\frac{0.21\sqrt{n}}{1.1}| \theta=1.21)\leq0.1$,

and $\frac{\sqrt{n}(\bar{X} - 1.21)}{1.1}$ has the standard normal distribution.

Hence $\frac{1.645}{1.1} -\frac{0.21\sqrt{n}}{1.1}=-1.282$ (from normal tables) $\implies n\approx212$.

You will integrate the distribution of your test statistic from $-\infty$ to some $\tau$ where $\tau$ is the threshold satisfying $\alpha=0.05$. The result of this integration should be $0.1$. From here you can find $n$. Basically, increasing $n$ will get your detection to $1$ and $\beta$ to $0$. Accordingly there should be some $n$ giving you $\beta=0.1$, which is told in the question as $212$.

• Thank you, presumably using the Central Limit Theorem will mean that all of the integration I can do by consulting normal tables. – user55225 Mar 12 '13 at 1:41
• that is correct, of course with some correct parameters. – Seyhmus Güngören Mar 12 '13 at 1:42