# Hypothesis Testing for Poisson rvs

Let $$x_1,\dots,x_m$$ be iid Poisson$$(\lambda_1)$$ and $$y_1,\dots,y_m$$ be iid Poisson$$(\lambda_2)$$ and independent of $$x_1,\dots,x_m$$. $$S_1=\displaystyle\sum_{i=1}^mx_i$$ and $$S_2=\displaystyle\sum_{i=1}^ny_i$$ are independent Poissons with parameters $$m\lambda_1$$ and $$n\lambda_2$$.
a) Show that the conditional distribution of $$S_1$$, given that $$S_1+S_2=S$$ is binomial with parameters $$(s,p)$$ where $$p=\frac{m\lambda_1}{m\lambda_1+n\lambda_2}$$b) Test the null hypothesis $$H_0:\theta=1$$ versus $$H_1:\theta<1$$, where $$\theta=\frac{\lambda_1}{\lambda_2}$$. What is $$p$$ under $$H_0$$? Should you reject the null hypothesis for large or small values of $$S_1$$?
c) Suppose that $$m=n=25$$ and $$S_1=6$$ and $$S_2=18$$. Compute the one-sided $$p$$-value for testing $$H_0:\theta=1$$ versus $$H_1:\theta<1$$ using the conditional binomial test suggested above.

I was able to figure out the first part, but need help figuring out parts b and c. Any help will be appreciated.

First of all

• $$X_1,...,X_n$$ are rv's thus Capital letter is required and obviously same observation for $$Y_i$$

Second:

you literally wrote $$y_1,...,y_m$$ are iid poisson with parameter $$\lambda_2$$; this means that the two random sample have the same size $$m$$ but then you show the sum of $$y_i$$ going to $$n$$, that is a different size from $$X-$$ sample

(clarify if the two random samples have the same size or not)

(b) it is evident that $$(p|\mathcal{H}_0)=\frac{m}{m+n}$$

(c) under the given data verify that $$\mathcal{H}_0:\theta=1$$ against the alternative $$\theta<1$$ means to verify that, in a Binomial

$$B(24;p)$$

the parameter $$p=0.5$$ against the alternative that $$p<0.5$$

Using the conditional distribution, as suggested, this is the drawing

As you can see, for $$S_1\leq 6$$ the pvalue is less than 2% (it's about $$1.133\%$$) thus, i.e. you can reject the null hypothesis with any significance level $$\alpha > p_{value}$$

In other words, you reject the null hypothesis with a significance level $$\alpha=5\%$$ but you cannot reject it with a significance level $$\alpha=1\%$$

This also answer to the second (b) question