Is there a Wald test to compare the means of two indepent Poisson distributions?

Question

I found a problem, which says

Let $X_1,...,X_{n_1}\sim Poisson(\lambda_1), Y_1,...,Y_{n_2}\sim Poisson(\lambda_2), i.i.d$ and independent of each other. $H_0:\lambda_1=\lambda_2, \ H_1: not \ H_0$.

Derive the Wald's test of size $\alpha$ for testing the hypothesis, when $n_1,n_2$ are large."

I've only learned about Wald's test for random samples from 1 distribution, but I don't know how to do it when comparing 2 distributions.

And I also couldn't understand what 'large' means, because even if $n_1,n_2$ both go to infinity, the speed can be different and there was no information about that.

Answer 1

Since to compare means of two iid distributions, you can compute the differences between respective random variables in both samples and generate new hypotheses as,

$H_0$: $\delta\ = 0\ $ and the alternate, $H_1$: $\delta\ \ne 0\ $.

And then you would get one distribution and then you can apply the wald's test on that distribution as,

computing $estimated \ \delta\\ $ using MLE and computing the standard normal RV,

$W\ = \frac{est.\ \delta -\delta_0}{est.\ se(\ est. \ \delta\ )}\ $

And finally compare it with the given $\alpha$ value for the test.