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

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.

• Does this answer your question? Likelihood-based hypothesis testing Commented Jan 9, 2020 at 2:04
• Commented Jan 9, 2020 at 2:07

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.

• That applies to paired data. This is instead a two-sample problem where the order of observations within sample is irrelevant. Commented Mar 1, 2021 at 12:42