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Suppose I have $X \sim \text{Poisson}(\lambda_x)$ and $Y \sim \text{Poisson}(\lambda_y)$ and they are independent. Suppose $H_0: \lambda_x =\lambda_y$ and $H_A: \lambda_x\ne\lambda_y$. My likelihood ratio is

$$\Lambda=\frac{\max_{\lambda_x=\lambda_y}\frac{e^{-\lambda_x}\lambda_x^x}{x!}\frac{e^{-\lambda_y}\lambda_y^y}{y!}}{\max_{\lambda_x\ne\lambda_y}\frac{e^{-(\lambda_x+\lambda_y)}(\lambda_x+\lambda_y)^{x+y}}{(x+y)!}}$$

The MLE for the null is $\hat{\lambda}=\frac{x+y}{2}$ and for the alternative, $\hat{\lambda}_x=x,\hat{\lambda}_x=y$. After canceling, I have

$$\Lambda = \frac{(\frac{x+y}{2})^{x+y}}{x^x y^y}$$

I want to now test whether this is statistically significant. But to do that, I usually use the result of Wilk's theorem which is

$$-2\log(\Lambda)\sim \chi^2_{d-d_0}$$

where $d = \dim(\Omega)$ and $d_0 = \dim(\Omega_0)$. Note, $\Omega_0$ is the set of possible values for the numerator maximization problem and $\Omega$ for the denominator.

Can I still apply this rule given that $n=2$?

I know that the sum of independent Poisson distributions is also Poisson distributed. Could I consider $X$ and $Y$ to be the sums of independent Poisson, which would be like having a large sample size?

kjetil b halvorsen
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Stan Shunpike
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1 Answers1

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You should not in general trust that asymptotic result when $n=2$, at least not without further information. If both $\lambda_x$ and $\lambda_y$ are large, so that the Poisson distributions can be approximated by normals, maybe. You could investigate it yourself by simulation.

But, in this case, why bother? There are non-asymptotic results you can use, see Checking if two Poisson samples have the same mean.

kjetil b halvorsen
  • 77,844