# Can Wilk's $-2\log(\Lambda)\sim \chi^2_d$ rule be used with a sample size $n=2$?

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?

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.