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Let $H_0 = \lambda_0$ and $H_1 = \lambda \neq \lambda_0$. Then

$$-2\log \lambda(y) = -2\log\frac{L(y|\lambda_0)}{L(y|\hat{\lambda})} = 2n \left(\bar{y} \log \left(\frac{\bar{y}}{\lambda_0}\right) + \lambda_0 - \bar{y}\right).$$

Then by Wilk's theorem when $H_0$ is true and $n$ is large

$$2n \left(\bar{Y} \log \left(\frac{\bar{Y}}{\lambda_0}\right) + \lambda_0 - \bar{Y}\right) \sim \chi^2_1$$.

Why is that?

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1 Answer 1

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First of all, we know from CLT that $$\sqrt{n}\Big(\bar{Y}-\lambda_0\Big)\longrightarrow_d \mathcal{N}(0,\lambda_0)$$ Set $g(y)=y\ln\big(\frac{y}{\lambda_0}\big)+\lambda_0 -y$. We have $g'(\lambda_0)=0$ so from the second order delta method $$2n\Big(g(\bar{Y})-g(\lambda_0)\Big)\longrightarrow_d \lambda_0 g''(\lambda_0)\chi^2_1$$ Since $g(\lambda_0)=0,g''(\lambda_0)=1/\lambda_0$ we get $2ng(\bar{Y})\longrightarrow _d\chi^2_1$ which is what you seek.

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