Let $X_1,\ldots, X_n\sim \operatorname{Poisson}(\lambda)$. Let $\lambda_w>0$ be given, I am trying to find the size $\alpha$ Wald test for $H_0$: $\lambda=\lambda_w$ vs $H_1$: $\lambda\neq \lambda_w$.
I got stuck with how to get start to compute. I think I need to go with computing the power function and use it (I am still trying...). But I was told that Wald test size is computed in a special way (which is faster also). Can any body help me on finding this size $\alpha$ Wald test?
$\endgroup$2 Answers
$\begingroup$The MLE of $\lambda$ is $\hat{\lambda} = \frac{\sum(X_i)}{n} = \bar{X}$. By asymptotic normality of the MLE:
$$\operatorname{se}=\sqrt{\operatorname{Var}(\hat{\lambda})}\approx\sqrt{\frac{1}{I_n(\lambda)}}$$
The following hold:
$$\frac{\hat{\lambda}-\lambda_w}{\operatorname{se}(\lambda_w)}\rightarrow N(0,1)$$
The Wald statistic test is
$$W = \frac{\hat{\lambda}-\lambda_w}{\operatorname{se}(\lambda_w)} \sim N(0,1)$$
where $I_n(\lambda) =nI(\lambda)=n\frac{1}{\lambda}$
$$\operatorname{se}(\lambda_w)=\sqrt{\frac{\lambda_w}{n}}$$
So
$$W =\frac{\hat{\lambda}-\lambda_w}{\operatorname{se}(\lambda_w)} = \sqrt{n}\frac{\hat{\lambda}-\lambda_w}{\sqrt{{\lambda_w}}} $$
$$\text{p-value}=2{\Phi}^{-1}(-|W|)$$
Reject $H_0$ if $\alpha > \text{p-value}=2{\Phi}^{-1}(-|W|)$
$\endgroup$ 1 $\begingroup$For large enough $\lambda$ the random variable $X \sim \operatorname{Pois}(\lambda)$ is approximately normal, so that $Z = (X - \lambda)/\sqrt{\lambda} \sim \operatorname{Norm}(0,1).$
I believe the Wald test rejects the null hypothesis at level $\alpha = 0.05$ when $|Z| > 1.96.$ For other values of $\alpha,$ adjust the critical value appropriately by using normal tables.
Because Poison $X$ is a discrete random variable taking nonnegative integer values, it is not generally possible to find a critical value in terms of an integer that exactly matches a desired significance level $\alpha.$ This difficulty is essentially ignored when using the normal approximation.
The Wald test at level $\alpha = 5\%$ can be 'inverted' to give a 95% confidence interval for $\lambda.$ This CI has the form $X + 2 \pm 1.96\sqrt{X + 1}.$ (Adding 2 to get the center and 1 to get the margin of error is somewhat in the same spirit as adding 2 imaginary successes and 2 imaginary failures to make the Agresti or 'plus-4' style of CI for the binomial success probability.)
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