Disclaimer: This is a question I couldn't solve. I feel it is ethical to state this here so you can evaluate how much of your solution will share through either statements or hints.

Let $X_{1},…,X_n $ be an $i.i.d.$ $Poiss(\lambda)$ distributed random variable for some unknown $\lambda . 0$. Now consider the following hypothesis test:

$H_0:λ=λ_0$ v.s. $H_1:λ≠λ_0$ where $λ_0>0$.

Question: What is the asymptotic p-value?

Here's what I know:

Let $\overline{X}_n$ be the average of the random variable $X$. Thanks to the Central Limit Theorem, I can write:

$\dfrac{|\overline{X}_n - \lambda_0|}{\sigma/\sqrt{n}} \xrightarrow{(d)} N(0,1)$ as $n$ goes to infinity.

Given it's a two-sided test, according to [1] I can write

$P_{\lambda_0}\left(\dfrac{|\overline{X}_n - \lambda_0|}{\sigma/\sqrt{n}} > z_{\alpha/2}\right)$ $\rightarrow \alpha$

$P_{\lambda_0}\left(|W| > z_{\alpha/2}\right)$ $\rightarrow \alpha$

where $W$ is the Wald Test and $z_{\alpha/2} = \Phi^{-1}(1 - \alpha/2)$

Also from [1], I can write

$p-value = P_{\lambda_0}(|W| > |z_{\alpha/2}|) = 2.\Phi(-|z_{\alpha/2}|)$

This is where I'm stuck. I know all of these statements are true, however, I cannot work with the knowledge I have to obtain the answer I need. I'm still re-reading about hypothesis testing and p-values to have a better grip on problems like this.


[1] Wasserman, Larry. All of statistics: a concise course in statistical inference. Springer Science & Business Media, 2004, second edition.

  • 1
    $\begingroup$ The p-value is the biggest $\alpha$ such that the hypothesis test doesn‘t reject, or the smallest $\alpha$ such that the test rejects. $\endgroup$
    – AlexR
    Jun 17, 2020 at 21:02

1 Answer 1


Significance level. Let $n = 100$ (sufficiently large that an asymptotic test has a chance) and $\lambda_0 = 10.$ Then it seems safe to use $Z = \frac{\bar X - \mu_0}{\sqrt{\mu_0/n}}$ as a test statistic, rejecting $H_0: \mu = 10$ against $H_a: \mu \ne 10,$ at the 5% level if $|Z| \ge 1.96.$ [Notice that if $X \sim \mathsf{Pois}(\lambda),$ then $E(X) = Var(X) = \lambda.]$

n = 100;  lam.0 = 10;  se = sqrt(lam.0/n)
a = replicate(10^6, mean(rpois(n,lam.0)))
z = (a-lam.0)/se
[1] 0.051348

This test has nearly the 5% level, as anticipated.

P-value. The P-value is the probability under $H_0$ that the test statistic would take a value farther from $0$ than observed. Using the R function pnorm (for which the default, without specifying parameters other than 0 or 1, is $\Phi)$ we get the following:

pv = pnorm(-abs(z)) + 1 - pnorm(abs(z))
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.0000  0.2549  0.5066  0.5000  0.7518  1.0000

In general, under $H_0$ for a continuous test statistic, the P-value has a standard uniform distribution. Our test statistic, based on Poisson data, is actually discrete with small increments. Among the $100\,000$ simulated tests, there were only 286 uniquely different values of the test statistic $Z.$

[1] 286

Thus, the histogram of our P-value is only roughly standard uniform. The important thing for testing at the 5% level is that the probability of the left-most bar is nearly 5%.

hist(pv, prob=T, col="skyblue2", main="Histogram of P-values")
  curve(dunif(x), 0, 1, add=T, col="red")

enter image description here

Power against specific alternatives. The power of this test against the particular alternative $\lambda = 12$ is above 99%.

n = 100;  lam.0 = 10;  se = sqrt(lam.0/n)
a = replicate(10^6, mean(rpois(n,12)))   # alternative 12
z = (a-lam.0)/se
[1] 0.99998

And its power against the particular alternative $\lambda = 9$ is about 90%.

n = 100;  lam.0 = 10;  se = sqrt(lam.0/n)
a = replicate(10^6, mean(rpois(n,9)))     # alternative 9
z = (a-lam.0)/se
[1] 0.899814

So for the parameters used in the simulation, the test seems to work as intended.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.