4
$\begingroup$

Problem

Suppose we are given $\text{Poisson}(\theta)$ model, and the null hypothesis is as follows:

$$ H_0 : \theta = 0.1 \ \ \text{vs}. \ \ H_1 : \theta < 0.1 $$

Suppose we take sample of $n=100$ from the model, $X_1, \ldots, X_{100}$.

Plot the power function $\gamma(\theta)$, with standard error, of the following test, using importance sampling.

Test : reject $H_0$ at $\alpha=0.05$, when

$$ \frac{\bar{X} - 0.1}{\sqrt{0.1/100}} < -1.645 $$

In the grammar of test function,

$$ \phi(X_1, \ldots, X_{100}) = \begin{cases} 1 & \mathrm{if} \frac{\bar{X} - 0.1}{\sqrt{0.1/100}} < -1.645 \\[7pt] 0 & \mathrm{if} \frac{\bar{X} - 0.1}{\sqrt{0.1/100}} \ge -1.645 \\[7pt] \end{cases} $$


Try

Power function is defined as

$$ \gamma(\theta) := \mathrm{E} \left[ \phi(X_1, \ldots, X_{100}) \right] $$

for a fixed $\theta$.

So my strategy is to first fix $\theta$, and evaluate the quantity

$$ \mathrm{E} \left[ \phi(X_1, \ldots, X_{100}) \right] = \Pr\left(\frac{\bar{X} - 0.1}{\sqrt{0.1/100}} < -1.645 \right) $$

where $100\bar{X} \sim \text{Poisson}(100\theta)$.


Question

I do not see any place to apply the importance sampling technique. How the importance sampling (which can give us standard error) can contribute my question?

Any help will be appreciated.

$\endgroup$
  • 1
    $\begingroup$ I added the self-study tag. Feel free to change it back if this isn't an exercise that was given to you by an instructor $\endgroup$ – Taylor Jun 1 '19 at 18:30
4
$\begingroup$

I will mention three ways to approximate the following $$ \mathrm{E} \left[ \phi(X_1, \cdots, X_{100}) \right] = Pr\left(\frac{\bar{X} - 0.1}{\sqrt{0.1/100}} < -1.645 \right). $$

  1. Use the CLT to justify $$ \Phi(-1.645). $$

2.

Second, simulate $N$ length $100$ data sets from your joint mass function to calculate $$ \frac{1}{N}\sum_{i=1}^N \phi(X^i_1, \cdots, X^i_{100}) $$

  1. Simulate $N$ length $100$ data sets from some proposal pmf $q(x_1, \ldots, x_{100})$ and calculate $$ \frac{1}{N}\sum_{i=1}^N\phi(X^i_1, \cdots, X^i_{100}) \frac{p(x^i_1, \ldots, x^i_{100})}{q(X^i_1, \cdots, X^i_{100})} $$ where $p$ is your true product-Poisson pmf.

Calculating approximate standard errors is probably easier for the last two, but you may also get Berry-Esseen bounds for the first one.

To check your answers, it might also be worth evaluating the true probability. This is possible using what you mentioned: that the sum of iid Poissons is also Poisson-distributed. The cdf is available in most statistical software packages: $$ Pr\left(\sum_j X_j < 10 -1.645\sqrt{10} \right). $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ (+1) for mentioning exact computations. $10 - 1.645\sqrt{10} \approx 4.8,$ but the next lower integer is $4$ for an actual significance level of about 3%. $\endgroup$ – BruceET Jun 1 '19 at 18:31
  • $\begingroup$ @BruceET thanks, and right, asymptotic $\alpha$ isn't the same as small sample $\alpha$. I'm just trying to help with power calculations using importance sampling, though, so I'm not fiddling with the rejection region. $\endgroup$ – Taylor Jun 1 '19 at 18:35
3
$\begingroup$

Here is a power curve for an exact test of $H_0: \lambda = 0.1$ vs. $H_a: \lambda \le 0.1,$ based on $n = 100$ observations from $\mathsf{Pois}(\lambda).$

Under $H_0,$ the total $T$ of the $n = 100$ observations $X_i \sim \mathsf{Pois}(0.1)$ has $T \sim \mathsf{Pois}(10).$ Because the Poisson distribution is discrete, a (nonrandomized) test at exactly level $\alpha = 0.05$ is not available.

The largest available level below 5% is $\alpha = 0.0293.$ So we will find the power of a test that rejects $H_0$ when $T \le 4$ or $\bar X = T/n \le 0.04.$ (See computations in R below.)

qpois(.05, 10)
[1] 5
ppois(5,10)
[1] 0.06708596
ppois(4,10)
[1] 0.02925269

The power of the test against alternative $\lambda_0 < 0.1$ is $P(T \le 4\,|\,\lambda_0).$

Here is a graph of the power function:

lam = seq(10, .01, by=-.01)/100
p.rej = ppois(4, 100*lam)
plot(lam, p.rej, type="l", ylim=0:1, main="Power Curve")
 abline(h=0:1, col="green2");  abline(v=c(0,.1), col="green2")

enter image description here

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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