# 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.

• 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 Jun 1, 2019 at 18:30

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: $$Pr\left(\sum_j X_j < 10 -1.645\sqrt{10} \right).$$

• (+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%. Jun 1, 2019 at 18:31
• @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. Jun 1, 2019 at 18:35

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")