Problem

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

$$
H_0 : theta = 0.1 vs. H_1 : theta < 0.1
$$

Suppose we take sample of $n=100$ from the model, $X_1, cdots, X_100$.

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

$$
fracbarX - 0.1sqrt0.1/100 < -1.645
$$

In the grammar of test function,

$$
phi(X_1, cdots, X_100) =
begincases
1 & mathrmif fracbarX - 0.1sqrt0.1/100 < -1.645 \[7pt]
0 & mathrmif fracbarX - 0.1sqrt0.1/100 ge -1.645 \[7pt]
endcases
$$

Try

Power function is defined as

$$
gamma(theta) := mathrmE left[ phi(X_1, cdots, X_100) right]
$$

for a fixed $theta$.

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

$$
mathrmE left[ phi(X_1, cdots, X_100) right] = Prleft(fracbarX - 0.1sqrt0.1/100 < -1.645 right)
$$

where $100barX sim Poisson(100theta)$.

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 will mention three ways to approximate the following
$$
mathrmE left[ phi(X_1, cdots, X_100) right] = Prleft(fracbarX - 0.1sqrt0.1/100 < -1.645 right).
$$

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

2.

Second, simulate $N$ length $100$ data sets from your joint mass function to calculate
$$
frac1Nsum_i=1^N phi(X^i_1, cdots, X^i_100)
$$

3. Simulate $N$ length $100$ data sets from some proposal pmf $q(x_1, ldots, x_100)$ and calculate
   $$
   frac1Nsum_i=1^Nphi(X^i_1, cdots, X^i_100) fracp(x^i_1, ldots, x^i_100)q(X^i_1, cdots, X^i_100)
   $$
   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:
$$
Prleft(sum_j X_j < 10 -1.645sqrt10 right).
$$

Here is a power curve for an exact test of $H_0: lambda = 0.1$ vs.
$H_a: lambda le 0.1,$ based on $n = 10$ observations from
$mathsfPois(lambda).$

Under $H_0,$ the total $T$ of the $n = 100$ observations $X_i sim mathsfPois(0.1)$ has
$T sim mathsfPois(10).$ 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 $bar X = T/n le 0.4.$ (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")