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


*

*Use the CLT to justify
$$
\Phi(-1.645).
$$
2.
Second, simulate $N$ length $100$ data sets from your joint mass function to calculate
$$
\frac{1}{N}\sum_{i=1}^N \phi(X^i_1, \cdots, X^i_{100})
$$


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


