$\newcommand{\var}{\operatorname{var}}$The problem is here:
Consider a nonnegative random variable X whose PDF is close to being exponential, of the form $$f_X(x) = g(x)e^{−x},$$ where $g(x)$ is a nonnegative function that satisfies $1/2 \leq g(x) \le 2$ for all $x,$ and $\int_{-\infty}^\infty g(x)e^{−x} \, dx = 1$. Let $\alpha$ be a large constant. We wish to estimate $p = P(X \geq \alpha)$ using Monte Carlo simulation. We assume that we are able to generate random variables drawn from the distribution of $X$, as well as from an exponential distribution.
The straightforward simulation method is to generate $n$ random samples, drawn from the distribution of $X,$ let $N$ be the number of samples that satisfy $X_i \geq a$, and form the estimate $\hat{P} = \frac{N}{n}$ Clearly, $E[\hat{P}] = p.$
a) Show that for large enough a, we have $\var(\hat{P})\geq \frac{e^{−a}}{(3n)}$
here is my attempt:
$N$ is random variable satisfy binomial distribution with parameter p, thus we have:$$\var(\hat{P})= \frac{1}{n^2}\var(N) = \frac{p(1-p)}{n}$$
the since $$ \frac{1}{2}\int_\alpha^\infty e^{−x} \, dx\leq \int_\alpha^\infty g(x)e^{−x} \, dx\leq 2\int_\alpha^\infty g(x)e^{−x} \, dx$$
thus:
$$ \frac{e^{-\alpha}}{2}\leq p \leq 2e^{-\alpha} $$
then I stuck here , how can I get $\frac{e^{−a}}{3n}$ as my lower bound?
then the question continuous:
Part (a) shows that the standard deviation of the estimation error $\hat{P}-p$ is of order $O(e^{−a/2})$, which is larger than the quantity p to be estimated by a $O(e^{a/2})$ factor. This is an instance of a general phenomenon: probabilities of rare events are difficult to estimate by simulation.
Consider now a random variable $Y$ whose PDF is exponential, with parameter $\lambda = 1/a$. with $$f_Y(x) = \frac{e^{-x/\alpha }}{\alpha}$$
we generate n random sample $Y_i$,drawn from the distribution of $Y,$ and estimate $p$ by $$Q = \frac{1}{n} \sum_{i=1}^n I_{Y_i \geq a} \frac{f_X(Y_i)}{f_Y(Y_i)} $$
b) show that $E[Q] = p$
this one is easy: $$E[Q] = \frac{1}{n} \sum_{i=1}^n \int_\alpha^\infty f_X(x) \, dx = p$$
c) Show that the standard deviation $\sigma_Q$ of Q is “comparable” to $p,$ in the sense that $\sigma_Q/p$ does not grow exponentially with $a.$
here is my attempt:
I use the fact that $$\var(X) = \var(X)\operatorname E[N]+\operatorname E[X]^2 \var[N]$$
thus: $$\var(Q) = \frac{1}{n} \var \left( \frac{f_X(Y_i)}{f_Y(Y_i)} \right) = \frac{1}{n} \left( \int_\alpha^\infty \frac{f_X(x)^2}{f_Y(x)} \, dx -p^2 \right) $$
then I stuck here again,it looks like that I can't get rid of $\alpha after integration what should I do next ..