Question

Let X and Y be independent exponential random variables with mean 1.

(a) Explain how we could use simulation to estimate E[eXY].

(b) Show how to improve the estimation approach in part (a) by using a control variate.

Short answer

(a) Estimator $M=\frac{1}{n}{\sum }_{i=1}^n{e}^{X_i{Y}_i}$
(b) Redefine estimator $\hat{M}=\frac{1}{n}{\sum }_{i=1}^n{e}^{X_i{Y}_i}+c\left(X_i{Y}_i-1\right)$and it is defined below.

Step-by-step solution

Part (a) Step 1: Given Information

We need to explain the simulation to estimate$E\left[e^{XY}\right]$.

Part (a) Step 2: Simplify

Considering to define $X_1,\dots ,X_n~X$and $Y_1,\dots ,Y_n~Y$.

Now, consider the statistic

$M=\frac{1}{n}\underset{i=1}{\overset{n}{}}{e}^{X_i{Y}_i}$

It is simple to show that $E\left(M\right)=E\left(e^{XY}\right)$so we have provided a good method of estimation of the mean.

Part (b) Step 1: Given Information

We need to give the solution to improve the estimation approach in part (a) by using a control variate.

Part (b) Step 2: Simplify

Consider

$\hat{M}=\frac{1}{n}\underset{i=1}{\overset{n}{}}{e}^{X_i{Y}_i}+c\left(X_i{Y}_i-1\right)$

Serve that it also holds

$E\left(\hat{M}\right)=\frac{1}{n}\underset{i=1}{\overset{n}{}}E\left(e^{X_i{Y}_i}+c\left(X_i{Y}_i-1\right)\right)=E\left(e^{XY}\right)$

as we have$E\left(c\left(X_i{Y}_i-1\right)\right)=0$. Let's minimize the variance. We have that

$\text{Var}\left(\hat{M}\right)=\frac{1}{n^2}\underset{i=1}{\overset{n}{}}\text{Var}\left(e^{X_i{Y}_i}+c\left(X_i{Y}_i-1\right)\right)$

Keeping that

$\text{Var}\left(e^{X_i{Y}_i}+c\left(X_i{Y}_i-1\right)\right)=\text{Var}\left(e^{X_i{Y}_i}\right)+c^2\text{Var}\left(X_i{Y}_i-1\right)+c\text{Cov}\left(e^{X_i{Y}_i},X_i{Y}_i-1\right)$

So, we have seen that the Variance is minimized when

$c=-\frac{\text{Cov}\left(e^{X_i{Y}_i},X_i{Y}_i-1\right)}{\text{Var}\left(X_i{Y}_i-1\right)}$