Simulation

**Sampras** · January 17th 2010, 01:28 PM

Suppose you have some random vector $\bold{X} = (X_1, \dots, X_n)$ with a density function $f(x_1, \dots, x_n)$ . Suppose you want to compute the expected value of some function of the random vector (e.g. $E[g(\bold{X})]$ ). We know that

$E[g(\bold{X})] = \int \int \cdots \int g(x_1, \dots, x_n) f(x_1, \dots, x_n) dx_1 \cdots dx_n$

What is a good way of developing approximation methods for computing $E[g(\bold{X})]$ ? If $g(\bold{X}) = \bold{X}$ then we are just computing the expected value of $\bold{X}$ . So then we can reduce this question (in this case) to what are some good ways of approximating the expected value of a random vector?

**CaptainBlack** · January 17th 2010, 02:04 PM

Originally Posted by Sampras

Suppose you have some random vector $\bold{X} = (X_1, \dots, X_n)$ with a density function $f(x_1, \dots, x_n)$ . Suppose you want to compute the expected value of some function of the random vector (e.g. $E[g(\bold{X})]$ ). We know that

$E[g(\bold{X})] = \int \int \cdots \int g(x_1, \dots, x_n) f(x_1, \dots, x_n) dx_1 \cdots dx_n$

What is a good way of developing approximation methods for computing $E[g(\bold{X})]$ ? If $g(\bold{X}) = \bold{X}$ then we are just computing the expected value of $\bold{X}$ . So then we can reduce this question (in this case) to what are some good ways of approximating the expected value of a random vector?

If you can sample from the distribution with density $f(x_1, \dots, x_n)$ then the mean of the function values for the sample is an unbiased estimator of the expectation of $g(\bold{X})$ .

CB

**Sampras** · January 17th 2010, 02:24 PM

Originally Posted by CaptainBlack

If you can sample from the distribution with density $f(x_1, \dots, x_n)$ then the mean of the function values for the sample is an unbiased estimator of the expectation of $g(\bold{X})$ .

CB

In other words, $\lim\limits_{n \to \infty} \frac{g(\bold{X}^{(1)}) + \cdots + g(\bold{X}^{(n)})}{n} = E[g(\bold{X})]$ ?

**CaptainBlack** · January 17th 2010, 07:47 PM

Originally Posted by Sampras

In other words, $\lim\limits_{n \to \infty} \frac{g(\bold{X}^{(1)}) + \cdots + g(\bold{X}^{(n)})}{n} = E[g(\bold{X})]$ ?

Yes

CB

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