# Expected value of a random variable from a bivariate distribution

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• February 25th 2009, 09:30 AM
notgoodatmath
Expected value of a random variable from a bivariate distribution
Hi,

I have two random variables, $X_1$ and $X_2$ that have joint pdf:

$f_{(x_1,x_2)}(x_1,x_2) = \left\{
\begin{array}{c l}
2 & 0 \le x_1 \le x_2 < 1 \\
0 & otherwise
\end{array}
\right.$

I can find the marginal pdf of $X_1$ and $X_2$

$\int_{x_2 = x_1}^{x_2=1} 2dx_2 = 2(1 - x_1)$ for $0 \le x_1 \le 1$

$\int_{x_1 = 0}^{x_1=x_2} 2dx_1 = 2x_2$ for $0 \le x_2 \le 1$

$X_1$ and $X_2$ are independent.

I can understand up till here. This is where I'm confused. How do you work out the expected values of each random variable from the marginal pdf's of each?

I know: The expected values of $X_1$ and $X_2$ are

$E[X_1] = \frac{1}{3}$

$E[X_2] = \frac{2}{3}$

Can anyone explain the working to get the expected values in more detail?
• February 25th 2009, 10:53 AM
Moo
Hello,
Quote:

Originally Posted by notgoodatmath
Hi,

I have two random variables, $X_1$ and $X_2$ that have joint pdf:

$f_{(x_1,x_2)}(x_1,x_2) = \left\{
\begin{array}{c l}
2 & 0 \le x_1 \le x_2 < 1 \\
0 & otherwise
\end{array}
\right.$

I can find the marginal pdf of $X_1$ and $X_2$

$\int_{x_2 = x_1}^{x_2=1} 2dx_2 = 2(1 - x_1)$ for $0 \le x_1 \le 1$

$\int_{x_1 = 0}^{x_1=x_2} 2dx_1 = 2x_2$ for $0 \le x_2 \le 1$

$X_1$ and $X_2$ are independent.

I can understand up till here. This is where I'm confused. How do you work out the expected values of each random variable from the marginal pdf's of each?

I know: The expected values of $X_1$ and $X_2$ are

$E[X_1] = \frac{1}{3}$

$E[X_2] = \frac{2}{3}$

Can anyone explain the working to get the expected values in more detail?

How can $X_1$ and $X_2$ be independent ? (Surprised)
If they were, the product of the marginal density should give the joint pdf !

Anyway, here is a formula :
$\mathbb{E}(h(X))=\int f(x)h(x) ~dx$, where f is the pdf of the rv X.

So $\mathbb{E}(X)=\int x f(x) ~dx$

Here, it gives :
$\mathbb{E}(X_1)=\int_0^1 2x_1(1-x_1) ~dx_1$
$\mathbb{E}(X_2)=\int_0^1 2x_2^2 ~dx_2$
(by letting h(x)=x)

Is it clearer now ?