# Thread: Expected value of a random variable from a bivariate distribution

1. ## Expected value of a random variable from a bivariate distribution

Hi,

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

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

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

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

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

$\displaystyle X_1$ and $\displaystyle X_2$ 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 $\displaystyle X_1$ and $\displaystyle X_2$ are

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

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

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

2. Hello,
Originally Posted by notgoodatmath
Hi,

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

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

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

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

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

$\displaystyle X_1$ and $\displaystyle X_2$ 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 $\displaystyle X_1$ and $\displaystyle X_2$ are

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

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

Can anyone explain the working to get the expected values in more detail?
How can $\displaystyle X_1$ and $\displaystyle X_2$ be independent ?
If they were, the product of the marginal density should give the joint pdf !

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

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

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

Is it clearer now ?