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Math Help - Expected value of a random variable from a bivariate distribution

  1. #1
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    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\{<br />
\begin{array}{c l}<br />
2 & 0 \le x_1 \le x_2 < 1 \\<br />
0 & otherwise <br />
\end{array}<br />
\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?
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  2. #2
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    Hello,
    Quote Originally Posted by notgoodatmath View Post
    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\{<br />
\begin{array}{c l}<br />
2 & 0 \le x_1 \le x_2 < 1 \\<br />
0 & otherwise <br />
\end{array}<br />
\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 ?
    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 ?
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