Thread: Uniform
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Old October 10th, 2008, 05:26 PM
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Originally Posted by brd_7 View Post
Let X have the uniform distribution (0,2) and let the conditional distribution of Y given X=x be uniform on (0,x^2).

a) Find the condition expectation and variance of Y given X=x. USe these to find the marginal expectation and variance of Y...

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The pdf of X is f_X(x) = \frac{1}{2} for 0 \leq x \leq 2 and zero elsewhere.

The conditional distribution of Y given X = x is f_Y(y | x) = \frac{1}{x^2} for 0 \leq y \leq x^2 and zero elsewhere.

Now apply the standard definitions and do the necessary calculations:

E(Y | X = x) = \int_0^{x^2} y \, f_Y(y | x) \, dy.

E(Y^2 | X = x) = \int_0^{x^2} y^2 \, f_Y(y | x) \, dy.

Var(Y | X = x) = E(Y^2 | X = x) - [E(Y | X = x)]^2.


E(Y) = E[E(Y | X = x)] = \int_{-\infty}^{+\infty} E(Y | X = x) \, f_X(x) \, dx = \int_0^2 E(Y | X = x) \, \frac{1}{2} \, dx.

To get Var(Y), read this: Law of total variance - Wikipedia, the free encyclopedia and do the necessary computation.
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