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mr fantastic · #6 October 11th, 2008, 08:51 PM

Quote:

Originally Posted by brd_7

Sorry, i did write down the answers, but realised they were wrong, made a careless mistake. Anyway.. I got the following..

x^2/2

x^4/3

x^4/12

1/3

With just Var(Y) to work out..

Many Thanks

So you need to calculate $E[Var(Y | X)] = E\left[\frac{X^4}{12}\right]$ and $Var[E(Y | X)] = Var\left[ \frac{X^2}{2}\right]$ .

$E\left[\frac{X^4}{12}\right] = \int_0^2 \left( \frac{x^4}{12}\right) \left(\frac{1}{2} \right) \, dx$ .

$Var\left[ \frac{X^2}{2}\right]$ :

You need the pdf of $U = \frac{X^2}{2}$ . This can be got by calculating G(u), the cdf of U. Then $g(u) = \frac{dG}{du}$ , where g(u) is the pdf of U.

Then $Var \left[ \frac{X^2}{2}\right] = Var (U) = E(U^2) - [E(U)]^2$ .

$G(u) = \Pr(U < u) = \Pr\left( \frac{X^2}{2} < u\right) = \Pr( -\sqrt{2u} < X < \sqrt{2u}) = \Pr(0 < X < \sqrt{2u})$ since $0 \leq x \leq 2$

$= \int_0^{\sqrt{2u}} \frac{1}{2} \, dx = \, ....$ for $0 \leq u \leq 2$ .

Therefore $g(u) = \frac{dG}{du} = \, .....$ for $0 \leq u \leq 2$ and zero elsewhere.

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brd_7
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