Alright, in this one, I'm given X~U(-1, 3) -1<x<3

Y = X

² Ok, so I know that the domain of y should be 0 < y< 9...right? Mr F says:Yes. Oh, and the pdf(x) = 1/4, -1 < x< 3 Mr F says: Yes. So I said, F(y) = P(Y <= y) = P(X² <= y) = P(X <= (y)^1/2 Mr F says: I think this is what you should have: $\displaystyle {\color{red}F(y) = \Pr (X^2 < y) = \Pr(-\sqrt{y} < X < \sqrt{y})=}$ $\displaystyle {\color{red} \left\{ \begin{array}{ll} \int_{\sqrt{y}}^{3} \frac{1}{4} \, dx + \int^{-\sqrt{y}}_{-1} \frac{1}{4} \, dx & 0 < y \leq 1 \\ & \\ \int_{\sqrt{y}}^{3} \frac{1}{4} \, dx & 1 < y < 9 \end{array} \right\} }$ Now do the integrations to get the cdf F(y) and then differentiate to get the pdf f(y).
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My question is...how do I know the domain for this f(y). Is it for the whole 0<y<9? Or just part? If just part of it, how do I get the other f(y). Mr F says: Note that there are three parts to the rule: 1. f(y) = 0 for values of y less than 0 or greater than 9. 2. f(y) = ...... when $\displaystyle {\color{red}0 \leq y \leq 1}$. 3. f(y) = ...... when $\displaystyle {\color{red}1 < y \leq 9}$. ---- Next question is how do I find the Cox(X,Y) over a continuous function? I know how to get marginal and mean, but not covariance. Mr F says: Cov(X, Y) = E(XY) - E(X) E(Y). You get E(X) and E(Y) in the usual way. To get E(XY) you need to find the joint distribution f(x,y) of X and Y (unless theres some easy way specific to the actual problem. See http://www.mathhelpforum.com/math-he...oach-time.html for example). Then $\displaystyle {\color{red}E(XY) = \int \int xy f(x,y) \, dx \, dy}$, with appropriate integral terminals of course. [snip]