# Math Help - Few quick questions...probability

1. ## Few quick questions...probability

Alright, in this one, I'm given X~U(-1, 3) -1<x<3
Y = X²

Ok, so I know that the domain of y should be 0 < y< 9...right?
Oh, and the pdf(x) = 1/4, -1 < x< 3

So I said, F(y) = P(Y <= y) = P(X² <= y) = P(X <= (y)^1/2

I integrated from (y)^1/2 to 3, dx/4 which gave me, F(y) = 3-(y)^1/2 / 4

So to get the pdf(y), I took the derivative, giving me 1/8*y^1/2

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).

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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.

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Let X and Y have the joint pmf defined by f(0,0) = f(1,2) = 0.2
f(0,1) = f(1,1) = 0.3

Depict the points and probabilities on a graph, giving the marginalpmf's in the margins.

Can someone show me what the graph would look like? I know of course where the points go, but I'm not sure what they mean about the marginal pmf in the margins.

2. Originally Posted by amor_vincit_omnia
Alright, in this one, I'm given X~U(-1, 3) -1<x<3
Y = X²

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:

${\color{red}F(y) = \Pr (X^2 < y) = \Pr(-\sqrt{y} < X < \sqrt{y})=}$ ${\color{red} \left\{ \begin{array}{ll} \int_{\sqrt{y}}^{3} \frac{1}{4} \, dx + \int^{-\sqrt{y}}_{-1} \frac{1}{4} \, dx & 0 < y \leq 1 \\ & \\ \int_{\sqrt{y}}^{3} \frac{1}{4} \, dx & 1 < y < 9 \end{array} \right\} }$

Now do the integrations to get the cdf F(y) and then differentiate to get the pdf f(y).

[snip]

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 ${\color{red}0 \leq y \leq 1}$.
3. f(y) = ...... when ${\color{red}1 < y \leq 9}$.

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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 ${\color{red}E(XY) = \int \int xy f(x,y) \, dx \, dy}$, with appropriate integral terminals of course.

[snip]
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