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Math Help - Few quick questions...probability

  1. #1
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    Few quick questions...probability

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

    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(X <= 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).

    ----

    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.

    --

    Then my last questions reads:

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

    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(X <= 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}.


    ----

    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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