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Math Help - Probability Bivariate Random Variable Problem

  1. #1
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    Probability Bivariate Random Variable Problem

    Let U=min(X,Y) and V=max(X,Y), where X and Y are independent R(0,1) random variables.

    1. Compute the cdf's (cumulative dist. function) F(u) and F(v) and hence derive the pdf's (probability density function) f(u) and f(v) of the random variables U and V respectively.

    2. Compute the joint cdf F(u,v) ], 0<=u<=v<=1, of the bivariate random variable (U,V) and use it to derive the pdf f(u,v).

    3. Are the random variables U and V independent? Why?

    Please help me! Thanks!
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  2. #2
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    Quote Originally Posted by maibs89 View Post
    Let U=min(X,Y) and V=max(X,Y), where X and Y are independent R(0,1) random variables.

    1. Compute the cdf's (cumulative dist. function) F(u) and F(v) and hence derive the pdf's (probability density function) f(u) and f(v) of the random variables U and V respectively.

    2. Compute the joint cdf F(u,v) ], 0<=u<=v<=1, of the bivariate random variable (U,V) and use it to derive the pdf f(u,v).

    3. Are the random variables U and V independent? Why?

    Please help me! Thanks!
    I'll do a little bit to give you some ideas:

    1. Consider F(v) = {\color{red}\Pr(0 < V \leq v) = 1 - }\Pr(V > v).

    If V > v then obviously X > v and Y > v.

    So F(v) = {\color{red}1 - }\Pr(X > v) \cdot \Pr(Y > v) = {\color{red}1 - }\left( \int_{v}^{1} 1 \, dx \right) \cdot \left( \int_{v}^{1} 1 \, dy \right) = ...... if 0 \leq v \leq 1 and zero otherwise.

    Now recall that the pdf of V is given by f(v) = \frac{dF}{dv} ......


    3. Does f(u, v) = f(u) f(v) ?

    Try the remaining bits again - post if you're still stuck after making a good attempt.
    Last edited by mr fantastic; May 3rd 2008 at 07:57 PM. Reason: Careless - The correction is in red
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  3. #3
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    I got f(u)=min(xy) and f(v)=max(xy). Then later when I tried to compute the pdfs by differentiating I got one for both. I think that is not the answer.

    I am even more confused on how to find the joint pdf. I know its f(u).f(v) but I don't know how to find part 1.

    Help!!
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    Quote Originally Posted by maibs89 View Post
    I got f(u)=min(xy) and f(v)=max(xy). Then later when I tried to compute the pdfs by differentiating I got one for both. I think that is not the answer.

    I am even more confused on how to find the joint pdf. I know its f(u).f(v) but I don't know how to find part 1.

    Help!!
    I have made a correction to my first reply. Please look at it. However, my original error is not the cause of your current difficulty .... I'm at a loss to understand how you could have got f(v) = 1. Please show your work (slightly revised to reflect my correction, of course).
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  5. #5
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    I think I know how to get the cdfs, and hence pdfs:

    F(v)= 1 - (1-v)^2 , 0<V<1
    so I just differentiate to get f(v).

    F(u)= u^2 , is it? Then the pdf f(u)=2u.

    I need to know how to find the joint cdfs. Then I will know how to find independence.

    Help!!!
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  6. #6
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    Quote Originally Posted by maibs89 View Post
    I think I know how to get the cdfs, and hence pdfs:

    F(v)= 1 - (1-v)^2 , 0<V<1
    so I just differentiate to get f(v).

    F(u)= u^2 , is it? Then the pdf f(u)=2u.

    Mr F says: Correct on both counts.

    I need to know how to find the joint cdfs. Then I will know how to find independence.

    Help!!!
    I'll give you some general direction:

    F(u, v) = \Pr(U \leq u ~ \text{and} ~ V \leq v | 0 \leq u \leq v \leq 1) = \Pr(U \leq u) \cdot \left( \, 1 - \Pr(V > v | v \geq u) \, \right).

    \Pr(V > v | v \geq u) = \frac{\Pr( V > v ~ \text{and} ~ V > u)}{\Pr(V > u)} = \frac{\Pr( V > v)}{\Pr(V > u)}

    and you can get Pr(V > u) from the pdf for V found in (a).

    Then pdf = f(u, v) = \frac{\partial^2 F(u, v)}{\partial u \partial v}.
    Last edited by mr fantastic; May 4th 2008 at 07:33 PM.
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  7. #7
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    So far I understand.

    How do I get Pr(V>V) and Pr(V>u)? What do yo mean by using the pdf in part a? Do I just
    1-integrate for both? What do I integrate?

    Help!!!
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  8. #8
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    Quote Originally Posted by maibs89 View Post
    So far I understand.

    How do I get Pr(V>V) and Pr(V>u)? What do yo mean by using the pdf in part a? Do I just
    1-integrate for both? What do I integrate?

    Help!!!
    \Pr(V > u) = \int_{u}^{1} f(v) \, dv. f(v) is known from part (a).
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  9. #9
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    I got the joint cdf as u^2 * (1-(v-1)^2/(u-1)^2)

    Is this correct?

    I am having difficulty integrating it to get the pdf and it looks complicated.

    If it is the joint cdf, how would I integrate it? Thanks!
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  10. #10
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    Quote Originally Posted by maibs89 View Post
    I got the joint cdf as u^2 * (1-(v-1)^2/(u-1)^2)

    Is this correct?

    I am having difficulty integrating it to get the pdf and it looks complicated.

    If it is the joint cdf, how would I integrate it? Thanks!
    The joint cdf looks OK.

    You don't integrate it, you take partial derivatives of it. Recall: f(u, v) = \frac{\partial^2 F(u, v)}{\partial u \partial v}.
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  11. #11
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    Quote Originally Posted by mr fantastic View Post
    I'll do a little bit to give you some ideas:

    1. Consider F(v) = {\color{red}\Pr(0 < V \leq v) = 1 - }\Pr(V > v).

    If V > v then obviously X > v and Y > v.

    So F(v) = {\color{red}1 - }\Pr(X > v) \cdot \Pr(Y > v) = {\color{red}1 - }\left( \int_{v}^{1} 1 \, dx \right) \cdot \left( \int_{v}^{1} 1 \, dy \right) = ...... if 0 \leq v \leq 1 and zero otherwise.

    Now recall that the pdf of V is given by f(v) = \frac{dF}{dv} ......


    3. Does f(u, v) = f(u) f(v) ?

    Try the remaining bits again - post if you're still stuck after making a good attempt.
    I'm big enough to admit my mistakes. I made a mistake here. The pdf's are actually the other way around.

    I only realised my mistake when answering the same question (more or less) asked by a different member at this thread: http://www.mathhelpforum.com/math-he...-question.html

    After answering it, I thought it seemed familiar so I searched and found this thread. I was baffled that I'd given two mutually exclusive answers. Reading this thread again I saw the mistake in it immediately.

    Sorry for any inconvenience. I did have the right ideas, just the wrong way around.
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