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Math Help - multivariate conditional probability help

  1. #1
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    multivariate conditional probability help

     f(x,y) = \left\{ \begin{array}{rcl}<br />
6(1-y) & \mbox{for} & 0 \leq x \leq y \leq 1 \\ <br />
0 & \mbox{for} & \mbox{other}<br />
\end{array}\right.

    so far I have:

    for the marginal density functions:

    \begin{array}{l} f_1(x) = \int_{x}^{1} 6-6y \ dy \ 0 \leq x \leq 1 \\ \\<br />
= 6y - 3y^2 \bigg{|}^{1}_{x} \\ \\<br />
= 3 - (6x- 3x^2) \\ \\ <br />
= 3(x-1)^2 \\ <br />
\end{array}

    f_1(x) = 3(x-1)^2 \ \ 0 \leq x \leq 1 \\

    \begin{array}{l} f_2(y) = \int_{1}^{0} 6-6y \ dx \ x \leq y \leq 1 \\ \\<br />
= 6x - 6xy \bigg{|}^{1}_{0} \\ \\<br />
= 6(1-y) \\ \\<br />
f_2(y) =  6(1-y) \ \ x \leq y \leq 1 \\<br />
\end{array}

    so my marginals are:

    f_1(x) = \left\{ \begin{array}{rcl}<br />
3(x-1)^2 & \mbox{for} & 0 \leq x \leq 1 \\ <br />
0 & \mbox{for} & \mbox{other}<br />
\end{array}\right.

    f_2(y) = \left\{ \begin{array}{rcl}<br />
6(1-y) & \mbox{for} & x \leq y \leq 1 \\ <br />
0 & \mbox{for} & \mbox{other}<br />
\end{array}\right.

    Find P \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right)

    for  X \leq \frac{3}{4} :

     \int^{3/4}_{0} 3(x-1)^2 \ dx = (x-1)^3 \bigg{|}^{3/4}_{0} = - \frac{1}{64} which is clearly wrong.

    and  Y \leq \frac{1}{2} :

     \int^{1/2}_{x} 6(1-y) \ dy = 6y-3y^2 \bigg{|}^{1/2}_{x} = \frac{9}{4} - 6x-3x^2 which doesn't make any sense.

    once you get the appropriate values do you just divide Y by X?

    P \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right) = \frac{32}{63} according to the solution in the back of the book.

    I have no clue what I'm doing wrong and any help clarifying this would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by lllll View Post
     f(x,y) = \left\{ \begin{array}{rcl}<br />
6(1-y) & \mbox{for} & 0 \leq x \leq y \leq 1 \\ <br />
0 & \mbox{for} & \mbox{other}<br />
\end{array}\right.

    so far I have:

    for the marginal density functions:

    \begin{array}{l} f_1(x) = \int_{x}^{1} 6-6y \ dy \ 0 \leq x \leq 1 \\ \\<br />
= 6y - 3y^2 \bigg{|}^{1}_{x} \\ \\<br />
= 3 - (6x- 3x^2) \\ \\ <br />
= 3(x-1)^2 \\ <br />
\end{array}

    f_1(x) = 3(x-1)^2 \ \ 0 \leq x \leq 1 \\

    \begin{array}{l} f_2(y) = \int_{1}^{0} 6-6y \ dx \ x \leq y \leq 1 \\ \\<br />
= 6x - 6xy \bigg{|}^{1}_{0} \\ \\<br />
= 6(1-y) \\ \\<br />
f_2(y) = 6(1-y) \ \ x \leq y \leq 1 \\<br />
\end{array}

    so my marginals are:

    f_1(x) = \left\{ \begin{array}{rcl}<br />
3(x-1)^2 & \mbox{for} & 0 \leq x \leq 1 \\ <br />
0 & \mbox{for} & \mbox{other}<br />
\end{array}\right.

    f_2(y) = \left\{ \begin{array}{rcl}<br />
6(1-y) & \mbox{for} & x \leq y \leq 1 \\ <br />
0 & \mbox{for} & \mbox{other}<br />
\end{array}\right.

    Find P \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right)

    for  X \leq \frac{3}{4} :

     \int^{3/4}_{0} 3(x-1)^2 \ dx = (x-1)^3 \bigg{|}^{3/4}_{0} = - \frac{1}{64} which is clearly wrong.

    and  Y \leq \frac{1}{2} :

     \int^{1/2}_{x} 6(1-y) \ dy = 6y-3y^2 \bigg{|}^{1/2}_{x} = \frac{9}{4} - 6x-3x^2 which doesn't make any sense.

    once you get the appropriate values do you just divide Y by X?

    P \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right) = \frac{32}{63} according to the solution in the back of the book.

    I have no clue what I'm doing wrong and any help clarifying this would be greatly appreciated.
    \Pr \left( Y \leq \frac{1}{2} \bigg{|} X \leq \frac{3}{4} \right) = \frac{\Pr\left( Y \leq \frac{1}{2}, X \leq \frac{3}{4} \right)}{\Pr\left( X \leq \frac{3}{4} \right)} = \frac{\int_{y=0}^{y=1/2} \int_{x=0}^{x=y} 6(1 - y) \, dx \, dy}{\int_{0}^{3/4} 3 (x - 1)^2 dx}

    The integral limits for the double integral are best got by first sketching the region 0 \leq x \leq y \leq 1 and then shading the part corresponding to 0 \leq y \leq \frac{1}{2} AND 0 \leq x \leq \frac{3}{4}.


    By the way:

    f_2(y) = \int_{0}^{y} 6-6y \ dx, \, \, x \leq y \leq 1 .......
    Last edited by mr fantastic; April 10th 2008 at 10:53 PM.
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  3. #3
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    By the way:

     \int^{3/4}_{0} 3(x-1)^2 \ dx = (x-1)^3 \bigg{|}^{3/4}_{0} = \left( \frac{3}{4} - 1 \right)^3 - (-1)^3 = \left( -1/4 \right)^3 + 1 = \frac{63}{64}, not - \frac{1}{64}.
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  4. #4
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    one more question regarding this problem:

    Find the conditional density function of X given Y=y? I really have no clue on how to do this, so if any body can help that would be greatly appreciated.
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  5. #5
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    Quote Originally Posted by lllll View Post
    one more question regarding this problem:

    Find the conditional density function of X given Y=y? I really have no clue on how to do this, so if any body can help that would be greatly appreciated.
    By definition: f(x | y) = \frac{f(x,y)}{f_2(y)}.

    And recall my earlier post: f_2(y) = \int_{0}^{y} 6-6y \ dx, \, \, x \leq y \leq 1 .... So f_2(y) = 6y(1 - y), \, \, x \leq y \leq 1 .....
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