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Math Help - Joint PDF and Mode questions

  1. #1
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    Joint PDF and Mode questions

    I have two questions:
    1. Let the joint PDF of X and Y be f(x,y)=2x for 0<x<1 and 0<y<1, 0 otherwise.
    I want to find P(X+Y<1 | X<1/2).
    First I find the individual distributions f_X(x)=2x and f_Y(y)=1. So P(X<1/2) is 1/4. Then I am stuck. I am not comfortable with the formula for conditional probability with multiple distributions. I am also interested if there is an intuitive answer because y does not appear in the joint PDF.

    2. the PDF of X is f(x)=kx if 0<x<(2/k)^(1/2). If the mode of X is (2^(1/2))/4 then the median is what?
    So I want to figure out k by using the mode. For continuous distributions I feel like I should take the derivate of the pdf and set it equal to 0 to find the mode, but this does not work, as f'(x)=k, hence is never 0. Instead my idea is that f(x) is maximized at (2/k)^(1/2) as k clearly must be positive. But I am not certain.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by robeuler View Post
    I have two questions:
    1. Let the joint PDF of X and Y be f(x,y)=2x for 0<x<1 and 0<y<1, 0 otherwise.
    I want to find P(X+Y<1 | X<1/2).
    P(a|b)=P(a \wedge b)/P(b)

    P(a \wedge b) is the integral of the joint distribution over the region defined by (a \wedge b) and P(b) is the integral of the joint distribution over the region defined by $$ b.

    In this case $$ a is X+Y<1 and $$ b is X<1/2, so:

    \displaystyle P((X+Y<1) \wedge (X<1/2))=\int_{x=0}^{1/2} \int_{y=0}^{1-x} f(x,y) dx dy

    and:

    \displaystyle P(X<1/2)=\int_{x=0}^{1/2} \int_{y=0}^{1} f(x,y) dx dy

    CB
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by robeuler View Post
    2. the PDF of X is f(x)=kx if 0<x<(2/k)^(1/2). If the mode of X is (2^(1/2))/4 then the median is what?
    So I want to figure out k by using the mode. For continuous distributions I feel like I should take the derivate of the pdf and set it equal to 0 to find the mode, but this does not work, as f'(x)=k, hence is never 0. Instead my idea is that f(x) is maximized at (2/k)^(1/2) as k clearly must be positive. But I am not certain.
    Sketch density, where is the mode? Now you know $$ k. Now compute the median from the definition that the median is $$ m such that:

    \displaystyle \int_{x=0}^m f(x)\;dx=0.5

    CB
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