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Thread: 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).
    $\displaystyle P(a|b)=P(a \wedge b)/P(b)$

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

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

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

    and:

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

    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 $\displaystyle $$ k$. Now compute the median from the definition that the median is $\displaystyle $$ m$ such that:

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

    CB
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