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Math Help - about jointly continuous random variable

  1. #1
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    about jointly continuous random variable

    Suppose X and Y are jointly continuous with joint density

    f(x,y)= (exp(-x/y)*exp(-y))/y for 0<x<oo and 0<y<oo

    f(x,y)= 0 otherwise

    Find P(X>1|Y=y).

    I first find the density functions of X and Y respectively.
    But I am in difficulty in integrating (exp(-x/y)*exp(-y))/y with respect to y. Can someoneone provides some hints?

    Also P(X>1|Y=y)=P(X>1 and Y=y)/P(Y=y)
    But for continuous random variable Y, P(Y=y)=0?
    Last edited by ldpsong; December 21st 2009 at 10:20 AM.
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  2. #2
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    Quote Originally Posted by ldpsong View Post
    Suppose X and Y are jointly continuous with joint density

    f(x,y)= (exp(-x/y)*exp(-y))/y for 0<x<oo and 0<y<oo

    f(x,y)= 0 otherwise

    Find P(X>1|Y=y).

    I first find the density functions of X and Y respectively.
    But I am in difficulty in integrating (exp(-x/y)*exp(-y))/y with respect to y. Can someoneone provides some hints?

    Also P(X>1|Y=y)=P(X>1 and Y=y)/P(Y=y)
    But for continuous random variable Y, P(Y=y)=0?
    Where is your single veriable distribution f(y)?
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  3. #3
    MHF Contributor matheagle's Avatar
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    To obtain P(X>1|Y=y) you need the conditional density f(x|y) and you integrate x from 1 to infinity.

    Both P(X>1 and Y=y) and P(Y=y) are zero.

    f(x|y)={f(x,y)\over f_Y(y)}
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  4. #4
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    Quote Originally Posted by novice View Post
    Where is your single veriable distribution f(y)?
    f(y)= S (exp(-x/y)*exp(-y))/y dx = exp(-y) <--------integration
    f(x)= S (exp(-x/y)*exp(-y))/y dy =?....

    I want to show that X and Y are independent,
    so P(X>1|Y=y) = P(X>1)
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  5. #5
    MHF Contributor matheagle's Avatar
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    SORRY, they are dependent by inspection.
    The joint denisty does not factor.
    This looks like a nasty integral.
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  6. #6
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    there is another question :
    Let X and Y be jointly continuous random variable with joint density function

    f(x,y) = kxy for 0<=x<=2 and 0<y<x
    f(x,y) =0 otherwise

    There are 4 questions. I have solved the first two.

    1, k=1/2
    2,density function of X: f(x)=x^3/4 for 0<=x<=2
    density function of Y: f(y)= y for 0<y<x

    I have not solved the 3rd and 4th questions yet

    3, Are X and Y independent?
    4,Find P(X>3/2|Y=1)



    It seems that the 2nd part is not correct.........
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  7. #7
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    Quote Originally Posted by matheagle View Post
    SORRY, they are dependent by inspection.
    The joint denisty does not factor.
    This looks like a nasty integral.
    integrate exp(-x/y)/y with respect to x from 1 to infinity is exp(-1/y)
    So P(X>1|Y=y)=exp(-1/y)

    Is it correct?
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  8. #8
    MHF Contributor matheagle's Avatar
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    any probability is between 0 and 1, inclusive.
    So you need to find the correct density and integrate.
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  9. #9
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by ldpsong View Post
    there is another question :
    Let X and Y be jointly continuous random variable with joint density function

    f(x,y) = kxy for 0<=x<=2 and 0<y<x
    f(x,y) =0 otherwise

    There are 4 questions. I have solved the first two.

    1, k=1/2
    2,density function of X: f(x)=x^3/4 for 0<=x<=2
    density function of Y: f(y)= y for 0<y<x

    I have not solved the 3rd and 4th questions yet

    3, Are X and Y independent?
    4,Find P(X>3/2|Y=1)



    It seems that the 2nd part is not correct.........

    These two are dependent rvs since y<x.

    To obtain k you need

    1=k\int_0^2\int_0^x xy dydx

    AND the mariginal density of Y CANNOT have x in it.
    It's a function of just y.
    The mariginal of Y has support from 0 to 2.
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  10. #10
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    Quote Originally Posted by matheagle View Post
    These two are dependent rvs since y<x.

    To obtain k you need

    1=k\int_0^2\int_0^x xy dydx

    AND the mariginal density of Y CANNOT have x in it.
    It's a function of just y.
    The mariginal of Y has support from 0 to 2.
    Do you means that this one
    " density function of Y: f(y)= y for 0<y<x "
    should be changed to this
    "density function of Y: f(y)= y for 0<y<2 "

    But I still found that density function cannot satisfy the condition ......
    Can I change it to:
    "density function of Y: f(y)= y for 0<y<2^(1/2) "
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