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Conditional distribution of X given Y calculation HELP , PLEASE watch

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    Hi there I have a question for my homework that I have been given. It involves calculating the conditional distribution of X given Y and then , with that answer, being able to establish what this distribution is (ie is it gamma, or exponential.. and then also the parameter). I think i have solved it using the alternative formula for bayes thereom which is f xgiveny = f ygivenx * f x which gets me xe^-2xy or maybe it is xe^-x^3 or even xe^-2x^2? I think it may be the second .......... but from this point, i am stuggling to establish what distribution this is ? It is multiple choice so I have the options of : M(X), Gamma (1, y+1), Gamma (2, y+1), M(X+1), M(Y+1) and M(y). Please if anybody knows this distribtuion please may you help me to understand how to work this out and / or which one it could be? Many thanks.
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    Hi y'all - I don't suppose anybody could please help? Thanks a bunch
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    (Original post by alex.kundert)
    Hi y'all - I don't suppose anybody could please help? Thanks a bunch
    Could you post an image of the full question, including the options. Your text is hard to read.
    You also say
    f(x|y) = f(y|x) * f(x)
    You forgot to divide the rhs by f(y)
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    (Original post by mqb2766)
    Could you post an image of the full question, including the options. Your text is hard to read.
    You also say
    f(x|y) = f(y|x) * f(x)
    You forgot to divide the rhs by f(y)
    Hi there,

    Please find attached. Any help would be majorly appreciated , thank you !
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    (Original post by alex.kundert)
    Hi there,

    Please find attached. Any help would be majorly appreciated , thank you !
    Been a bit since I've done this, but since X \sim M(1) then f_X(x) = \displaystyle \int_{\mathbb{R}} f_{X,Y}(x,y) .dy = e^{-x}.

    And since Y | X \sim M(x) then we have f_{X,Y}(y|x) = \dfrac{f_{X,Y}(x,y)}{f_X(x)} = xe^{-xy}

    We can use this information to determine what f_{X,Y}(x,y) must be.

    Indeed, we can then use this to determine what f_Y(y) must be by integrating marginally along x \in (0, \infty).

    Hence we seek what f_{X,Y}(x|y) = \dfrac{f_{X,Y}(x,y)}{f_Y(y)} is.
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    (Original post by RDKGames)
    Been a bit since I've done this, but since X \sim M(1) then f_X(x) = \displaystyle \int_{\mathbb{R}} f_{X,Y}(x,y) .dy = e^{-x}.

    And since Y | X \sim M(x) then we have f_{X,Y}(y|x) = \dfrac{f_{X,Y(x,y)}}{f_X(x)} = xe^{-xy}

    We can use this information to determine what f_{X,Y}(x,y) must be.

    Indeed, we can then use this to determine what f_Y(y) must be by integrating marginally along x \in (0, \infty).

    Hence we seek what f_{X,Y}(x|y) = \dfrac{f_{X,Y}(x,y)}{f_Y(y)} is.
    Hi but I think ultimately as per the question what we want to achieve in the end is f(x given y) not f(y). I think I know that f(x,y) is xe^-3x but from there I am not sure where to go?
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    (Original post by RDKGames)
    Been a bit since I've done this, but since X \sim M(1) then f_X(x) = \displaystyle \int_{\mathbb{R}} f_{X,Y}(x,y) .dy = e^{-x}.

    And since Y | X \sim M(x) then we have f_{X,Y}(y|x) = \dfrac{f_{X,Y}(x,y)}{f_X(x)} = xe^{-xy}

    We can use this information to determine what f_{X,Y}(x,y) must be.

    Indeed, we can then use this to determine what f_Y(y) must be by integrating marginally along x \in (0, \infty).

    Hence we seek what f_{X,Y}(x|y) = \dfrac{f_{X,Y}(x,y)}{f_Y(y)} is.
    Ah , I see what you mean now after your edit... How do we go about solving f(y) do you know? I know you mention integrating, but what do we integrate?
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    (Original post by alex.kundert)
    Ah , I see what you mean now after your edit... How do we go about solving f(y) do you know? I know you mention integrating, but what do we integrate?
    f_{X,Y}(x,y)
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    (Original post by alex.kundert)
    I think I know that f(x,y) is xe^-3x but from there I am not sure where to go?
    Don't think I agree with this. Show your working out?
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    (Original post by RDKGames)
    Don't think I agree with this. Show your working out?
    Yes, I think it should actually be xe^-2xy after we have multipled them together. So I think that after integration this becomes e^-y(xe^x - e^x)

    Any thoughts on where we go from there ? In order to establish the distribution?
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    (Original post by alex.kundert)
    Yes, I think it should actually be xe^-2xy after we have multipled them together. So I think that after integration this becomes e^-y(xe^x - e^x)

    Any thoughts on where we go from there ? In order to establish the distribution?
    Isn't it
    xe^(-x(y+1))
    for the joint?
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    (Original post by alex.kundert)
    Yes, I think it should actually be xe^-2xy after we have multipled them together.
    I still don't agree.

    We have  \dfrac{f_{X,Y}(x,y)}{f_X(x)} = xe^{-xy} hence \dfrac{f_{X,Y}(x,y)}{e^{-x}} = xe^{-xy}
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    Hi all resolved now. The answer is xe^(-x(y+1)) as suggested by mqb and this is a gamma (2, y+1) distribution which is equal to f( x given y) . Thanks all !!
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    (Original post by alex.kundert)
    Hi all resolved now. The answer is xe^(-x(y+1)) as suggested by mqb and this is a gamma (2, y+1) distribution which is equal to f( x given y) . Thanks all !!
    Agreed.
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    (Original post by RDKGames)
    Agreed.
    Awesome, thanks for all your help again buddy

    I'll make sure to leave the post up this time for people to hopefully benefit from!!! Always appreciate your help, honestly
 
 
 
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