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    Hi, I have a couple of questions.

    If you know the probability mass function of X and Y and then work out conditional mass function of XlY and YlX, how do you work out: E(Y-XlX<Y)?
    What im thinking is writing out all the values of y transformed to y-x and only considering x<y then doing the mean of this data. is this right?

    also, does anyone know about moment generating functions? i know that M_X=E(e^(tx)) but what does this do? why do we want to know M_X?

    thanks!!
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    (Original post by number23)
    Hi, I have a couple of questions.

    If you know the probability mass function of X and Y and then work out conditional mass function of XlY and YlX, how do you work out: E(Y-XlX<Y)?
    What im thinking is writing out all the values of y transformed to y-x and only considering x<y then doing the mean of this data. is this right?
    Depending on the amount of pmf's it's a fair bit of work, but yes, your methodology would work. Assuming you meant all the values of y transformed by all the values of x.

    Notice that X<Y can be rewritten Y-X > 0.

    So if you restrict your calculations to just those combinations satisfying that criterion it will more than halve your working.

    also, does anyone know about moment generating functions? i know that M_X=E(e^(tx)) but what does this do? why do we want to know M_X?

    thanks!!
    Not something I've studied, unfortunately.
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    (Original post by ghostwalker)
    Depending on the amount of pmf's it's a fair bit of work, but yes, your methodology would work. Assuming you meant all the values of y transformed by all the values of x.

    Notice that X<Y can be rewritten Y-X > 0.

    So if you restrict your calculations to just those combinations satisfying that criterion it will more than halve your working.



    Not something I've studied, unfortunately.
    great thanks, i was wondering if there was a specific formula i needed to remember. so say if x could be: 1 or 2. y: could be 1 or 2. then y-x can be: 0, -1, 1, 0
    but we just consider: 0, 1, 0 hence e(x)=1/3 ??
    thanks
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    (Original post by number23)
    great thanks, i was wondering if there was a specific formula i needed to remember. so say if x could be: 1 or 2. y: could be 1 or 2. then y-x can be: 0, -1, 1, 0
    but we just consider: 0, 1, 0 hence e(x)=1/3 ??
    thanks
    Well it's a strict inequality so we are only interested in one of those 4 combinations.
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    (Original post by ghostwalker)
    Well it's a strict inequality so we are only interested in one of those 4 combinations.
    so is 1/3 not correct?
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    (Original post by number23)
    also, does anyone know about moment generating functions? i know that M_X=E(e^(tx)) but what does this do? why do we want to know M_X?

    thanks!!
    M_X(t)\\ \\=E(e^{tx})\\ \\=E(1+tX+\frac{t^2 X^2}{2}+\frac{t^3 X^3}{3!}...)\\ \\=E(1)+E(tX)+E(t^2X^2/2)+E(t^3X^3/6)+...\\ \\=1+tE(X)+\frac{t^2}{2}E(X^2)+ \frac {t^3}{6} E(X^3)+...\\ \\M'(t)=E(X)+tE(X^2)+ \frac{t^2}{2} E(X^3) +...\\ \\M'(0)=E(X)\\ \\M''(t)=E(X^2)+tE(X^3)+...\\ \\M''(0)=E(X^2)

    A nice idea, I think.
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    (Original post by number23)
    so is 1/3 not correct?
    Nope. There is only one combination that satisfies Y>X.
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    (Original post by number23)
    also, does anyone know about moment generating functions? i know that M_X=E(e^(tx)) but what does this do? why do we want to know M_X?
    Well, differentiating M_X n times w.r.t. t and setting t = 0 will give you the n-th moment (hence the name).

    Another very useful property is that if X and Y are independent, then M_{X+Y}(t) = M_X(t)M_Y(t).
 
 
 
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