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Math Help - Quick Help with Joint Probability Mass function

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    Quick Help with Joint Probability Mass function

    Trying to do this question.

    Finding K i get 1/25 but I know this is wrong.

    MY method P(1,1) + P(1,3) + P(2,3) = 1

    Thanks in advance for your help.
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    Hello,

    No, you're all correct !


    Note : please next time, put your attachments as .jpg, not .bmp, as it's more painful to open !
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    Cool thanks! But I thought that P(1,1) = 1/3 and if K= 1/25

    Then P(1,1) = 2/25?

    Sorry about the pic format I usually change it but forgot this time.
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    Quote Originally Posted by Niall101 View Post
    Cool thanks! But I thought that P(1,1) = 1/3 and if K= 1/25

    Then P(1,1) = 2/25?

    Sorry about the pic format I usually change it but forgot this time.
    Why would P(1,1)=1/3 ? Because there are only 3 points ?
    This is only correct if we assume that the distribution is uniform.
    Which is not necessarily the case.

    Yes, P(1,1)=2/25
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    Cool Thanks so much! Im glad i was on the right track.

    Does that same idea hold for when Im getting the marginal function of X

    I get 2x^2 / 25 + 10/25 which is getting X over all values of Y where Y can be 1 or 3.

    But is it also possible say for x = 1 and y = 50 for example which would still give a value for X even though the joint would give 0?
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    Perhaps it'll be clearer if you see it written this way :

    Let \mathcal{X} be the domain of X, namely {1,2}
    Let \mathcal{Y} be the domain of Y, namely {1,3}

    Then for any x\in\mathbb{R}, we have :
    f_X(x)=P(X=x)=\sum_{y\in\mathcal{Y}} P(x,y)=P(x,1)+P(x,3)

    Which exactly gives what you got :
    2x^2 / 25 + 10/25
    if x is in \mathcal{X}

    If x\neq 1,2, we still have f_X(x)=P(x,1)+P(x,3), but then these values are 0, according to the definition of P(.,.)
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    Hey, thanks a lot for your help I appreciate it.
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