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    Mean of discrete distributions




    Hi All,

    Need your help on understanding the following.

    Suppose we have a conditional probability distribution P(C|A,B) over three binary random variables A,B,C. The distribution looks like the following

    A B P(C|A,B)

    0 0 .2 .8
    0 1 .1 .9
    1 0 .5 .5
    1 1 .9 .1

    If we condition on A=0, B=0 then we get the distribution P(C|A=0, B=0) = {.2, .8}. My question is, is the mean of this distribution .8 assuming P(C=0|A=0, B=0) = .2 and P(C=1|A=0, B=0) = .8?

    Thank you very much.

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    Re: Mean of discrete distributions

    The conditional mean of C given that A=0,B=0 is 0.8
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    Re: Mean of discrete distributions

    Thanks Dason for your reply. Here is my next questions ...

    Suppose we have a second distribution over the binary random variables A,B and D. Here the variables A and B are the same as the previous variables. So the example has all together four variables A,B,C,D and two distributions. The distribution is as follows.

    A B P(D|A,B)
    0 0 .3 .7
    0 1 .6 .4
    1 0 .5 .5
    1 1 .1 .9

    Given A=0, B=0 the variables C and D are independent. I am not sure how to get the mean of the two conditioned distributions P(C|A=0, B=0) = {.2, .8} and P(D|A=0,B=0)={.3, .7}. In other words I am interested in the mean of the distribution P(C,D|A=0,B=0) = P(C|A=0,B=0) x P(D|A=0,B=0).

    Thanks a lot.

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    Re: Mean of discrete distributions

    What exactly do you mean when you say you want the mean of that distribution? Are you talking about the mean of a function of those random variables or are you just talking about the multivariate mean (in which case it's just a vector of the means)...
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    Re: Mean of discrete distributions


    Not a function of those random variables but just the multivariate mean. So, the mean of a multivariate distribution is going to be a vector of mean?

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