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Discrete random variables

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    I don't understand this symmetry thing... how did they find E(x) without working out the unknowns??Name:  615b000a8c30519cabe540c789a172e7.png
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    Remember that the probablity will equal to one.Therefore, a+b+a=1. From here you will get simultaneous equations, find the value of a and b, then work your way to find E(X).

    (Original post by FamilyFirst)
    I don't understand this symmetry thing... how did they find E(x) without working out the unknowns??Name:  615b000a8c30519cabe540c789a172e7.png
Views: 31
Size:  16.7 KB
    undercxver has correctly pointed out that you can solve this question by solving the simultaneous equations that fall out from the information you are given.

    However, I take it that you're puzzled by the gnomic "by symmetry" in the answer for E[X]. The point here is that the probability distribution is symmetric about the value X=2. Whenever a distribution is symmetric about a value you can immediately write down the value of the mean (*). Why?

    To make this easier to see, think about the case where X takes the values -1, 0 and 1 and the distribution is still a, b, a. Then E[X] = (-1).a +(0).b + (1).a = 0. Can you see how the cancellation occurs? Now extend this to the case where X takes on more values: -3,-2,-1,0,1,2,3, for example, but where the distribution is still symmetric. Do you see that the part of the expectation sum where X is negative cancels with the part where it is positive?

    Now see if you can generalize this to where the distribution is symmetrical about a non-zero value of X.

    (*) for the experts: provided that the mean exists.
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