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    Properties of discrete random variables




    Hello,

    would some one with more experience mind to confirm or correct me:

    For discrete random variables I know:

    E[aX+b] = aE[X]+b and Var(aX+b)=a^2Var(X)+b

    if there are more than two constants, is correct to assume that this holds?

    E[abX+c] = abE[X]+c and Var(abX+c) = (ab)^2Var(x)+b

    Many Thanks
    jc

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    Re: Properties of discrete random variables

    The variance of a constant is 0.

    V(aX+b) = a^2 V(X)

    V(abX + c) = (ab)^2 V(X)

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    Re: Properties of discrete random variables

    Thanks for that

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    Re: Properties of discrete random variables

    It's easy to see that
    V(abX + c) = (ab)^2 V(X) from
    V(dX + c) = (d)^2 V(X)
    just by letting ab = d. Since a and b are constants then d = ab is a constant so we get the result for the two constants as a direct consequence.

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    Re: Properties of discrete random variables


    Hi,

    I have to catch up on that, there is still some confusion.

    I have a random varialbe X with E[X]=7.45 min. On a daily basis I am charged 0.3 per min and fix costs of 3.

    So for a day E[X] = 0.3*7.45 + 3 = 5.235

    But for a year (300 days)?

    My intution would tell me 300*(0.3*7.45+3)=1570.5,

    but following the formula E[abX+c] = abE[X]+c strictly, I would get :

    0.3*300*7.45+3 = 673.5, which seems to low to me.

    I am going off in a totally wrong direction?

    Thanks a lot for comments.

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