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Math Help - Can someone just explain this part of the solution to me....

  1. #1
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    Can someone just explain this part of the solution to me....

    It says

    Suppose P(X=a)=p and P(X=b)=1-p
    So clearly (X-b)/(a-b) is a Bernoulli Random variable. Find Var(X).

    The solution says:

    p(1-p)=Var((X-b)/(a-b))=1/(a-b)^2Var(X-b)=1/(a-b)^2*Var(X)

    So

    Var(X)=(a-b)^2*p(1-p)

    What I don't understand is how they did the part where I have underlined the equal sign (i.e. how to get out the 1/(a-b)^2 from that). Thanks.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by zhupolongjoe View Post
    It says

    Suppose P(X=a)=p and P(X=b)=1-p
    So clearly (X-b)/(a-b) is a Bernoulli Random variable. Find Var(X).

    The solution says:

    p(1-p)=Var((X-b)/(a-b))=1/(a-b)^2Var(X-b)=1/(a-b)^2*Var(X)

    So

    Var(X)=(a-b)^2*p(1-p)

    What I don't understand is how they did the part where I have underlined the equal sign (i.e. how to get out the 1/(a-b)^2 from that). Thanks.
    For any constant a, \text{Var}(aX)=a^2 \text{Var}(X)

    We indeed have :
    \text{Var}(X)=E(X^2)-[E(X)]^2

    So \text{Var}(aX)=E(a^2X^2)-[E(aX)]^2

    But for any constant b, E(bX)=bE(X)

    So \text{Var}(aX)=a^2 E(X^2)-a^2 [E(X)]^2=a^2 [E(X^2)-[E(X)]^2]=a^2 \text{Var}(X)
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  3. #3
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    Ah thanks. I don't recall that formula from class so I was confused.
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