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Math Help - expected value and variance (continuous random variables)

  1. #1
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    expected value and variance (continuous random variables)

    Hello,
    I would like your help!

    X1, X2, are independent continuous random variables with E[x]=2, V(X)=9.

    Yi=(0.5^i)Xi, i=1,2,
    Tn=Y1+Y2++Yn
    An=(1/n) Tn

    E and V of Yn, Tn, An ???

    E[Yn] = E[(0.5^i)Xi] = (0.5^i)E[Xi] = 2(0.5^i)
    V[Yn] = V[(0.5^i)Xi] = [(0.5^i)^2] V[Xi] = 9 [(0.5^i)^2]
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  2. #2
    MHF Contributor matheagle's Avatar
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    YOU mean Yi not Yn
    Then sum to get Tn, which is a partial geometric sum from calc 2.
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  3. #3
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    Thanks, matheagle!

    Is the following solution correct ?

    E(Tn) = E(Y1)+E(Y2)+…+E(Yn) = 2*SUM(0.5^n)

    V(Tn) = V(Y1)+V(Y2)+…+V(Yn)= 9*SUM((0.5^n)^2)


    E(An)=E((1/n)* Tn)= 1/n *E(Tn) = 2/n* SUM(0.5^n)

    V(An)= V((1/n)* Tn) =(1/n)^2 *V(Tn) = ((1/n)^2 )*(9*SUM((0.5^n)^2))
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  4. #4
    MHF Contributor matheagle's Avatar
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    T_n=\sum_{i=1}^n\left(1/2\right)^iX_i

    E(T_n) =\sum_{i=1}^n\left(1/2\right)^iE(X_i) =2\sum_{i=1}^n\left(1/2\right)^i

    V(T_n) =\sum_{i=1}^n\left(1/4\right)^iV(X_i) =9\sum_{i=1}^n\left(1/4\right)^i
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  5. #5
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    thanks for your response, matheagle!

    The same solution I was thinking, but I could not insert the summation symbol, so I used the word "SUM" (from 0 to n) !!! Maybe I confused you, I apologize.

    Anyway, thanks again.
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  6. #6
    MHF Contributor matheagle's Avatar
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    yea, but not only did I use TeX, which you should learn
    But I switched the powers so you can obtain the geometric sum.
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  7. #7
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    I know it was my fault!

    I appreciate your help.
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