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Math Help - expected value

  1. #1
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    expected value

    Suppose X1 ~ Binomial(n = 15; p = 1=3) and X2 ~ Poisson(lambda = 4) and X1 and X2 are independent. Let Y = 3X1 + 4X2.

    Then E[Y] = 3E[X1] + 4E[X2]. But Var(Y) = E(Y^2)- E(Y)^2. How do I find E(Y^2)?
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  2. #2
    Member Last_Singularity's Avatar
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    Quote Originally Posted by lord12 View Post
    Suppose X1 ~ Binomial(n = 15; p = 1=3) and X2 ~ Poisson(lambda = 4) and X1 and X2 are independent. Let Y = 3X1 + 4X2.

    Then E[Y] = 3E[X1] + 4E[X2]. But Var(Y) = E(Y^2)- E(Y)^2. How do I find E(Y^2)?
    It is difficult to calculate variance using E[Y^2].

    Instead, I recommend that you try this:
    var(aX_1 + bX_2) = a^2 var(X_1) + b^2 var(X_2) + 2 ab cov(X_1,X_2)

    Of course, if X_1,X_2 are independent, the covariance term goes to zero.

    So for independent variables X_1,X_2,
    var(aX_1 + bX_2) = a^2 var(X_1) + b^2 var(X_2)

    Since a,b are given as 3,4 respectively, simply recall the variance expressions for a binomial and Poisson distribution.
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