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Math Help - Distributions of Functions of Random variables - variance question

  1. #1
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    Distributions of Functions of Random variables - variance question

    I have a problem that I'm stuck on.

    var(x)=8100
    var(y)=10,000
    var(x+y)=20,000

    The question is: calculate var(x+500+(1.08)y).

    I tried just doing var(x) + 1.08^2(var(y)) thinking that they were independent...
    The part I'm really confused about is why var(x+y) is different from var(x)+var(y).

    I know the answer is 21816, but I have no idea how to get that.

    any help is much appreciated, thanks!!
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    Re: Distributions of Functions of Random variables - variance question

    For any rvs x,y

    Var(x+y) = Var[x] + Var[y] - Cov[x,y]

    without know something about the means of x and y I wasn't able to get very far with this. Are they zero mean?
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    Re: Distributions of Functions of Random variables - variance question

    Thanks for looking at this!

    Mean is not included in the question, but it does come from a section about independent random variables- not sure if that helps.

    The exact wording is this: "In considering medical insurance for a certain operation, let X equal the amount (in dollars) paid for the doctor and let Y equal the amount paid to the hospital. In the past, the variances have been Var(X) = 8100, Var(Y) = 10,000, and Var(X + Y) = 20,000. Due to increased expenses, it was decided to increase the doctor's fee by $500 and increase the hospital charge Y by 8 percent. Calculate the variance of X+500+(1.08)Y, the new total claim.

    All of the other questions involved independent random variables, and were much more straightforward than this one. Unfortunately, I can't find any similar examples or problems in the book. Thanks again for any help!
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    Re: Distributions of Functions of Random variables - variance question

    Quote Originally Posted by optometryorchidsmelly View Post
    Thanks for looking at this!

    Mean is not included in the question, but it does come from a section about independent random variables- not sure if that helps.

    The exact wording is this: "In considering medical insurance for a certain operation, let X equal the amount (in dollars) paid for the doctor and let Y equal the amount paid to the hospital. In the past, the variances have been Var(X) = 8100, Var(Y) = 10,000, and Var(X + Y) = 20,000. Due to increased expenses, it was decided to increase the doctor's fee by $500 and increase the hospital charge Y by 8 percent. Calculate the variance of X+500+(1.08)Y, the new total claim.

    All of the other questions involved independent random variables, and were much more straightforward than this one. Unfortunately, I can't find any similar examples or problems in the book. Thanks again for any help!
    my intuition.

    the 500 does nothing to the variance.

    the 1.08 should increase the variance of Y by (1.08)2 and the covariance[x,y] by 1.08

    Var(x+y) = Var(x) + Var(y) - Cov(x,y)

    20000 = 8100 + 10000 - Cov(x,y)

    Cov(x,y) = -1900

    The new Cov(x,y) will be 1.08*1900 = -2052

    The new Var(y) = 1.082 * 10000 = 11664

    The new variance is thus

    Var[x+y] = 8100 + 11664 + 2052 = 21816

    I'd have to read up on this to be 100% certain of this solution but I'm fairly sure this is how it's done.
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    Re: Distributions of Functions of Random variables - variance question

    Ok, this makes a lot of sense. I think the part that I hadn't gotten at first was the 1.08*1900.

    Thank you so much!!!
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