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?
I have a problem that I'm stuck on.
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!!
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!
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.