# Distributions of Functions of Random variables - variance question

• Dec 8th 2013, 12:00 PM
optometryorchidsmelly
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!!
• Dec 8th 2013, 02:08 PM
romsek
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?
• Dec 8th 2013, 02:39 PM
optometryorchidsmelly
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!
• Dec 8th 2013, 03:16 PM
romsek
Re: Distributions of Functions of Random variables - variance question
Quote:

Originally Posted by optometryorchidsmelly
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.
• Dec 8th 2013, 04:48 PM
optometryorchidsmelly
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!!!