Actuarial Outpost > MFE Variance of weighted normal variables
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#1
06-19-2017, 02:57 PM
 mistersunnyd Member SOA Join Date: Aug 2016 Studying for how to find a job Posts: 104
Variance of weighted normal variables

I know this seems like a P question, but it appeared as a practice problem for MFE. Thus, I thought I'd ask here.
If I have two jointly normally distributed RVs, X and Y, with correlation rho, what is the variance of Z, with Z being aX - bY?
I always thought the formula for Var[aX-bY] = a^2Var[X]+b^2Var[Y] - 2abCov[X,Y], but the solution had a plus instead of a minus in front of 2abCov[X,Y]. Does anyone know why this is?
#2
06-19-2017, 03:11 PM
 jooni Member CAS Join Date: Jan 2017 Posts: 74

Which problem are you looking at?

I don't know about the formula for Var(aX-bY), but if you just use Var(aX + bY) with b being negative, then essentially you would get the minus sign before the 2abCov term, wouldn't you?
#3
06-19-2017, 04:42 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 30,881

Quote:
 Originally Posted by mistersunnyd I know this seems like a P question, but it appeared as a practice problem for MFE. Thus, I thought I'd ask here. If I have two jointly normally distributed RVs, X and Y, with correlation rho, what is the variance of Z, with Z being aX - bY? I always thought the formula for Var[aX-bY] = a^2Var[X]+b^2Var[Y] - 2abCov[X,Y], but the solution had a plus instead of a minus in front of 2abCov[X,Y]. Does anyone know why this is?
Your equation is right. As you have described the problem, the solution is wrong. Have you checked the errata, if errata exists?
#4
06-19-2017, 04:48 PM
 bujayherster Member CAS SOA Join Date: May 2016 Location: Southbury, CT Studying for MFE College: Union Collge Alumnus Posts: 64

Quote:
 Originally Posted by mistersunnyd I know this seems like a P question, but it appeared as a practice problem for MFE. Thus, I thought I'd ask here. If I have two jointly normally distributed RVs, X and Y, with correlation rho, what is the variance of Z, with Z being aX - bY? I always thought the formula for Var[aX-bY] = a^2Var[X]+b^2Var[Y] - 2abCov[X,Y], but the solution had a plus instead of a minus in front of 2abCov[X,Y]. Does anyone know why this is?

Isn't it because b = (-1), so with your equation:

Var[aX-bY] = Var[aX+(-b)Y]
= (a^2)Var[X]+(-b^2)Var[Y] - 2a(-b)Cov[X,Y]
= a^2Var[X]+b^2Var[Y] + 2abCov[X,Y]
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#5
06-19-2017, 05:09 PM
 ToBeAnActuaryOrNotToBe Member SOA Join Date: May 2014 Favorite beer: The ones in red solo cups. Posts: 671

Quote:
 Originally Posted by bujayherster Isn't it because b = (-1), so with your equation: Var[aX-bY] = Var[aX+(-b)Y] = (a^2)Var[X]+(-b^2)Var[Y] - 2a(-b)Cov[X,Y] = a^2Var[X]+b^2Var[Y] + 2abCov[X,Y]
Nope. Var[aX+bY] = a^2Var[X] +b^2Var[Y] + 2abCov[X,Y].
The original equation without a different b does not have a minus in front of the Covariance term. There is an error with the practice problem OP had.
#6
06-19-2017, 05:23 PM
 tkt Member CAS SOA Join Date: Jun 2011 Location: Des Moines College: Drake University Posts: 478

Not sure if this post is referring to an Adapt problem. We have come across a similar forum post in Adapt's discussion forum today. What's in the Adapt solution (see the attached) is correct.

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#7
06-19-2017, 07:03 PM
 mistersunnyd Member SOA Join Date: Aug 2016 Studying for how to find a job Posts: 104

Ah never mind. It seems I made b negative while also putting a negative before the entire covariance term. Thanks for the help guys!

 Tags covariance, variance