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How to weight two normally distributed random variables to minimize the variance of their sum

Posted on 2011-03-05
Medium Priority
Last Modified: 2012-05-11
I have 2 normally distributed random variables, A and B, with variances a^2 and b^2, respectively, and I calculate the weighted sum of A and B as: C = (1-k)A + kB, where k is a constant, 0<k<1.

What value of k will minimize the variance of C?
Question by:Milewskp
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LVL 37

Expert Comment

by:Neil Russell
ID: 35043768
How about you at least TRY to do your homework first and then ask the questions that you really need MORE help explaining?
LVL 37

Expert Comment

ID: 35044209
You know (from your book) that if A is normal with a variance of a^2, then kA has a variance of k^2a^2. Also Z=X+Y gives Z a variance of x^2+y^2. So work out the equation for C.
LVL 27

Expert Comment

ID: 35044533
But according to the given link the variance of the total is equal to the sum of the variances. Therefore, if you can  weight them, you will get the min variance if
k = 1 or 0
In your notation, if the variance of A is less use k=0 and if the variance of B is less use k = 1
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Author Comment

ID: 35044773
OK people,
Green Acres was a TV show, Paul was in a band before Wings, and 'Drive-In' used to mean a theatre or fast food joint. Anyone still think I'm a 'student' with 'homework'?
LVL 37

Expert Comment

ID: 35044922
Even if you aren't a student with homework, you will learn better and be more valuable to your employer (or yourself) on whatever this is for if you can learn to do it yourself.
In the future, if you think a question might look like homework (we get a TON of those in the Math & Science zone, perhaps more than half of the questions), let us know in the question what it's for so we can treat it accordingly.
Anyway, the variance of C is going to be (1-k)^2a^2 + k^2b^2 so that's the function that needs to be minimized. Do you know anything about a and b?
LVL 37

Accepted Solution

TommySzalapski earned 2000 total points
ID: 35044936
Oh, I got it. It's minimized when k = a^2/(a^2 + b^2)

Author Comment

ID: 35046175
Thanks Tommy,
That's what I got (dC/Dk = 0, so k = a^2/(a^2 + b^2)). I just needed a verification. Stay tuned for the next (related) question.

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