x
• Status: Solved
• Priority: Medium
• Security: Public
• Views: 700

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

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?
0
Milewskp
1 Solution

How about you at least TRY to do your homework first and then ask the questions that you really need MORE help explaining?
0

Commented:
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.
http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables
0

Commented:
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
0

Author Commented:
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'?
0

Commented:
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?
0

Commented:
Oh, I got it. It's minimized when k = a^2/(a^2 + b^2)
0

Author Commented:
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.
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.