how do i find the variance of a weighed sum of random variables?
http://stats.stackexchange.com/question ... variables
the question in this link suggests it's the sum of the (w_i*std_i)^2 given that correlation is 0
the first answer suggests it's the square of the sum of the (w_i*std_i)
my intuition tells me it's the sum of w_i*std_i^2, since iirc variances linearly add.
what's correct?
variance of weighed sum of random variables
Moderators: gmalivuk, Moderators General, Prelates

 Posts: 481
 Joined: Wed Sep 21, 2011 3:44 am UTC
Re: variance of weighed sum of random variables
The variances do linearly add, but the weighting coefficient scales the standard deviation, not the variance. So each variance is effectively weighted by the square of the weighting coefficient, giving you the sum of (w_i*std_i)^2 for the total variance.
Re: variance of weighed sum of random variables
let sum = 2*A = A + A
if variances linearly add
var(sum) = var(A) + var(A) = 2*var(A)
but if weights scale standard deviation
std(sum) = std(A) + std(A) = 2*std(A)
var(sum) = sqrt(std(sum)) = sqrt(2)*sqrt(std(A)) = sqrt(2)*var(A)
this is a contradiction though
edit: whoops i see where i messed up now because var(A) + var(A) =/= 2*var(A)
it all makes sense now
if variances linearly add
var(sum) = var(A) + var(A) = 2*var(A)
but if weights scale standard deviation
std(sum) = std(A) + std(A) = 2*std(A)
var(sum) = sqrt(std(sum)) = sqrt(2)*sqrt(std(A)) = sqrt(2)*var(A)
this is a contradiction though
edit: whoops i see where i messed up now because var(A) + var(A) =/= 2*var(A)
it all makes sense now

 Posts: 481
 Joined: Wed Sep 21, 2011 3:44 am UTC
Re: variance of weighed sum of random variables
>) wrote:edit: whoops i see where i messed up now because var(A) + var(A) =/= 2*var(A)
Well, var(A)+var(A)=2*var(A), since var(A) is just another variable. I think what you mean is that var(A+A) = var(2*A) =/= 2*var(A) = var(A)+var(A). Variances add linearly only with uncorrelated variables, but A is perfectly correlated with itself. It comes out that var(A+A)=4*var(A)
Re: variance of weighed sum of random variables
The addition rule for variances is var(aX + bY) = a² var(X) + b² var(Y) + 2ab cov(X, Y), which you can then extend to an arbitrary number of such variables. So they don't really add linearly, it just looks like it if you're only directly adding two completely uncorrelated variables.
pollywog wrote:I want to learn this smile, perfect it, and then go around smiling at lesbians and freaking them out.Wikihow wrote:* Smile a lot! Give a gay girl a knowing "Hey, I'm a lesbian too!" smile.
Re: variance of weighed sum of random variables
I just realized what I really needed wasn't what I thought I was looking for here.
What I actually want is the variance of the weighed average of two probability mass functions X and Y. In other words, the variance of the probability mass function Z = aX+bY, given a+b = 1.
I'm not exactly sure what the difference is between this and just the variance of aX + bY but I think there is one.
Suppose X was just 0 with p = 1 and Y was just 1 with p = 1.
The variance of X/2+Y/2 is 0, obviously, but what I want is the variance of Z = {0 with p = 0.5, 1 with p = 0.5}, which is 1/4.
I'm not sure how this can be done. Or even what the name for this is.
I'm willing to make any necessary assumptions about X and Y (normality or whatever else)
What I actually want is the variance of the weighed average of two probability mass functions X and Y. In other words, the variance of the probability mass function Z = aX+bY, given a+b = 1.
I'm not exactly sure what the difference is between this and just the variance of aX + bY but I think there is one.
Suppose X was just 0 with p = 1 and Y was just 1 with p = 1.
The variance of X/2+Y/2 is 0, obviously, but what I want is the variance of Z = {0 with p = 0.5, 1 with p = 0.5}, which is 1/4.
I'm not sure how this can be done. Or even what the name for this is.
I'm willing to make any necessary assumptions about X and Y (normality or whatever else)
Re: variance of weighed sum of random variables
What you're looking for is the variance of a mixture of X and Y.
Re: variance of weighed sum of random variables
Thanks, that's the word I was looking for, and http://stats.stackexchange.com/a/16609 gives a nice expression for it.
Who is online
Users browsing this forum: curiosityspoon and 14 guests