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I know the typical variance formula for correlated random variables, but can't seem to find the variance for a linear combination of uncorrelated random variables.

Now there are a few things regarding uncorrelated variables that obviously play into this: - Two random variables are said to be uncorrelated if their Cov(X,Y)=0 - The variance of the sum of uncorrelated random variables is the sum of their variances

But what about the variance itself for a linear combination of these r.v.'s?

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    $\begingroup$ What is the question? You seem to know everything already $\endgroup$
    – Aksakal
    Sep 19, 2016 at 20:36
  • $\begingroup$ Is there a formula for the variance of a linear combination of uncorrelated random variables? I know the rules regarding variances, but can't seem to find the formula for a linear combination of these r.v.'s anywhere. Sorry, I edited my question to include "linear combination" $\endgroup$
    – coderX
    Sep 19, 2016 at 20:43
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    $\begingroup$ It is the sum of the individual variances $\endgroup$
    – Carl
    Sep 19, 2016 at 20:43
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    $\begingroup$ for two variables, see stats.stackexchange.com/a/123963/805 or more generally en.wikipedia.org/wiki/Variance#Basic_properties (then taking advantage of the fact that when the covariances are 0 the terms including them will drop out) $\endgroup$
    – Glen_b
    Sep 20, 2016 at 4:25
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    $\begingroup$ Possible duplicate of Use the properties of linear combinations to derive means and standard deviations $\endgroup$
    – Tim
    Feb 27, 2017 at 19:11

3 Answers 3

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Let $Z$ be a linear combination of two random uncorrelated variables $X$ and $Y$ so that:

$$Z=aX+bY$$

Then:

$$var(Z)=a^2·var(X)+b^2·var(Y)$$

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There's some confusion. "Variance" is not a property of a pair of variables, it's a property of a random variable. If your r.v. happens to be the sum of two others, then there is a formula for that variance as a function of the other two. It depends on the correlation, and if that correlation is zero, then plug in zero, and there you go. To take from Pere's answer, if $$ Z = aX+bY $$ then $$\newcommand{\var}{{\rm var}} \var(Z)=a^2·\var(X)+b^2·\var(Y) + 2·{\rm cov}(X, Y) $$

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    $\begingroup$ Isn't a factor $a.b$ needed before the covariance term? $\endgroup$ Feb 25, 2017 at 17:48
  • $\begingroup$ @f.g., you shouldn't edit someone's post to correct it. Instead, leave a comment, downvote, post your own answer, etc. $\endgroup$ Feb 27, 2017 at 18:54
  • $\begingroup$ That was my own answer. "Filipe" was just the profile I was using on another machine. (for boring reasons). $\endgroup$
    – f.g.
    Feb 28, 2017 at 20:13
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If $X_1, ..., X_K$ are all uncorrelated with each other, then

$$ {\rm var} \left( \sum_{i=1}^K a_i X_i \right) = \sum_{i=1}^{K} a_i^2 {\rm var}(X_i) $$

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  • $\begingroup$ This was already said in the two previous answers... $\endgroup$
    – Tim
    Feb 27, 2017 at 19:03
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    $\begingroup$ The previous two answers pre-specified the case where $K=2$ $\endgroup$ Feb 27, 2017 at 19:04
  • $\begingroup$ Yes, but extending it to K>2 cases is pretty obvious since it follows from the same general properties of variance... $\endgroup$
    – Tim
    Feb 27, 2017 at 19:06
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    $\begingroup$ Tim, all of the answers here are pretty obvious. This is a question with an obvious answer that takes about 3 seconds to google. And it's a duplicate. And the OP gives the answer in the statement of this question. I don't know why you're singling me out here. The question was already bumped by gung so I went ahead and added this answer because the OP asked about the variance of a linear combination, without pre-specifying that there are only two. I added something beyond what was already said (unlike the previous two answers, frankly). ~Fin~ $\endgroup$ Feb 27, 2017 at 19:10
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    $\begingroup$ If I'd searched and found a question that had been inactive for 6 months and bumped it just to add this answer, I guess I could understand your point of view, but I didn't. The fact is, it popped up at the top of the queue, I read it, and I added a piece of information that no one else had mentioned yet. Was it a profound piece of information? No. But, it could be useful to a future reader. Whether or not the OP accepted an answer seems immaterial. $\endgroup$ Feb 27, 2017 at 19:18

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