Seems a simple enough question, and I presume that, if Yi are normally distributed,

Var(Sum(Yi)) = Sum(Var(Yi))

This feels like I'm jumping to the wrong conclusion though.

Any help would be wonderful.

Thanks a lot!

  • 5
    $\begingroup$ Your presumption is incorrect. The variance of a sum is the sum of the variances if the random variables are uncorrelated; normality has nothing to do with the issue. $\endgroup$ Commented May 21, 2016 at 2:43
  • 1
    $\begingroup$ en.wikipedia.org/wiki/Variance#Basic_properties $\endgroup$
    – Glen_b
    Commented May 21, 2016 at 3:48

2 Answers 2


It is indeed a simple question so let's explore it:

$\text{var}(\sum_i Y_i) = E[(\sum_i Y_i-E[\sum_i Y_i])^2] = E[(\sum_i Y_i)^2] - E[\sum_i Y_i]^2$

So far this is just standard results. Let's move on to the fun stuff:

$E[(\sum_i Y_i)^2] = E[\sum_{i,j} Y_i Y_j] = \sum_{i,j} E[Y_i Y_j]$

$E[\sum_i Y_i]^2 = \sum_{i,j} E[Y_i] E[Y_j]$

Separating into two pieces we get

$ \sum_{i,j} E[Y_i Y_j]-\sum_{i,j} E[Y_i Y_j] = \sum_{i=j} (E[Y_i Y_i] - E[Y_i]^2) + 2 \sum_{i<j} (E[Y_i Y_j] - E[Y_i]E[Y_j])$

Now we are ready to turn this into a friend of ours:

$E[Y_i Y_j] - E[Y_i]E[Y_j] = \text{cov}(Y_i,Y_j)$

$E[Y_i Y_i] - E[Y_i]^2 = \text{var}(Y_i)$

Thus: $\text{var}(\sum_i Y_i) = \sum_i\text{var}(Y_i) + 2\sum_{i<j} \text{cov}(Y_i,Y_j)$

Now you can see that this property relies on the covariance in the $Y_i$'s


Let $X$ and $Y$ be two random variables. It can be shown from the definition of variance that: $$Var(X + Y) = Var(X) + 2Cov(X,Y) + Var(Y)$$ Hence $Var(X+Y) = Var(X) + Var(Y)$ if and only if the covariance $Cov(X,Y) = 0$

A sufficient (but not necessary) condition for $Cov(X,Y) = 0$ is that $X$ and $Y$ are independent.


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