# Variance of a series of IID's vs a multiple of a random variable

For any random variable, X, Var(aX) = a^2*Var(X), which is easy to demonstrate.

Suppose you have a series of IID's, and want to find the variance. So, in that case for example, Var(X+X+X+X+X) = Var(X) + Var(X) + Var(X) + Var(X) + Var(X) = 5Var(X) since there is no covariance involved. But isn't Var(X+X+X+X+X) = Var(5X) = 25Var(X)?

Follow-up, when doing the variance of a sum of dependent random variables would you add two times every possible pairwise covariance to the individual variances?

• I think reading this Wikipedia article en.wikipedia.org/wiki/Variance would answer all your questions and give you a better understanding of simple concepts such as the variance – Stelios Kounis Jul 7 at 19:54
• This is a very common mistake. – leonbloy Jul 7 at 20:41

Follow-up, when doing the variance of a sum of dependent random variables would you add two times every possible pairwise covariance to the individual variances?

Yes. And the covariance of $$X$$ and $$X$$ is the variance: $$\operatorname{Cov}(X,X)=\operatorname{Var}(X)$$, where $$X$$ fully depend on itself. That´s why

\begin{align} \operatorname{Var}(X+X)& =\operatorname{Var}(X)+\operatorname{Var}(X) + 2\cdot \operatorname{Cov}(X,X) \\[6pt] &=2\cdot \operatorname{Var}(X) + 2 \cdot \operatorname{Var}(X) \\[6pt] &=4\cdot \operatorname{Var}(X) \end{align}

If $$X_1,...,X_5$$ are independent, then $$V(X_1+X_2+X_3+X_4+X_5) = V(X_1)+V(X_2)+V(X_3)+V(X_4)+V(X_5)$$ For obvious reasons, $$X$$ is not independent from $$X$$, so it is your second formula that is correct: $$V(X+X+X+X+X)=V(5X)=25V(X)$$

• I guess it really just comes down to the fact that multiplying the same random variable by 5 is not the same as having 5 independent samples from the same underlying distribution. Yeah? That was my presumption initially, but just wanted to verify. – OzarkNathan Jul 7 at 19:59
• You are correct, those are two very different things. – charlus Jul 7 at 20:03

$$\newcommand{\v}{\operatorname{var}}\newcommand{\c}{\operatorname{cov}}$$

would you add two times every possible pairwise covariance

Here is a way to think about that. First, suppose you know that $$\v(X+Y) = \v(X) + \v(Y) + 2\c(X,Y).$$ That can be applied repeatedly, as follows: \begin{align} & \v(X_1+X_2+X_3+X_4+X_5) \\[6pt] = {} & \v(X_1+X_2) + \v(X_3+X_4+X_5) \tag1 \\ & {} + 2\c(X_1+X_2,\,X_3+X_4+X_5) \tag2 \\[10pt] & \text{The first term on line (1) is:} \\[8pt] & \v(X_1+X_2) = \v(X_1)+\v(X_2) + 2\c(X_1,X_2). \\[8pt] & \text{The second term on line (1) is:} \\[8pt] & \v(X_3+X_4+ X_5) \\[4pt] = {} & \v(X_3)+ \v(X_4+X_5) \\ & {} + 2\c(X_3,\,X_4+X_5). \\[10pt] & \text{The thing on line (2) is:} \\[8pt] & \c(X_1+X_2,\,\,X_3+X_4+X_5) \\[8pt] = {} & \c(X_1,\,\,X_3+X_4+X_5) + \c(X_2,\,\,X_3+X_4+X_5) \tag3 \\[8pt] = {} & \c(X_1,X_3) + \c(X_1,X_4) + \cdots \end{align} Keep going like that until you see just how many covariances you need, and which ones.