Does the variance of a sum equal the sum of the variances?

Question

Is it (always) true that $$\mathrm{Var}\left(\sum\limits_{i=1}^m{X_i}\right) = \sum\limits_{i=1}^m{\mathrm{Var}(X_i)} \>?$$

Answer 1

The answer to your question is "Sometimes, but not in general".

To see this let $X_1, ..., X_n$ be random variables (with finite variances). Then,

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = E \left( \left[ \sum_{i=1}^{n} X_i \right]^2 \right) - \left[ E\left( \sum_{i=1}^{n} X_i \right) \right]^2$$

Now note that $(\sum_{i=1}^{n} a_i)^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} a_i a_j $, which is clear if you think about what you're doing when you calculate $(a_1+...+a_n) \cdot (a_1+...+a_n)$ by hand. Therefore,

$$ E \left( \left[ \sum_{i=1}^{n} X_i \right]^2 \right) = E \left( \sum_{i=1}^{n} \sum_{j=1}^{n} X_i X_j \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i X_j) $$

similarly,

$$ \left[ E\left( \sum_{i=1}^{n} X_i \right) \right]^2 = \left[ \sum_{i=1}^{n} E(X_i) \right]^2 = \sum_{i=1}^{n} \sum_{j=1}^{n} E(X_i) E(X_j)$$

so

$$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} \big( E(X_i X_j)-E(X_i) E(X_j) \big) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j)$$

by the definition of covariance.

Now regarding Does the variance of a sum equal the sum of the variances?:

• If the variables are uncorrelated, yes: that is, ${\rm cov}(X_i,X_j)=0$ for $i\neq j$, then $$ {\rm var} \left( \sum_{i=1}^{n} X_i \right) = \sum_{i=1}^{n} \sum_{j=1}^{n} {\rm cov}(X_i, X_j) = \sum_{i=1}^{n} {\rm cov}(X_i, X_i) = \sum_{i=1}^{n} {\rm var}(X_i) $$

• If the variables are correlated, no, not in general: For example, suppose $X_1, X_2$ are two random variables each with variance $\sigma^2$ and ${\rm cov}(X_1,X_2)=\rho$ where $0 < \rho <\sigma^2$. Then ${\rm var}(X_1 + X_2) = 2(\sigma^2 + \rho) \neq 2\sigma^2$, so the identity fails.

• but it is possible for certain examples: Suppose $X_1, X_2, X_3$ have covariance matrix $$ \left( \begin{array}{ccc} 1 & 0.4 &-0.6 \\ 0.4 & 1 & 0.2 \\ -0.6 & 0.2 & 1 \\ \end{array} \right) $$ then ${\rm var}(X_1+X_2+X_3) = 3 = {\rm var}(X_1) + {\rm var}(X_2) + {\rm var}(X_3)$

Therefore if the variables are uncorrelated then the variance of the sum is the sum of the variances, but converse is not true in general.

Answer 2

$$\text{Var}\bigg(\sum_{i=1}^m X_i\bigg) = \sum_{i=1}^m \text{Var}(X_i) + 2\sum_{i\lt j} \text{Cov}(X_i,X_j).$$

So, if the covariances average to $0$, which would be a consequence if the variables are pairwise uncorrelated or if they are independent, then the variance of the sum is the sum of the variances.

An example where this is not true: Let $\text{Var}(X_1)=1$. Let $X_2 = X_1$. Then $\text{Var}(X_1 + X_2) = \text{Var}(2X_1)=4$.

Answer 3

Yes, if each pair of the $X_i$'s are uncorrelated, this is true.

See the explanation on Wikipedia