# When is the variance of the sum of random variables greater than the sum of the variances?

My professor asked my class to 'qualitatively' analyze the two scenarios with the assumption that there is no previous knowledge held in the concept of covariance (as we have not covered that chapter yet): For two dependent random variables X and Y,

1.) Var(X+Y) > Var(X) + Var(Y) 2.) Var(X+Y) < Var(X) + Var(Y)

I know that if two random variables are positively correlated then the variance of their sums will grow faster than the sum of their respective variances. However, when I stated this to my professor, she said "forget about correlation." So my question is, what can we say about the above inequalities while assuming no knowledge of covariance and correlation? I am unsure how to approach this question 'qualitatively' Any help would be appreciated.

• Like you, I would have started with $\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y) + 2\text{Cov}(X,Y)$ and then based my answer on the covariance. Will look further into this. Oct 19, 2018 at 19:39
• If the variables are independent, the variance of the sum is the sum of the variances. The question is missing some further information. Oct 19, 2018 at 20:13
• Hi Xi'an, thank you. I noticed the error as i wrote for two 'independent' random variables. For this question, I am only concerned with X and Y being dependent. Oct 19, 2018 at 20:24

$$\newcommand{\v}{\operatorname{var}}$$The most extreme case in which that happens is when they're both the same random variable. $$\v(X+X) = \v(X)+\v(X)+2\operatorname{cov}(X,X) = 4\v(X) > \v(X)+\v(X).$$ Generally you have $$\v(X+Y) = \v(X)+\v(Y) + 2\operatorname{cov}(X,Y).$$ If the covariance is positive, then the variance of the sum exceeds the sum of the variances.