# Minimum variance for sum of three random variables

I have been working on the following problem:

Given you have $$\text{Var}(X) = 1$$, $$\text{Var}(Y) = 4$$, and $$\text{Var}(Z) = 25$$, what is the minimum possible variance for the random variable $$W = X + Y + Z$$, or $$\text{min}\{ \text{Var}(X+Y+Z)\}$$?

My first thought is to complete the variance-covariance expansion as follows: $$\text{Var}(X + Y + Z) = \text{Var}X + \text{Var}Y + \text{Var}Z +2[\text{Cov}(X,Y) + \text{Cov}(Y,Z) + \text{Cov}(X,Z)]$$

Then to use the Cauchy-Schwarz inequality to determine the minimum covariance for each of the covariance terms (i.e. $$|\text{Cov}(X,Y)| \le \sqrt{\text{Var}(X)\text{Var}(Y)}$$ ). However, I am obtaining a negative potential minimum, which leads me to think that the lower bound could be zero?

$$\text{Var}(X+Y+Z) = 1 + 4 + 25 + 2[-2 - 5 - 10] = 30 - 34$$ ???

The other thought is that using Cauchy-Schwarz in this way is not correct and my approach is wrong.

My next thought is to consider the expansion as $$\text{Var}[(X+Y), Z]$$, but was not sure how to proceed by considering the sum of 2 variables $$(X+Y)$$ and $$Z$$. Perhaps use the fact that $$\text{Cov}(X+Y,Z) = \text{Cov}(X,Y) + \text{Cov}(X,Z)$$?

Any thoughts on how to proceed are appreciated.

• Are you asking to solve the problem allowing the unknown covariances to be any values consistent with mathematical restrictions on covariances? Note that these are more restrictive than what you have inferred from the Cauchy-Schwarz Inequality. – whuber Oct 23 '14 at 18:27
• There is no assumption of independence for the three variables, so you are correct in that I need to determine the restrictions on the covariances, and as expected, I cannot simply use the Cauchy-Schwarz inequality to determine these restrictions. Based on your link, should I weight the correlations so that the determinant of the matrix is at least 0 and in such a way that I obtain the minimum covariance? – skijunkie Oct 23 '14 at 18:48
• $x = -z/5, y = -2z/5$ where $\mathbb{E}z = 0$ for simplicity of exposition gets you to $\text{Var}(x+y+z) = 4$ while preserving the variances. Intuitively (why this is a comment) I think this is the best you can do, and the implication for the covariance matrix is clear. A proof-by-contradiction of $\min \text{Var}(x+y+z) = 4$ might work, basing it on the covariance matrix having to be nonnegative-definite. – jbowman Oct 23 '14 at 19:20
• Yes, applying the determinant restriction is correct. In fact, if you write the covariance matrix in the form $$\pmatrix{1&0&0\\0&2&0\\0&0&5}\pmatrix{1&\rho&\sigma\\ \rho&1&\tau\\ \sigma&\tau&1}\pmatrix{1&0&0\\0&2&0\\0&0&5}$$ you can directly apply the formulas from the other thread. They will show that you need only consider the cases $\pm\rho=\pm\sigma=\pm\tau=1$, from which you easily obtain the minimum of $4$ in your example (as well as a general formula for any three variables). Solutions for more than three variables are more challenging to obtain! – whuber Oct 23 '14 at 20:44
• More generally, semidefinite programming can be used to obtain bounds on variances/covariances involving many random variables. – Brian Borchers Oct 23 '14 at 23:27