2

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.

SecretAgentMan
  • 1,463
  • 10
  • 30
skijunkie
  • 471
  • 4
  • 7
  • Are you asking to solve the problem allowing the unknown covariances to be any values consistent with [mathematical restrictions on covariances](http://stats.stackexchange.com/questions/5747)? 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
  • 1
    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

0 Answers0