The determinant of the sum of a positive definite matrix with a symmetric singular matrix

284 Views Asked by At

I came across this problem when trying to prove the following: If you have two binary random variables $A$, $B$ that are positively correlated with probability $p$ and otherwise are uncorrelated, prove that they are correlated on average. Correlation here is defined as $P(a=0, b=0)P(a=1, b=1) - P(a=0,b=1)P(a=1,b=0)$ as per the Pearson correlation measure. Notice that this measure can be calculated directly from the joint probability matrix $\rho_{AB}$, which is a $2\times 2$ matrix with elements $[\rho_{AB}]_{i,j} = P(a = i, b= j)$. The correlation measure is just the determinant of this matrix.

In the scenario described, the joint distribution is therefore the convex sum of a positive definite matrix (the joint distribution matrix for when they are correlated, multiplied by $p$) and a symmetric singular matrix multiplied by $(1-p)$ (this is singular as $(1-p)$ of the time the two variables are uncorrelated, giving a Pearson correlation of 0, and it is symmetric as when they are correlated $P(a=i, b=j) = P(a=i)P(b=j)$.

My question is this - it is obvious from this case that when two random variables are positively correlated $p$ of the time, and uncorrelated $(1-p)$ of the time, then their probability distribution is correlated. This demands that a positive definite matrix summed with a symmetric singular matrix is a positive definite matrix, under the following constraints.

  1. The matrices are positive with rows that sum to 1
  2. (If needs be) the matrices are $2\times 2$ dimensional

How can I prove this? And is this a generic feature, or does the proof need to use things like the fact that the matrices rows sum to 1 etc.

1

There are 1 best solutions below

2
On

This is not even true for numbers: $1+(-2)=-1$.