Consider nondegenerate real-valued random variables $X$ and $Y$ with finite mean and variance. We can prove the inequality
$$\text{Var}(X)\cdot\text{Var}(Y)-\text{Cov}(X,Y)^2\geq0$$
in a variety of ways:
(1) We can recognize left-hand side as the determinant of the variance-covariance matrix of the random vector $(X, Y)$, and then invoke the fact that variance-covariant matrices are positive semidefinite.
(2) The variance and covariance are invariant under shifting by a constant, so by mean-centering $X$ and $Y$, the inequality reduces to $E(X^2)E(Y^2)-E(XY)^2\geq0$, which follows from Cauchy-Schwarz for random variables.
(3) We can try to prove the inequality “directly” by unraveling the left-hand side as
$$\big(E(X^2)-E(X)^2\big)\big(E(Y^2)-E(Y)^2\big)-\big(E(XY)-E(X)E(Y)\big)^2$$
which upon simplifying gives (if I haven’t made an error)
$$\color{blue}{E(X^2)E(Y^2)-E(XY)^2}+\color{red}{2E(XY)E(X)E(Y)-E(X^2)E(Y)^2-E(Y^2)E(X)^2}$$
According to the argument in (2), using shift-invariance of variance and covariance, the red term should vanish, but I’m not seeing why. What algebraic cleverness am I missing?
Best Answer
You have made no error in your calculations. However, you have misinterpreted the implication of shift invariance. It's not that the red term vanishes, but rather than if you replace $X$ and $Y$ in the blue portion with $X-\mathbb EX$ and $Y-\mathbb EY$ respectively, then when you simplify it out you will be left with the red and blue expressions together.
By the way, whenever $\textrm{Var}(Y)\not=0$ one has $$ \textrm{Var} (X)\cdot \textrm{Var} (Y)-\textrm{Cov}(X,Y)^2=\frac{\textrm{Var}\bigl(\textrm{Var}(Y)\cdot X-\textrm{Cov}(X,Y)\cdot Y\bigr)}{\textrm{Var}(Y)}. $$ I think this fits the bill of a "slick algebraic proof" of the type you are looking for.