cauchy schwarz inequality

Let \(u,v\) be vectors in an inner product space. Then, \(|\langle u, v \rangle|^2 \leq \langle u, u \rangle \dot \langle v, v \rangle\)

1. probability

Define the inner product between two random variables \(X,Y\) to be \(E[XY]\).

Then, Cauchy Schwarz gives: \(|E[XY]|^2 \leq E[X^2]E[Y^2]\)

From this, we can get the covariance inequality: \[\begin{align*} |Cov(X,Y)|^2 &= |E[(X-\mu)(Y-\nu)]|^2 \\ &\leq E[(X-\mu)^2]E[(Y-\nu)^2] \\ &= Var(X)Var(Y) \\ \end{align*}\] So, \(Cov(X,Y)^2 \leq Var(X)Var(Y)\)