Inequality of expectation of a quadratic form

Question

I was reading a proof in a paper. Let $X$ and $Y$ be two possibly correlated $K$-dimensional random vectors. Suppose $\mathrm{E}(YY^T)=I$, where $I$ is an $K\times K$ identity matrix. By Cauchy-Schwarz inequality, we have $\left\{\mathrm{E}(X^TY)\right\}^2 \leq \mathrm{E}(X^TX)\mathrm{E}(Y^TY)$, and $\mathrm{E}(Y^TY)$ here equals to $K$. Beyond this, the author claims the following inequality: $$ \mathrm{E}\left\{(X^TY)^2\right\} \leq \mathrm{E}(X^TX)\mathrm{E}(Y^TY). $$ I don't see the reason as in general we have only $\mathrm{E}\left\{(X^TY)^2\right\} \geq \left\{\mathrm{E}(X^TY)\right\}^2$. This claim is analogous to the quadratic form as we can write $\mathrm{E}\left\{(X^TY)^2\right\}=\mathrm{E}\left\{(X^TYY^TX)\right\}$. When $YY^T$ is constant, it is easy $$ \mathrm{E}\left\{X^TAX\right\} \leq \lambda_\max(A)\mathrm{E}(X^TX), $$ where $\lambda_\max(A)$ is the greatest eigenvalue of $A$ (we also have the formula for $\operatorname{E}\left\{X^TAX\right\}$ in terms of the trace of $A\operatorname{var}(X)$). Is this claim right? How to show it?

Answer 1

It doesn't even work for scalar random variables. For example, if $X=Y$ then the proposed inequality reads $$\mathbb{E}(X^4)\leq [\mathbb{E}(X^2)]^2.$$ Subtracting gives $$\mbox{Var}(X^2)=\mathbb{E}(X^4)-[\mathbb{E}(X^2)]^2\leq 0,$$ which can only happen if $X^2$ is almost surely constant.