Show that $|EX| \le E|X|$ where $X \in R^d$

Question

The hint was to use Cauchy-Schwarz, so

$$|X'EX| \le |X||EX| $$ $$E|X'EX| \le E|X||EX| $$

I guess I need to convert the left hand side into a product like

$$|EX'||EX| $$

but I don't know how to proceed.

Answer 1

No idea how Cauchy-Schwarz is supposed to help here, but, assuming one is considering the Euclidean norm $\|\cdot\|$ in $\mathbb R^d$, one can start from the identity, valid for every $x$ in $\mathbb R^d$, $$\|x\|=\sup_{\|t\|=1}(t\cdot x)$$ Now, for every fixed $t$ such that $\|t\|=1$, $$t\cdot X\leqslant\|X\|$$ hence $$t\cdot E(X)=E(t\cdot X)\leqslant E(\|X\|)$$ and taking the supremum over $t$ yields $$\|E(X)\|=\sup_{\|t\|=1}(t\cdot E(X))\leqslant E(\|X\|)$$