# Bessel's inequality for expected value

Let $X_1, X_2,\ldots$ be independent random variables with expected value $\mathbb{E}[X_i]=0$ and variance $V[X_i]=1$. Let $Y$ be another random variable, such that $\mathbb{E}[Y^2] < \infty$.

I have to prove the following inequality:

$$\mathbb{E}[Y^2] \geq \sum_{i=1}^{n} \mathbb{E}[YX_i]$$

In my probability theory lecture we defined $\langle X,Y\rangle:=E[X\cdot Y]$ as a semi inner product on $\{ X\text{ random variable on }(\Omega, P)\text{ such that }\mathbb{E}[X^2]< \infty \}$. If we leave aside that I'm not really on a Hilbert space (yet), my random variables $\{X_n\}_{n \in \mathbb{N}}$ kind of form an orthonormal set, and I thought about my functional analysis lecture. There we proved following bessel's inequality:

Let $\{x_n\}_{n \in \mathbb{N}}$ be an orthonormal set on a Hilbert space $H$. For any $y \in H$ holds:

$$\|y\|^2 \geq \sum_{i=1}^n |\langle y,x_i\rangle |^2$$

So I'm confused now, because in my upper inequality I'm missing the square on the right side.

Does the inequality still hold?

I neither managed to come up with a counter example, nor modify the fairly easy proof the get git of the square, I'd really appreciate if someone could give a hint... Thank you very much in advance!

• Are you sure you mean it when you say that $E[Y^2]$ is less than or equal to $\infty$? There really are too different cases to be considered here if you are really are meaning what you say. – Dilip Sarwate Jun 19 '14 at 12:49
• I'm sorry, of course it's supposed to be less than Infinity, I will edit it right now!! Thanks for pointing it out! – user121314 Jun 19 '14 at 12:51

Here's a counter-example. Assume $E(Y)=0.$ For $n=1$ the inequality is
$$\sigma_Y^2 \ge Cov(X,Y) \text{ since } E(X)=0.$$
$$\sigma_Y^2 \ge \rho \sigma_X \sigma_Y$$
$$\sigma_Y \ge \rho$$