What is the expectation value of the product of two random variables each with expectation zero?

Question

Consider two random variables $X$ and $Y$ such that $\mathbb{E}[X]=\mathbb{E}[Y]=0$. I want to show that $$(\mathbb{E}[XY])^2\leq \mathbb{E}[X^2]\mathbb{E}(Y^2).$$

I've considered the following so far:

\begin{equation} \begin{split} (\mathbb{E}[XY])^2 & = \Big(\sum_i\sum_jx_iy_jP(X=x_i, Y=y_j)\Big)^2 \\ & = ... + \sum_i\sum_jx_i^2y_j^2P(X=x_i, Y=y_j)^2 ... \\ \implies (\mathbb{E}[XY])^2 & \geq \sum_i\sum_jx_i^2y_j^2P(X=x_i, Y=y_j)^2. \end{split} \end{equation}

Also,

$$\mathbb{E}(X^2)\mathbb{E}(Y^2) = \sum_ix_i^2P(X=x_i)\sum_jy_j^2P(Y=y_j).$$

I can't see whether or not this has actually got me anywhere. Some help would be appreciated.

Answer 1

$Var(X) = \mathbb{E}(X-\mathbb{E}X)^2 = \mathbb{E} [X ^ 2]$, same for the $Y$, $Var(Y) = \mathbb{E} [Y ^ 2]$. Also recall that $$ f_X(x)=\int_{\mathbb{R}}f_{X,Y}(x,y)dy $$ and $$ \int_{\mathbb{R^2}}f_{X,Y}(x,y)dxdy = \int_{\mathbb{R}} dx\int_{\mathbb{R}}f_{X,Y}(x,y)dy = \int_{\mathbb{R}}f_X(x)dx \, . $$ So, by the Cauchy-Schwartz inequality \begin{align} (\mathbb{E}XY)^2 &= \left( \int_{\mathbb{R}^2}xyf(x,y)dxdy \right) ^ 2\\ &\le \int_{\mathbb{R}^2}x ^ 2 f(x,y)dxdy \int_{\mathbb{R}^2}y ^ 2 f(x,y)dxdy\\ &\le\int_{\mathbb{R}} x ^ 2f(x)dx \int_{\mathbb{R}} y ^ 2f(y)dx\\ &= \mathbb{E}[X^2]\mathbb{E}[Y^2]. \end{align}

Basically, you don't even need these integrals, the conclusion is just a straightforward application of the inequality.

Another approach. Define the inner product $\langle X, Y \rangle =\mathbb{E}XY $, hence by Cauchy-Schwartz inequality you have $$ (\mathbb{E}XY)^2 = |\langle X, Y \rangle | ^ 2 \le \langle X, X \rangle \langle Y, Y \rangle = \mathbb{E}( X ^ 2 ) \mathbb{E} (Y ^ 2) . $$

Answer 2

We have that $E(XY)=\sigma_{XY}$, the covariance between $X$ and $Y$, $E(X^2)=\sigma^2_X$ and $E(Y^2)=\sigma^2_Y$ their variances. Now $\frac{\sigma_{XY}}{\sigma_X\sigma_Y}$, the correlation between them, is known to be a number between $-1$ and $1$.

Answer 3

Note that: $$-1\le Corr(X,Y)=\frac{Var(X,Y)}{\sqrt{Var(X)\cdot Var(Y)}}\le 1 \iff \\ -\sqrt{Var(X)\cdot Var(Y)}\le Var(X,Y)\le \sqrt{Var(X)\cdot Var(Y)} \iff \\ 0\le [Var(X,Y)]^2\le Var(X)\cdot Var(Y) \iff \\ 0\le \left(\mathbb E[(X-\mathbb E(X))(Y-\mathbb E(Y))]\right)^2\le \mathbb E[(X-\mathbb E(X))^2]\cdot \mathbb E[(Y-\mathbb E(Y))^2] \iff \\ \left(\mathbb E[XY]-\mathbb E(X)\mathbb E(Y)\right)^2\le (\mathbb E(X^2)-(\mathbb E(X))^2)(\mathbb E(Y^2)-(\mathbb E(Y))^2) \iff \\ (\mathbb E(XY))^2\le \mathbb E(X^2)\mathbb E(Y^2).$$