Proving the correlation $\rho$ is always $|\rho|\leq 1$ and finding when $\rho=-1$

Question

Let $\xi$ and $\eta$ be random variables. $\mathbb{D}\xi >0$ and $\mathbb{D}\eta>0$ Let $\rho=\rho(\xi,\eta)=\frac{\mathbb{E}(\xi-\mathbb{E}\xi)(\eta-\mathbb{E}\eta)} {\sqrt{\mathbb{D}\xi \mathbb{D}\eta}}$.

I need to show that

1) $|\rho|\leq 1$

2) $\rho=-1$ when $\frac{\eta -\mathbb{E}\eta}{\sqrt{\mathbb{D}\eta}}=-\frac{\xi -\mathbb{E}\xi}{\sqrt{\mathbb{D}\xi}}$

For 1) I think I should use $|\mathbb{E}\xi\eta|\leq \sqrt{\mathbb{E}\xi^2\mathbb{E}\eta^2}$ but actually I don't know how to do it correctly.

For 2) I have no Ideas. How should I start and what should I use?

Answer 1

For 1) you can apply the inequality you want to use on $\xi-E\xi$ and $\eta-E\eta$ instead of $\xi$ and $\eta$. To follow the statement from this, you will have to use that

$E((\xi-E\xi)^2)=E(\xi^2-2\xi E(\xi)+E(\xi)^2)=E(\xi^2)-2E(\xi)E(\xi)+E(\xi)^2=E(\xi^2)-E(\xi)^2=D\xi.$

Answer 2

1) Since zero-mean random variables comprise a vector space on which covariance is an inner product, $|\rho|\le 1$ is a restatement of that inner product space's Cauchy-Schwarz inequality. If this isn't intuitive, rewrite your favourite proof of the inequality, including its saturation condition (which we'll need in a moment), in the language of expectations, covariances etc.

2) Since $\rho=-1$ requires the unit vectors $\frac{\xi-\Bbb E\xi}{\sqrt{\Bbb D\xi}},\,\frac{\eta-\Bbb E\eta}{\sqrt{\Bbb D\eta}}$ to be "antiparallel", their ratio is $-1$ as required.

Answer 3

1) Let $z=(\xi,\eta)'$, then the variance matrix $V(z)=E[(z-Ez)(z-Ez)']$ is positive semidefinite and so $det(V(z)) \geq 0$, which implies $|\rho| \leq 1$.

2) If $a=-b$ then $a^2=-ab$ and $E[a^2]=-E[ab]$, and so $\rho \equiv E[ab] = -E[a^2] = -1$, by the definition of the variance of $a$.