coefficent of correlation smaller than 1 proof problem

Question

Hi! I have recently found a proof for that the correlation must be smaller than $1$.

I have two questions:

1. Why must the discriminant be negative? It seems that it is related to treating $a$ as unknown and solving the quadratic equation. And looking at the discriminant. But I don't get why the discriminant has to be smaller than or equal $0$?

2. Why do they use the $\{a(x-\mu_x)+(y-\mu_y)\}^2$ to prove in general that coefficent of correlation is $\leq 1$?

Source: A. Papoulis, "Probability, random variables, and stochastic processes", Third Edition, Chap 7. p. 152-153

Answer 1

1. A quadratic expression $f(x) = ax^2 + bx + c$ that is always positive cannot have any real roots. The graph of $y = f(x)$ is a parabola, and $f(x) > 0$ means it lies entirely above the $x$-axis. Thus, it never intersects or touches the $x$-axis, which means it has no real roots. This is so only when the discriminant is negative. In the case where $f(x) \ge 0$, there can be at most one real root (where the parabola touches the $x$-axis), so the discriminant can be $0$ or negative.
2. This is an indirect method of proof.

In the spirit of simplicity, here is a rewriting of the same proof:

Let $X$ and $Y$ be two random variables, and let $a$ be a positive real number. Now, $aX + Y$ is a random variable, so its variance $V(aX + Y) \ge 0$ (variance is always non-negative). Using the properties of variance:

$V(aX + Y) \ge 0\Rightarrow\\ a^2V(X) + 2a\text{Cov}(X, Y) + V(Y^2) \ge 0$

Now, as the LHS is a quadratic function of $a$ that is always non-negative, its graph is a parabola that lies entirely above the $x$-axis. Thus, it either has complex roots, or has at most one real root, which implies that the discriminant is non-positive:

$[2\text{Cov}(X, Y)]^2 - 4a^2V(X)V(Y) \le 0 \Rightarrow\\ [\text{Cov}(X, Y)]^2 \le V(X)V(Y) \Rightarrow\\ \left[\dfrac{\text{Cov}(X, Y)}{V(X)V(Y)}\right]^2 \le 1 \Rightarrow\\ \rho^2 \le 1 \Rightarrow\\ -1 \le \rho \le 1$