I am currently studying data analysis and statistics on my own and have come across proof that I do not fully understand. I have looked at alternative proofs that the correlation coefficient lies in the interval $~[-1,1]~$ (mostly with the Cauchy-Schwarz inequality), but I want to understand this specific proof as well.

"To this end, suppose that $t$ is some real number that we will choose later,and consider the obvious inequality

$$E((V+tW)^2)≥0~, ~~~~\text{where}~~~~ V=X−μ_X~~~~\text{and}~~~~ W=Y−μ_Y~.$$

Expanding out the left-hand-side, and using the linearity of expectation, we find that

$$E(V^2) + 2tE(V W) +t^2E(W^2)≥0~.$$

Note that the left-hand-side is just a quadratic polynomial in $t$. Now, clearly we have that

$$E(V^2) =σ^2_X,E(W2) =σ^2_Y,$ and $E(V W) = Cov(X, Y)~.$$

And so, our polynomial inequality becomes

$$σ^2_Yt^2+ 2Cov(X, Y)t+σ^2_X≥0~.$$

From this inequality we find that the only way the left-hand-side could be $0$ is if the polynomial has a double-root (i.e. it touches the x-axis in a single point), which could only occur if the discriminant is $0$. So, the discriminant must always be negative or $0$, which means that

$$4Cov(X, Y)^2−4σ^2_Xσ^2_Y≤0~.$$

In other words,

$$\frac{Cov(X, Y)^2}{σ^2_Xσ^2_Y}≤1~,$$

provided, of course, that the denominator does not vanish."

I understand all the steps except the bold one. Specifically, I would like to know how the author arrived from

$$σ^2_Yt^2+ 2Cov(X, Y)t+σ^2_X≥0$$


$$4Cov(X, Y)^2-4σ^2_Xσ^2_Y≤0~.$$

So I would find more details of the calculation path very helpful. I guess my understanding problem is based on the fact that I can't see where a double root is used here or what exactly the discriminant is. I try hard, but I have a background in psychology, which often makes me realize that I lack proficiency in reading mathematical proofs.

Many thanks in advance!

  • $\begingroup$ You recieved an answer to your question. Is it what you needed? If so, you should upvote and accept it. $\endgroup$ – 5xum Jul 30 at 8:35
  • $\begingroup$ Yes, it has helped a lot! $\endgroup$ – Sol Jul 30 at 10:50
  • $\begingroup$ Then... why did you not accept the answer? $\endgroup$ – 5xum Aug 1 at 9:48

A polynomial $at^2 + bt + c$ has two real roots if and only if it has a positive discriminant, i.e. if $b^2-4ac> 0$.

If a polynomial has two real roots, then its sign between the two roots is different from the sign outside the roots.

This means that if $at^2+bt+c\geq 0$ for all values of $t$, meaning the polynomial's sign doesn't change, which means the polynomial cannot have two real roots, therefore, $b^2-4ac>0$ is not true, and $b^2-4ac\leq 0$ must be true.


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