# Correlation Coefficient Squared is Less Than or Equal to One

Problem Statement: Let $$Y_1$$ and $$Y_2$$ be jointly distributed random variables with finite variances. Let $$\rho$$ denote the correlation coefficient of $$Y_1$$ and $$Y_2.$$ Using the inequality $$[E(Y_1Y_2)]^2\le E\!\left(Y_1^2\right) E\!\left(Y_2^2\right),$$ show that $$\rho^2\le 1.$$

This is essentially Exercise 5.111b in Mathematical Statistics with Applications, 5th Ed., by Wackerly, Mendenhall, and Scheaffer.

My Work So Far: We have that $$\newcommand{\Cov}{\operatorname{Cov}}$$ \begin{align*} \rho^2 &=\frac{(\Cov(Y_1,Y_2))^2}{V(Y_1)V(Y_2)}\\ &=\frac{(E(Y_1Y_2)-E(Y_1)E(Y_2))^2} {\left(E\left(Y_1^2\right)-(E(Y_1))^2\right)\left(E\!\left(Y_2^2\right)-(E(Y_2))^2\right)}\\ &=\frac{(E(Y_1Y_2))^2-2E(Y_1Y_2)E(Y_1)E(Y_2)+(E(Y_1))^2(E(Y_2))^2} {E\!\left(Y_1^2\right)E\!\left(Y_2^2\right)-E\!\left(Y_1^2\right)(E(Y_2))^2-(E(Y_1))^2E\!\left(Y_2^2\right)+(E(Y_1))^2(E(Y_2))^2}\\ &\le\frac{E\!\left(Y_1^2\right)E\!\left(Y_2^2\right)-2E(Y_1Y_2)E(Y_1)E(Y_2)+(E(Y_1))^2(E(Y_2))^2} {E\!\left(Y_1^2\right)E\!\left(Y_2^2\right)-E\!\left(Y_1^2\right)(E(Y_2))^2-(E(Y_1))^2E\!\left(Y_2^2\right)+(E(Y_1))^2(E(Y_2))^2}. \end{align*} Now I can see that there are two terms common to the numerator and denominator, but I'm very unsure of where to go next.

My Question: What are good next steps? Or is this even the right trail to follow?

• If instead you were to start over by applying the inequality to the two variables $Y_1-E[Y_1]$ and $Y_2-E[Y_2],$ you could complete this exercise in a single short line of work: divide by the right hand side and notice that the definition of $\rho$ appears on the left hand side.
– whuber
Commented Aug 13, 2020 at 21:40
• Excellent, thanks! Commented Aug 14, 2020 at 17:51

That inequality is an application of the Cauchy–Schwarz inequality:

$$|\langle \mathbf{u},\mathbf{v}\rangle| ^2 \leq \langle \mathbf{u},\mathbf{u}\rangle \cdot \langle \mathbf{v},\mathbf{v}\rangle,$$ where $$\langle\cdot,\cdot\rangle$$ is the inner product.

For random variables $$Y_1$$ and $$Y_2$$, the expected value of their product is an inner product:

$$\langle \mathbf{Y_1},\mathbf{Y_2}\rangle:=E[Y_1Y_2]$$

Therefore

\begin{aligned}Cov(Y_1,Y_2)^2 &= E[(Y_1 - E[Y_1])(Y_2 - E[Y_2])]^2\\ &=\langle Y_1 - E[Y_1], Y_2 - E[Y_2] \rangle ^2\\ &\leq \langle Y_1 - E[Y_1], Y_1 - E[Y_1] \rangle \langle Y_2 - E[Y_2], Y_2 - E[Y_2] \rangle\\ &= E[(Y_1-E[Y_1])^2] E[(Y_2-E[Y_2])^2]\\ &= Var(Y_1) Var(Y_2) \end{aligned}

• This works, too, although it's not technically using the inequality directly. Perhaps indirectly? Commented Aug 14, 2020 at 17:51
• It looks to me, Adrian, like it is using the inequality directly--it merely has been written in a different notation.
– whuber
Commented Aug 14, 2020 at 18:25