Correlation Bounds: A Cauchy-Schwarz Proof

Why Correlation Has a Fixed Bound

Covariance measures how two variables vary together, but its magnitude depends on the units used. Correlation standardizes covariance by the two standard deviations, making it dimensionless. That standardization is not arbitrary: a fundamental inequality ensures the result cannot be smaller than −1 or greater than 1, provided both variables have finite, positive variances.

The proof comes from applying the Cauchy–Schwarz inequality to the centered variables $X-E[X]$ and $Y-E[Y]$. Centering matters because correlation concerns deviations from the means, rather than the raw values. The finite-variance condition matters because the standard deviations in the denominator must exist and be nonzero.

Definition: Let $X$ and $Y$ be random variables with finite, positive variances. Their correlation is $$ \operatorname{Corr}(X,Y)=\frac{\operatorname{Cov}(X,Y)}{\sqrt{\operatorname{Var}(X)}\sqrt{\operatorname{Var}(Y)}}. $$ It is defined only when $0<\operatorname{Var}(X)<\infty$ and $0<\operatorname{Var}(Y)<\infty$.

The Cauchy–Schwarz Argument

For random variables $U$ and $V$ with finite second moments, Cauchy–Schwarz states that the square of their expected product is no larger than the product of their second moments. To see why, consider the expected square $\operatorname{E}[(U-tV)^2]$, which cannot be negative for any real number $t$. When $\operatorname{E}[V^2]>0$, this expression is a quadratic in $t$ that is nonnegative for every $t$, so its discriminant cannot be positive. This gives the inequality below. If $\operatorname{E}[V^2]=0$, then $V=0$ with probability 1 and the inequality holds as well.

\bigl(\operatorname{E}[UV]\bigr)^2 \leq \operatorname{E}[U^2]\operatorname{E}[V^2]

Now take $U=X-E[X]$ and $V=Y-E[Y]$. These centered variables have finite second moments when $X$ and $Y$ have finite variances. Their expected product is the covariance, and their second moments are the variances. Substitution therefore bounds the square of the covariance by the product of the variances.

\begin{aligned} \bigl(\operatorname{Cov}(X,Y)\bigr)^2 &\leq \operatorname{Var}(X)\operatorname{Var}(Y),\\ \frac{\bigl(\operatorname{Cov}(X,Y)\bigr)^2} {\operatorname{Var}(X)\operatorname{Var}(Y)} &\leq 1,\\ \left|\operatorname{Corr}(X,Y)\right|&\leq 1. \end{aligned}

Theorem: If $X$ and $Y$ have finite, positive variances, then $$ -1\leq \operatorname{Corr}(X,Y)\leq 1. $$ Finite variances alone are not enough to define correlation: if either variance is zero, the denominator is zero. If a variance is not finite, the standard correlation formula is not justified by this result.

The proof has a useful structure: first apply Cauchy–Schwarz to centered variables; then divide by the positive product of their standard deviations. The positivity condition justifies that division. It also shows why the result is about correlation, not about covariance: covariance itself is not restricted to the interval from −1 to 1.

Check the variances.
Confirm that both variances are finite and strictly positive before using the correlation formula.

Find the covariance.
Use the joint distribution or a valid covariance calculation to obtain $\operatorname{Cov}(X,Y)$.

Standardize.
Divide the covariance by $\sqrt{\operatorname{Var}(X)}\sqrt{\operatorname{Var}(Y)}$, not by the variances themselves.

Check the result.
A computed correlation outside $[-1,1]$ signals an arithmetic or modeling error, or a failure of the conditions needed to define correlation.

Worked Calculations

Worked Example: Correlation From Equally Likely Outcomes

A hypothetical quality-control test produces one of three equally likely pairs of measurements, $(X,Y)=(0,1),(2,0),(4,5)$. Find the correlation. Both measurements are recorded in the same arbitrary units.

Method. Calculate the means, variances, and covariance from the three outcomes, then standardize the covariance. This direct calculation avoids treating the covariance as though it were already a correlation.

Each outcome has probability $1/3$, so $\operatorname{E}[X]=(0+2+4)/3=2$ and $\operatorname{E}[Y]=(1+0+5)/3=2$.

For $X$, $\operatorname{E}[X^2]=(0^2+2^2+4^2)/3=20/3$, giving $\operatorname{Var}(X)=20/3-2^2=8/3$. For $Y$, $\operatorname{E}[Y^2]=(1^2+0^2+5^2)/3=26/3$, giving $\operatorname{Var}(Y)=26/3-2^2=14/3$.

The expected product is $\operatorname{E}[XY]=(0\cdot1+2\cdot0+4\cdot5)/3=20/3$. Thus $\operatorname{Cov}(X,Y)=20/3-(2)(2)=8/3$, and

\operatorname{Corr}(X,Y) =\frac{8/3}{\sqrt{(8/3)(14/3)}} =\frac{2}{\sqrt{7}} \approx 0.7559.

Check. The variances are finite and positive, and $0.7559$ is within the required bounds. A common mistake would be to report $8/3$, the covariance, as the correlation; standardization is essential.

Worked Example: A Continuous Variable and Its Square

Let $X$ have a uniform distribution on $[-2,2]$, and define $Y=X^2$. Find their correlation. The variables are dependent because $Y$ is determined by $X$, but dependence does not prevent calculating correlation.

Method. Use the moments of the uniform distribution and the definition of covariance. In particular, retain the centering term when calculating the variance of $Y$.

For a uniform variable on $[-2,2]$, symmetry gives $\operatorname{E}[X]=0$ and $\operatorname{E}[X^3]=0$. Direct integration gives $\operatorname{E}[X^2]=4/3$ and $\operatorname{E}[X^4]=16/5$. Therefore, $\operatorname{Var}(X)=4/3$, while $\operatorname{E}[Y]=\operatorname{E}[X^2]=4/3$.

Since $Y^2=X^4$, $\operatorname{Var}(Y)=\operatorname{E}[X^4]-(\operatorname{E}[X^2])^2 =16/5-16/9=64/45$. Also, $\operatorname{Cov}(X,Y)=\operatorname{E}[XY]-\operatorname{E}[X]\operatorname{E}[Y] =\operatorname{E}[X^3]-0=0$. Both variances are finite and positive, so $\operatorname{Corr}(X,Y)=0/\sqrt{(4/3)(64/45)}=0$.

Check. Zero lies within the Cauchy–Schwarz bounds. A common mistake is to calculate $\operatorname{Var}(Y)$ as $\operatorname{E}[Y^2]$ and omit $(\operatorname{E}[Y])^2$; that would give $16/5$, not the correct variance $64/45$.

Worked Example: Correlation of a Loss and a Capped Payment

In a hypothetical insurance model, a loss $X$, measured in thousands of dollars, takes the values $0,2,4,6$ with equal probability. The payment is $Y=\min(X,4)$, also in thousands of dollars. Find the correlation between loss and payment.

Method. First compute each variable’s mean and variance and their covariance from the four outcomes. Then divide by the product of standard deviations. This method avoids confusing a covariance, which has squared units here, with a dimensionless correlation.

The corresponding payment values are $0,2,4,4$. Thus $\operatorname{E}[X]=(0+2+4+6)/4=3$ and $\operatorname{E}[Y]=(0+2+4+4)/4=5/2$. For the loss, $\operatorname{E}[X^2]=(0+4+16+36)/4=14$, so $\operatorname{Var}(X)=14-3^2=5$. For the payment, $\operatorname{E}[Y^2]=(0+4+16+16)/4=9$, so $\operatorname{Var}(Y)=9-(5/2)^2=11/4$.

The product expectation is $\operatorname{E}[XY]=(0+4+16+24)/4=11$. Therefore, $\operatorname{Cov}(X,Y)=11-(3)(5/2)=7/2$. The correlation is

\operatorname{Corr}(X,Y) =\frac{7/2}{\sqrt{5(11/4)}} =\frac{7}{\sqrt{55}} \approx 0.9439.

Check. Both variances are positive, and the result is between −1 and 1. Reporting $7/2$ as the correlation would be wrong: that is the covariance, and it is measured in squared thousands of dollars.

Using the Bound as an Exam Check

The bound is especially useful as a diagnostic after a multi-step calculation. If a proposed correlation is $1.08$, for example, the answer cannot be correct under the stated finite, positive-variance assumptions. Recheck the covariance, the two variances, and the square roots in the denominator. Do not “fix” an out-of-range value by rounding it into the interval.

The bound does not imply that every number in the interval is attainable for every specified model. It is a necessary restriction on correlation whenever correlation is defined. It also does not assert that $X$ and $Y$ are independent; it only limits the value of their standardized covariance.

Common exam trap: Cauchy–Schwarz bounds the covariance in magnitude by the product of the standard deviations: $$ |\operatorname{Cov}(X,Y)| \leq \sqrt{\operatorname{Var}(X)}\sqrt{\operatorname{Var}(Y)}. $$ Dividing by that positive product gives the correlation bound. Dividing by the variances instead, or forgetting to center the variables when finding covariance, does not give the correlation formula.

Check Your Understanding

For each problem, verify that the variances needed for correlation are finite and positive, then apply the appropriate calculation.

A device generates equally likely pairs $(X,Y)=(1,4),(3,1),(5,2)$. Find their correlation. (Answer: $-\sqrt{3/7}\approx -0.6547$)
Let $X$ be uniform on $[-2,2]$ and $Y=X^2$. Find $\operatorname{Cov}(X,Y)$ and $\operatorname{Corr}(X,Y)$. (Answer: $\operatorname{Cov}(X,Y)=0$, $\operatorname{Corr}(X,Y)=0$)
A hypothetical loss $X$, in thousands of dollars, is equally likely to be $0,2,4,$ or $6$, and a contract pays $Y=\min(X,4)$. Find $\operatorname{Cov}(X,Y)$ and $\operatorname{Corr}(X,Y)$. (Answer: $\operatorname{Cov}(X,Y)=7/2$, $\operatorname{Corr}(X,Y)=7/\sqrt{55}\approx0.9439$)
A proposed model has $\operatorname{Var}(X)=9$, $\operatorname{Var}(Y)=16$, and $\operatorname{Cov}(X,Y)=-13$. Can these summaries belong to random variables with finite variances? (Answer: No; the bound requires $|\operatorname{Cov}(X,Y)|\leq\sqrt{9}\sqrt{16}=12$)
Suppose $\operatorname{Var}(X)=25$, $\operatorname{Var}(Y)=4$, and $\operatorname{Cov}(X,Y)=-6$. Set $X'=3X+7$ and $Y'=-2Y+1$. Find $\operatorname{Corr}(X',Y')$. (Answer: $0.6000$)

What You'll Learn