The Correlation Bound And Its Proof

Why Correlation Has a Fixed Range

Correlation puts covariance on a standardized scale: it describes the direction and strength of linear co-movement without retaining the units of the two random variables. For risk analysis, this makes it useful when comparing dependence across losses measured on different scales. Its range is not an empirical convention or a modeling preference. It follows from a basic fact: the expectation of a squared random variable cannot be negative.

The result requires finite, positive variances. If either variable has zero variance, its standard deviation is zero and its correlation is undefined; the bound does not assign a correlation in that case.

Definition: For random variables $X$ and $Y$ with finite, positive variances, their standard deviations are $\sigma_X=\sqrt{\operatorname{Var}(X)}$ and $\sigma_Y=\sqrt{\operatorname{Var}(Y)}$. Their Pearson correlation is $$ \rho_{X,Y}=\frac{\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y}. $$ The value is dimensionless. Its sign matches the sign of the covariance.

The Bound From Squared Linear Combinations

Write $U=X-E[X]$ and $V=Y-E[Y]$, so $E[U]=E[V]=0$, $\operatorname{Var}(U)=\sigma_X^2$, $\operatorname{Var}(V)=\sigma_Y^2$, and $E[UV]=\operatorname{Cov}(X,Y)$. Because a square is nonnegative for every outcome, $E[(U/\sigma_X-V/\sigma_Y)^2]$ and $E[(U/\sigma_X+V/\sigma_Y)^2]$ must both be nonnegative. Expanding these expectations gives the bound directly.

\begin{aligned} 0 &\leq E\left[\left(\frac{U}{\sigma_X}-\frac{V}{\sigma_Y}\right)^2\right] =1+1-2\rho_{X,Y}=2(1-\rho_{X,Y}),\\ 0 &\leq E\left[\left(\frac{U}{\sigma_X}+\frac{V}{\sigma_Y}\right)^2\right] =1+1+2\rho_{X,Y}=2(1+\rho_{X,Y}). \end{aligned}

The first inequality implies $\rho_{X,Y}\leq 1$; the second implies $\rho_{X,Y}\geq -1$. Thus $-1\leq\rho_{X,Y}\leq1$. The standardization matters: each centered variable is divided by its own standard deviation, making its second moment equal to one. The cross-product expectation is then precisely the correlation.

Theorem: If $X$ and $Y$ have finite, positive variances, then $-1\leq\rho_{X,Y}\leq1$. Equality at $+1$ holds if and only if $$ \frac{X-E[X]}{\sigma_X}=\frac{Y-E[Y]}{\sigma_Y}\quad\text{almost surely}, $$ and equality at $-1$ holds if and only if $$ \frac{X-E[X]}{\sigma_X}=-\frac{Y-E[Y]}{\sigma_Y}\quad\text{almost surely}. $$

To justify the equality conditions, a nonnegative random variable with expectation zero must equal zero almost surely. Therefore, if $\rho_{X,Y}=1$, the first squared difference in the derivation has expectation zero and must itself be zero almost surely. The standardized centered variables are then equal in every outcome except possibly on an event of probability zero. For $\rho_{X,Y}=-1$, apply the same reasoning to the squared sum. Equivalently, at either endpoint one centered variable is an exact constant multiple of the other almost surely: the multiple is positive at $+1$ and negative at $-1$. This is a much stronger condition than merely tending to move in the same or opposite directions.

Worked Correlation Calculations

Worked Example: Correlation in a Four-Scenario Loss Model

A hypothetical insurer considers two losses, $X$ and $Y$, measured in thousands of dollars. The joint model has four equally likely scenarios, each with probability $1/4$: $(X,Y)=(0,0),(0,2),(2,0),(4,2)$. Calculate the correlation and verify that it satisfies the bound.

Method. Compute the means, variances, and covariance from the joint outcomes, then standardize the covariance. This avoids treating a covariance as if it were already a correlation.

The means are

E[X]=\frac{0+0+2+4}{4}=\frac{6}{4}=1.5,\qquad E[Y]=\frac{0+2+0+2}{4}=\frac{4}{4}=1.

For $X$, $E[X^2]=(0+0+4+16)/4=5$, so $\operatorname{Var}(X)=5-(1.5)^2=2.75=11/4$. For $Y$, $E[Y^2]=(0+4+0+4)/4=2$, so $\operatorname{Var}(Y)=2-1^2=1$. The standard deviations are $\sigma_X=\sqrt{11}/2$ and $\sigma_Y=1$.

The product $XY$ takes values $0,0,0,8$, so $E[XY]=8/4=2$. Hence

\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=2-(1.5)(1)=0.5, \qquad \rho_{X,Y}=\frac{0.5}{(\sqrt{11}/2)(1)} =\frac{1}{\sqrt{11}}\approx0.3015.

The result is positive and lies between $-1$ and $1$, as required. The relationship is not an equality case: the four outcomes do not place the centered losses in an exact constant-multiple relationship. The displayed decimal is rounded to four places.

Worked Example: Recognizing Perfect Negative Correlation

A hypothetical portfolio has three equally likely loss scenarios. Its losses $X$, in thousands of dollars, are $1,2,3$; the corresponding losses $Y$ are $6,4,2$. Calculate the correlation and explain why the lower bound is attained.

Method. Calculate the centered values and check whether they satisfy the equality condition, while also verifying the correlation from covariance and standard deviations.

The means are $E[X]=(1+2+3)/3=2$ and $E[Y]=(6+4+2)/3=4$. Thus the centered values are $(-1,0,1)$ for $X$ and $(2,0,-2)$ for $Y$. The variances are

\operatorname{Var}(X)=\frac{1+0+1}{3}=\frac{2}{3}, \qquad \operatorname{Var}(Y)=\frac{4+0+4}{3}=\frac{8}{3}.

The products of the centered values are $-2,0,-2$, giving

\operatorname{Cov}(X,Y)=\frac{-2+0-2}{3}=-\frac{4}{3}, \qquad \sigma_X\sigma_Y =\sqrt{\frac{2}{3}\cdot\frac{8}{3}} =\sqrt{\frac{16}{9}}=\frac{4}{3}.

Therefore $\rho_{X,Y}=(-4/3)/(4/3)=-1$. The equality condition is visible directly: $Y-E[Y]=-2(X-E[X])$ in every scenario. The centered losses are exact negative multiples, not just losses with opposite movements on average. The calculation checks because the correlation equals the lower bound.

Worked Example: Checking Whether Proposed Risk Summaries Are Possible

Two hypothetical losses have standard deviations of $3$ and $4$ thousand dollars. A report gives their covariance as $10$ thousand-dollars squared. Check the implied correlation, then assess a separate proposed covariance of $13$ thousand-dollars squared for the same standard deviations.

Method. Apply the correlation bound to the covariance after dividing by the product of standard deviations. The covariance must be no greater in absolute value than that product.

For covariance $10$, the correlation would be

\rho=\frac{10}{(3)(4)}=\frac{10}{12}=\frac{5}{6}\approx0.8333.

This value is within the permitted range. For these standard deviations, any valid covariance $c$ must satisfy

-1\leq\frac{c}{12}\leq1, \qquad\text{so}\qquad -12\leq c\leq12.

A proposed covariance of $13$ would imply $\rho=13/12\approx1.0833$, which exceeds the upper bound. The reported summaries cannot all describe the same pair of random variables with finite, positive variances. This is a consistency check, not a way to adjust an impossible covariance into a valid one. The correlation from covariance $10$ is rounded to four decimal places.

A Quadratic Form of the Same Argument

The squared-combination proof also gives a useful way to inspect covariance directly. For any real number $t$, the squared difference between one centered loss and $t$ times the other is nonnegative. Taking expectations produces a quadratic that must remain nonnegative for every $t$.

E[(U-tV)^2] =\operatorname{Var}(X)-2t\operatorname{Cov}(X,Y)+t^2\operatorname{Var}(Y)\geq0 \quad\text{for every real }t.

When $\operatorname{Var}(Y)>0$, choose $t=\operatorname{Cov}(X,Y)/\operatorname{Var}(Y)$, the value that minimizes this quadratic. Substitution gives

\operatorname{Var}(X) -\frac{\operatorname{Cov}(X,Y)^2}{\operatorname{Var}(Y)} \geq0, \qquad \operatorname{Cov}(X,Y)^2 \leq\operatorname{Var}(X)\operatorname{Var}(Y).

Dividing by the positive product of standard deviations yields $|\rho_{X,Y}|\leq1$, the same bound. This version emphasizes that the squared covariance cannot exceed the product of the variances. Equality occurs when the minimizing squared difference is zero almost surely, again requiring an exact linear relationship between the centered variables.

Common Exam Trap and Risk Interpretation

A frequent calculation error is to divide covariance by the product of variances rather than the product of standard deviations. In the four-scenario example, that wrong calculation is $0.5/(2.75\cdot1)\approx0.1818$, not the correct $0.3015$. Correlation standardizes by $\sqrt{\operatorname{Var}(X)}\sqrt{\operatorname{Var}(Y)}$, so its units cancel. Dividing by the variances gives the wrong scale and generally retains units.

Another error is to interpret a correlation near $+1$ or $-1$ as proof of an exact linear relationship. Only the endpoint values force the corresponding equality case. Likewise, correlation at an endpoint is not defined if one variable is constant, because the required standard deviation is zero. In risk aggregation, the bound is a basic validation check for inputs and calculations; it does not by itself specify a joint loss model or determine tail behavior.

Key takeaway: With finite, positive variances, correlation is covariance divided by the product of standard deviations and must lie in $[-1,1]$. It reaches either endpoint exactly when the centered variables are proportional almost surely, with a positive multiplier at $+1$ and a negative multiplier at $-1$.

Check Your Understanding

For each problem, use the stated assumptions and check whether the result is consistent with the correlation bound.

Two losses have standard deviations $2$ and $5$, and covariance $-5$. Calculate their correlation. (Answer: $-0.5000$)
Two hypothetical losses are equally likely to take paired values $(1,2),(2,1),(3,2),(2,3)$. Calculate their means, variances, covariance, and correlation. (Answer: $E[X]=E[Y]=2$, both variances $0.5$, covariance $0$, and correlation $0$)
For losses with standard deviations $2.5$ and $1.2$, state the full allowable range of their covariance. (Answer: from $-3$ to $3$, inclusive)
Suppose $\operatorname{Var}(X)=4$, $\operatorname{Var}(Y)=9$, and $\operatorname{Cov}(X,Y)=6$. Find the correlation and state what the moment values imply about the centered variables. (Answer: $\rho_{X,Y}=1$; $Y-E[Y]=1.5(X-E[X])$ almost surely)
A proposed joint model has finite variances $\operatorname{Var}(X)=16$, $\operatorname{Var}(Y)=25$, and covariance $-21$. Is this set of summaries possible? If not, give the largest permissible absolute covariance. (Answer: No; the largest permissible absolute covariance is $20$)

What You'll Learn