How Transformations Change Covariance And Correlation

Why Transformations Matter

Risk variables are often recorded in different units, adjusted by fixed amounts, or converted into net amounts after a deterministic change. An actuary may need to express losses in thousands rather than dollars, subtract a fixed allowance, or reverse the direction of a variable to represent a benefit or offset. These changes affect covariance and correlation in related but distinct ways.

The central distinction is that covariance changes with scale, while correlation is unaffected by positive rescaling. Adding constants changes neither measure. A negative scale factor reverses the sign of both covariance and correlation, provided the other variable is not also reversed. The rules let us update a dependence calculation efficiently while keeping units and signs clear.

Definition: Let $X$ and $Y$ be random variables with finite second moments. For constants $a,b,c,d$, define $U=aX+c$ and $V=bY+d$. Then $$ \operatorname{Cov}(U,V)=ab\operatorname{Cov}(X,Y). $$ If $X,Y,U,V$ have positive variances, their Pearson correlations satisfy $$ \operatorname{Corr}(U,V)=\operatorname{sign}(ab)\operatorname{Corr}(X,Y), $$ where $\operatorname{sign}(ab)$ is $1$ for $ab>0$ and $-1$ for $ab<0$. If $a=0$ or $b=0$, the corresponding transformed variable is constant, so its correlation is undefined.

Deriving the Transformation Rules

The covariance of variables with finite second moments is $\operatorname{Cov}(X,Y)=E[(X-E[X])(Y-E[Y])]$. The constants disappear when we subtract the transformed means: $E[U]=aE[X]+c$ and $E[V]=bE[Y]+d$. Therefore,

\begin{aligned} \operatorname{Cov}(U,V) &=E[(U-E[U])(V-E[V])]\\ &=E[(aX+c-aE[X]-c)(bY+d-bE[Y]-d)]\\ &=E[ab(X-E[X])(Y-E[Y])]\\ &=ab\operatorname{Cov}(X,Y). \end{aligned}

This derivation shows why adding a constant has no effect: it shifts every possible value and the mean by the same amount, leaving each deviation from the mean unchanged. Multiplication by a constant scales each deviation, so the product of the two deviations is multiplied by $ab$.

For correlation, use $\operatorname{Corr}(X,Y)=\operatorname{Cov}(X,Y)/(\sigma_X\sigma_Y)$, where $\sigma_X=\sqrt{\operatorname{Var}(X)}$ and $\sigma_Y=\sqrt{\operatorname{Var}(Y)}$. The variance of a transformed variable is $\operatorname{Var}(aX+c)=a^2\operatorname{Var}(X)$, so its standard deviation is $|a|\sigma_X$. Consequently, when $a$ and $b$ are nonzero,

\operatorname{Corr}(U,V) =\frac{ab\operatorname{Cov}(X,Y)} {|a|\sigma_X\,|b|\sigma_Y} =\frac{ab}{|a||b|}\operatorname{Corr}(X,Y).

Thus the magnitude of correlation is unchanged, and its sign is reversed exactly when one transformation has a negative scale factor and the other has a positive one. If both scale factors are negative, the two sign reversals cancel. These statements require nonzero variances; correlation is not defined for a constant random variable.

Applying the Rules to Loss Variables

When using the rules, keep the transformation of each variable explicit. The constants $c$ and $d$ may change the means, but they do not appear in the covariance formula. The coefficients $a$ and $b$ determine the covariance multiplier and, through their signs, whether correlation is preserved or reversed.

Worked Example: A Loss and a Positive Linear Transformation

A hypothetical loss $X$ is $0$ with probability $0.8$ and $10$ with probability $0.2$. Let $Y=3X+2$. Find $\operatorname{Cov}(X,Y)$, then find the covariance and correlation of $U=4-X$ and $V=2Y-1$.

Method. First calculate the mean and variance of $X$, then use the transformation rules. This makes the sign of each scaling factor visible and avoids reversing the order in the covariance identity.

The mean is $E[X]=0(0.8)+10(0.2)=2$. The second moment is $E[X^2]=0^2(0.8)+10^2(0.2)=20$, so $\operatorname{Var}(X)=20-2^2=16$. Since $Y=3X+2$, its scale factor relative to $X$ is $3$, and

\operatorname{Cov}(X,Y)=3\operatorname{Var}(X)=3(16)=48.

For the second pair, $U=4-X=-X+4$ has scale factor $-1$, while $V=2Y-1$ has scale factor $2$ relative to $Y$. Therefore,

\operatorname{Cov}(U,V)=(-1)(2)\operatorname{Cov}(X,Y)=(-2)(48)=-96.

Because $Y=3X+2$ is a positive linear transformation of $X$, their correlation is $1$. The product of the scale factors for $U$ and $V$, relative to $X$ and $Y$, is negative, so $\operatorname{Corr}(U,V)=-1$.

Check. $U$ takes values $4$ and $-6$, with mean $2$ and standard deviation $4$. $V$ takes values $3$ and $63$, with mean $15$ and standard deviation $24$. Their covariance is $0.8(4-2)(3-15)+0.2(-6-2)(63-15)=-19.2-76.8=-96$. Dividing by $4(24)=96$ gives correlation $-1$, as required.

Worked Example: Two Variables with Imperfect Correlation

A hypothetical joint loss model gives each of the following four outcomes probability $0.25$: $(X,Y)=(1,2),(1,6),(5,2),(5,10)$. For $U=2X+7$ and $V=-3Y+4$, calculate the original covariance and correlation, then the transformed covariance and correlation.

Method. Calculate the means, second moments, and cross-moment from the four equally likely outcomes. Then apply the scale factors $2$ and $-3$. This separates the effect of translation from the effect of scaling.

The mean of $X$ is $(1+1+5+5)/4=3$. Its second moment is $(1+1+25+25)/4=13$, so $\operatorname{Var}(X)=13-3^2=4$. The mean of $Y$ is $(2+6+2+10)/4=5$. Its second moment is $(4+36+4+100)/4=36$, so $\operatorname{Var}(Y)=36-5^2=11$.

The cross-moment is $E[XY]=(1(2)+1(6)+5(2)+5(10))/4=68/4=17$. Thus $\operatorname{Cov}(X,Y)=17-(3)(5)=2$, and

\operatorname{Corr}(X,Y)=\frac{2}{\sqrt{4}\sqrt{11}} =\frac{1}{\sqrt{11}}\approx 0.3015.

For $U=2X+7$ and $V=-3Y+4$, the product of the scale factors is $-6$. Hence $\operatorname{Cov}(U,V)=(-6)(2)=-12$. Their variances are $\operatorname{Var}(U)=2^2(4)=16$ and $\operatorname{Var}(V)=(-3)^2(11)=99$. Therefore,

\operatorname{Corr}(U,V)=\frac{-12}{\sqrt{16}\sqrt{99}} \approx -0.3015.

Check. The transformation retains the magnitude $0.3015$ and reverses the sign because exactly one scale factor is negative. The offsets $7$ and $4$ have no effect on either covariance or correlation.

Worked Example: Adding Baselines and Reversing a Loss Measure

A hypothetical three-state model has probabilities $0.5,0.3,0.2$, with corresponding loss pairs $(X,Y)=(0,20),(100,0),(300,200)$, measured in loss units. Define $U=1000+X$ and $V=500-2Y$. Find the covariance and correlation of $X,Y$, then those of $U,V$.

Method. Compute the original moments directly and use the transformation rules for the new pair. The negative coefficient on $Y$ represents a reversal of direction; the positive baseline of $500$ does not affect dependence.

The means are $E[X]=0(0.5)+100(0.3)+300(0.2)=90$ and $E[Y]=20(0.5)+0(0.3)+200(0.2)=50$. The cross-moment is $E[XY]=0(0.5)+0(0.3)+60000(0.2)=12000$, so $\operatorname{Cov}(X,Y)=12000-(90)(50)=7500$.

For the variances, $E[X^2]=10000(0.3)+90000(0.2)=21000$, giving $\operatorname{Var}(X)=21000-90^2=12900$. Also, $E[Y^2]=400(0.5)+40000(0.2)=8200$, giving $\operatorname{Var}(Y)=8200-50^2=5700$. Therefore,

\operatorname{Corr}(X,Y) =\frac{7500}{\sqrt{12900}\sqrt{5700}} =\frac{7500}{\sqrt{73530000}} \approx 0.8746.

The transformation coefficients are $a=1$ and $b=-2$. Thus $\operatorname{Cov}(U,V)=(1)(-2)(7500)=-15000$, while $\operatorname{Var}(U)=12900$ and $\operatorname{Var}(V)=(-2)^2(5700)=22800$. The transformed correlation is

\operatorname{Corr}(U,V) =\frac{-15000}{\sqrt{12900}\sqrt{22800}} \approx -0.8746.

Check. Since $\sqrt{22800}=2\sqrt{5700}$, the transformed denominator is twice the original denominator, while the transformed covariance is negative and twice the original covariance in magnitude. The correlation therefore has the same magnitude and opposite sign.

Common Mistake: Reversing the Covariance Difference

A frequent error is to calculate $E[X]E[Y]-E[XY]$ instead of $E[XY]-E[X]E[Y]$. Those expressions have opposite signs. The first worked example provides a direct check: $E[X]=2$, $Y=3X+2$, and $E[Y]=3(2)+2=8$. Also, $E[XY]=E[3X^2+2X]=3(20)+2(2)=64$. Hence

\operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]=64-(2)(8)=48.

The incorrect subtraction gives $(2)(8)-64=-48$. It conflicts with the fact that $Y$ increases whenever $X$ increases, and with the transformation rule $ \operatorname{Cov}(X,3X+2)=3\operatorname{Var}(X)=48$. A useful check is to apply both the defining covariance formula and the transformation rule when the arithmetic is short.

Do not multiply covariance by a translation. The constants $c$ and $d$ do not appear in $ab\operatorname{Cov}(X,Y)$.
Do not use signed factors in a variance. Variance uses $a^2$; covariance uses the product $ab$.
Do not assume correlation keeps its sign. Exactly one negative scale factor reverses the sign.
Do not report a correlation for a constant variable. If a scale factor is zero, the transformed standard deviation is zero and the correlation denominator vanishes.

Key takeaway: For $U=aX+c$ and $V=bY+d$, translations leave covariance and correlation unchanged; covariance is multiplied by $ab$; and, when both transformed variables have positive variance, correlation keeps its magnitude and changes sign exactly when $ab<0$.

Check Your Understanding

For each problem, identify the scale factors before applying the transformation rules.

Suppose $\operatorname{Var}(X)=9$ and $Y=-2X+5$. Find $\operatorname{Cov}(X,Y)$. (Answer: $-18$)
Suppose $\operatorname{Corr}(X,Y)=0.25$, $U=3X+8$, and $V=-4Y+1$. Find $\operatorname{Corr}(U,V)$. (Answer: $-0.25$)
Suppose $\operatorname{Cov}(X,Y)=-6$, $U=4X+9$, and $V=Y-7$. Find $\operatorname{Cov}(U,V)$. (Answer: $-24$)
Suppose $\operatorname{Cov}(X,Y)=5$, $\operatorname{Var}(X)=4$, and $\operatorname{Var}(Y)=25$. For $U=2X-1$ and $V=-Y+8$, find the transformed covariance and correlation. (Answer: covariance $-10$; correlation $-0.5000$)
Suppose $X,Y$ have finite variances and $\operatorname{Cov}(X,Y)=12$. Set $U=7$ and $V=-2Y+3$. State the covariance and whether the correlation of $U,V$ is defined. (Answer: covariance $0$; correlation is undefined because $U$ is constant)

What You'll Learn