Standardizing Correlated Normal Variables

Why Standardize a Correlated Pair?

A bivariate normal model describes two jointly distributed measurements, each with its own center and scale and with a specified dependence between them. A threshold such as “the first measurement exceeds 150 while the second is at most 65” is expressed in the original units. Standardizing converts both thresholds to standard-deviation units, making the event easier to describe and compare across scales.

The essential point is that standardizing the two variables separately does not remove their dependence. The resulting variables each have mean 0 and variance 1, but they retain the original correlation. This tutorial focuses on that conversion and on expressing events accurately; it does not treat a joint probability as a product of marginal probabilities.

Definition: Suppose $(X,Y)$ is a bivariate normal pair with $E[X]=\mu_X$, $E[Y]=\mu_Y$, positive standard deviations $\sigma_X$ and $\sigma_Y$, and correlation $\rho$. The standardized variables are $$ Z_X=\frac{X-\mu_X}{\sigma_X}, \qquad Z_Y=\frac{Y-\mu_Y}{\sigma_Y}. $$ Each has a standard normal marginal distribution, and their correlation is $\rho$. Thus $(Z_X,Z_Y)$ is a standard bivariate normal pair with correlation $\rho$.

Here $\sigma_X=\sqrt{\operatorname{Var}(X)}$ and $\sigma_Y=\sqrt{\operatorname{Var}(Y)}$. The denominator is a standard deviation, not a variance. Standardization subtracts the variable’s own mean and divides by its own standard deviation; using the same denominator for both variables is generally incorrect.

Why the Correlation Is Preserved

The preservation of correlation follows directly from the covariance definition. Subtracting constants does not change covariance, and dividing by positive constants scales it by their product. Therefore,

\begin{aligned} \operatorname{Cov}(Z_X,Z_Y) &=\operatorname{Cov}\left(\frac{X-\mu_X}{\sigma_X}, \frac{Y-\mu_Y}{\sigma_Y}\right)\\ &=\frac{\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y}\\ &=\rho. \end{aligned}

Because each standardized variable has variance 1, its standard deviation is 1. Consequently, its covariance with the other standardized variable is also its correlation. This proves that the standardized pair has correlation $\rho$, not 0 in general. The argument uses positive standard deviations; if a variance were zero, that component could not be standardized by this formula.

The bivariate normal model also ensures that the standardized pair remains bivariate normal. One way to see this is to substitute $x=\mu_X+\sigma_Xz_X$ and $y=\mu_Y+\sigma_Yz_Y$ into the bivariate normal density. For $|\rho|<1$, the original density is

f_{X,Y}(x,y)= \frac{1}{2\pi\sigma_X\sigma_Y\sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)} \left[ \left(\frac{x-\mu_X}{\sigma_X}\right)^2 -2\rho\left(\frac{x-\mu_X}{\sigma_X}\right) \left(\frac{y-\mu_Y}{\sigma_Y}\right) +\left(\frac{y-\mu_Y}{\sigma_Y}\right)^2 \right]\right).

Under this substitution, the area element changes by $dx\,dy=\sigma_X\sigma_Y\,dz_X\,dz_Y$, which cancels the two standard-deviation factors in the density. The resulting density is

f_{Z_X,Z_Y}(z_X,z_Y)= \frac{1}{2\pi\sqrt{1-\rho^2}} \exp\left(-\frac{z_X^2-2\rho z_Xz_Y+z_Y^2}{2(1-\rho^2)}\right).

The standardized density still contains $\rho$. That term records the dependence that remains after the locations and scales have been removed. The density formula assumes $|\rho|<1$; at the boundary values $\rho=1$ or $\rho=-1$, the pair is degenerate and has no two-dimensional density. The standardization formulas and covariance calculation still make sense when both variances are positive.

Converting Thresholds and Events

For a real threshold $a$, the conversion for $X$ is

X\le a \quad\Longleftrightarrow\quad Z_X\le \frac{a-\mu_X}{\sigma_X}, \qquad X>a \quad\Longleftrightarrow\quad Z_X>\frac{a-\mu_X}{\sigma_X}.

Since $\sigma_X>0$, dividing by it does not reverse the inequality. The same rule applies to $Y$. An interval condition becomes an interval with both endpoints standardized using that variable’s own mean and standard deviation. For example, $a<X\le b$ becomes $(a-\mu_X)/\sigma_X<Z_X\le(b-\mu_X)/\sigma_X$.

When both conditions are imposed at once, convert each one separately and retain the same pair $(Z_X,Z_Y)$. If the original correlation is nonzero, the standardized variables are not independent. In particular, a joint event such as $Z_X\le c$ and $Z_Y\le d$ cannot generally be evaluated by multiplying the two marginal probabilities.

Worked Example: Two Different Threshold Directions

Method. Standardize each threshold with its own marginal mean and standard deviation, then translate the two event conditions separately. This avoids using a variance in place of a standard deviation.

A hypothetical portfolio model has jointly normal annual claim measures $X$ and $Y$, with $E[X]=120$, $\operatorname{Var}(X)=400$, $E[Y]=80$, $\operatorname{Var}(Y)=225$, and $\operatorname{Corr}(X,Y)=0.40$. Find the standardized form of the event $X>150,\;Y\le65$, and state the correlation after standardization.

First take square roots of the variances:

\sigma_X=\sqrt{400}=20, \qquad \sigma_Y=\sqrt{225}=15.

For the $X$ threshold, the standardized value is

\frac{150-120}{20}=\frac{30}{20}=1.5.

Thus $X>150$ is equivalent to $Z_X>1.5$. For the $Y$ threshold,

\frac{65-80}{15}=\frac{-15}{15}=-1.

Thus $Y\le65$ is equivalent to $Z_Y\le-1$. The full event is therefore $Z_X>1.5,\;Z_Y\le-1$. Its standardized pair has correlation $0.40$, not 0. The result checks out: both cutoffs were centered on their respective means, scaled by their respective standard deviations, and the inequality directions stayed the same.

Worked Example: A Lower Tail Paired With an Upper Tail

Method. Convert each condition independently and keep the correlation attached to the standardized pair. This avoids treating two standardized variables as independent merely because both have standard normal marginals.

In a hypothetical testing process, jointly normal readings $X$ and $Y$ have means 50 and 70, standard deviations 8 and 12, and correlation $-0.60$. Express the event $X\le42,\;Y>88$ in standardized units. Also find the covariance before and after standardization.

The standardized variables are $Z_X=(X-50)/8$ and $Z_Y=(Y-70)/12$. The $X$ cutoff becomes

\frac{42-50}{8}=\frac{-8}{8}=-1,

so $X\le42$ becomes $Z_X\le-1$. The $Y$ cutoff becomes

\frac{88-70}{12}=\frac{18}{12}=1.5,

so $Y>88$ becomes $Z_Y>1.5$. The event is $Z_X\le-1,\;Z_Y>1.5$.

Before standardization, covariance is correlation times the product of standard deviations:

\operatorname{Cov}(X,Y)=(-0.60)(8)(12)=-57.6.

After standardization, the covariance is

\operatorname{Cov}(Z_X,Z_Y)=\frac{-57.6}{(8)(12)} =\frac{-57.6}{96}=-0.60.

Since both standardized variances are 1, the standardized correlation is also $-0.60$. The signs and magnitudes agree with the preservation result, providing a check on the calculation.

Worked Example: Translating Standardized Cutoffs Back to Original Units

Method. Rearrange the standardization equation to recover each raw cutoff, then apply the event directions. This prevents confusing a standardized value with a value in the original measurement units.

A hypothetical pair of jointly normal operational measurements has means $\mu_X=200$ and $\mu_Y=500$, standard deviations $\sigma_X=25$ and $\sigma_Y=40$, and correlation $0.25$. The standardized event is $-0.8<Z_X\le1.2,\;Z_Y>1.25$. Express it in terms of $X$ and $Y$.

From $Z_X=(X-200)/25$, the lower endpoint $-0.8$ corresponds to

$$ X=200+25(-0.8)=200-20=180. $$

The upper endpoint $1.2$ corresponds to

$$ X=200+25(1.2)=200+30=230. $$

Therefore, $-0.8<Z_X\le1.2$ is equivalent to $180<X\le230$. For $Y$, the standardized cutoff $1.25$ gives

$$ Y=500+40(1.25)=500+50=550. $$

The full raw event is $180<X\le230,\;Y>550$. The standardized correlation is still $0.25$. Checking the endpoints by substitution gives $(180-200)/25=-0.8$ and $(230-200)/25=1.2$, as required.

Common Mistake: Using Variance as the Scale

A frequent error is to divide a threshold difference by the variance rather than the standard deviation. In the first example, the incorrect calculation for the $X$ cutoff would be

\frac{150-120}{400}=0.075.

That is not a standard score: dividing a difference measured in the units of $X$ by a quantity measured in squared units does not produce a dimensionless number. The correct calculation uses $\sqrt{400}=20$, giving $(150-120)/20=1.5$.

A second trap is to multiply marginal probabilities for a joint event. Standardization gives each component a standard normal marginal distribution; it does not make the components independent. Unless independence has been established—for a bivariate normal pair with nonzero marginal variances, zero correlation is the relevant condition—the joint event must still be treated as an event involving the correlated pair.

Key takeaway: Standardize each variable by subtracting its own mean and dividing by its own positive standard deviation. Translate each threshold with that variable’s parameters, preserve the inequality direction, and carry the original correlation into the standardized pair.

Check Your Understanding

For each problem, convert the event to standardized units or recover the raw event as requested. Retain the correlation of the pair.

A hypothetical jointly normal pair has means $\mu_X=30,\mu_Y=100$, standard deviations $\sigma_X=5,\sigma_Y=20$, and correlation $0.30$. Express $X>40,\;Y\le80$ in standardized units. (Answer: $Z_X>2,\;Z_Y\le-1$, with correlation $0.30$.)
For a hypothetical bivariate normal pair, $\operatorname{Var}(X)=144$, $\operatorname{Var}(Y)=49$, $\operatorname{Cov}(X,Y)=-42$, $E[X]=10$, and $E[Y]=20$. Find the standard deviations, correlation, and standardized form of $X\le4,\;Y>27$. (Answer: $\sigma_X=12,\;\sigma_Y=7,\;\rho=-0.50;\;Z_X\le-0.5,\;Z_Y>1$.)
A hypothetical jointly normal pair has means 75 and 40, standard deviations 10 and 8, and correlation $-0.20$. Convert $65<X\le95,\;Y<32$ to standardized units. (Answer: $-1<Z_X\le2,\;Z_Y<-1$, with correlation $-0.20$.)
A hypothetical pair has means $\mu_X=300,\mu_Y=120$, standard deviations $\sigma_X=30,\sigma_Y=15$, and correlation $0.45$. Convert $-1.5\le Z_X<0.5,\;Z_Y\ge2$ to raw units. (Answer: $255\le X<315,\;Y\ge150$, with correlation $0.45$.)
A hypothetical jointly normal pair has variances 256 and 100 and covariance 48. Its means are 500 and 200. Express $X<476,\;Y\ge220$ in standardized units, and state why the standardized variables cannot be treated as independent. (Answer: $\sigma_X=16,\;\sigma_Y=10,\;\rho=48/(16\cdot10)=0.30;\;Z_X<-1.5,\;Z_Y\ge2$. They are not independent because their bivariate normal correlation is nonzero.)

What You'll Learn