Valid Correlations and Degenerate Bivariate Normals

Why the Endpoints Matter

In a bivariate normal model, the correlation describes the direction and strength of linear association between two variables. Its value cannot be chosen independently of the variances and covariance: some proposed parameter combinations are mathematically impossible. There is also an important distinction between a valid correlation and a nondegenerate bivariate normal model. A correlation of $1$ or $-1$ is possible, but it forces an exact linear relationship between the variables.

This distinction matters when checking a model, interpreting a covariance matrix, or deciding whether a two-dimensional density can describe the pair. The discussion below assumes both variables have positive, finite standard deviations. If either variable has variance zero, its correlation with the other variable is undefined, so the endpoint results do not apply.

Definition: For random variables $X$ and $Y$ with positive finite standard deviations $\sigma_X$ and $\sigma_Y$, their correlation is $$ \rho_{X,Y}=\frac{\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y}. $$ A bivariate normal pair is called nondegenerate when its covariance matrix is positive definite. In that case, its probability distribution is not confined to a line in the plane.

The Feasible Correlation Range

Every valid correlation must satisfy $-1\leq \rho_{X,Y}\leq 1$. For a nondegenerate bivariate normal pair, both inequalities are strict. The covariance matrix makes this condition visible:

\Sigma= \begin{pmatrix} \sigma_X^2 & \rho_{X,Y}\sigma_X\sigma_Y\\ \rho_{X,Y}\sigma_X\sigma_Y & \sigma_Y^2 \end{pmatrix}.

For positive $\sigma_X$ and $\sigma_Y$, positive definiteness requires the determinant to be positive. Its value is

\det(\Sigma) =\sigma_X^2\sigma_Y^2-(\rho_{X,Y}\sigma_X\sigma_Y)^2 =\sigma_X^2\sigma_Y^2(1-\rho_{X,Y}^2).

Since $\sigma_X^2\sigma_Y^2>0$, the determinant is positive exactly when $1-\rho_{X,Y}^2>0$, or equivalently $-1<\rho_{X,Y}<1$. If the correlation equals either endpoint, the determinant is zero and the covariance matrix is singular. A singular covariance matrix describes a bivariate normal pair confined to a line, not a nondegenerate pair.

Theorem: If $X$ and $Y$ have positive variances, any valid correlation satisfies $-1\leq\rho_{X,Y}\leq1$. For a nondegenerate bivariate normal pair, $-1<\rho_{X,Y}<1$. At $\rho_{X,Y}=1$ or $\rho_{X,Y}=-1$, the pair is supported on a straight line.

Why the Bound Holds and What Happens at Equality

The bound can be justified directly from variance. Standardize the variables by setting $U=(X-E[X])/\sigma_X$ and $V=(Y-E[Y])/\sigma_Y$. Then $E[U]=E[V]=0$, $\operatorname{Var}(U)=\operatorname{Var}(V)=1$, and $\operatorname{Cov}(U,V)=\rho_{X,Y}$. A variance cannot be negative, so

0\leq \operatorname{Var}(U-V)=2(1-\rho_{X,Y}), \qquad 0\leq \operatorname{Var}(U+V)=2(1+\rho_{X,Y}).

The first inequality gives $\rho_{X,Y}\leq1$, and the second gives $\rho_{X,Y}\geq-1$. This proves the correlation bound without assuming that the pair is bivariate normal.

At an endpoint, one of these variances is zero. A random variable with variance zero equals its mean with probability one. Consequently, if $\rho_{X,Y}=1$, then $U=V$ with probability one; if $\rho_{X,Y}=-1$, then $U=-V$ with probability one. Rewriting those standardized equalities gives the exact relationships

\rho_{X,Y}=1 \quad\Longrightarrow\quad Y=E[Y]+\frac{\sigma_Y}{\sigma_X}\bigl(X-E[X]\bigr) \quad\text{with probability one},

\rho_{X,Y}=-1 \quad\Longrightarrow\quad Y=E[Y]-\frac{\sigma_Y}{\sigma_X}\bigl(X-E[X]\bigr) \quad\text{with probability one}.

Thus the pair can vary, but only along one line: the line has positive slope at correlation $1$ and negative slope at correlation $-1$. In a bivariate normal model, the coordinates may each have positive variance while the pair still has no spread away from that line. The model is then degenerate and cannot have a nondegenerate two-dimensional density.

Worked Example: Testing a Proposed Covariance

A proposed bivariate normal model has $E[X]=12$, $E[Y]=20$, $\operatorname{Var}(X)=4$, $\operatorname{Var}(Y)=9$, and $\operatorname{Cov}(X,Y)=7$. Determine whether the parameters are feasible. Then find the largest possible absolute covariance for these variances and check whether a covariance of $5$ would give a nondegenerate model.

Method. Convert covariance to correlation using the product of the standard deviations, then use the correlation range and covariance-matrix determinant to classify the model.

The standard deviations are $\sigma_X=\sqrt{4}=2$ and $\sigma_Y=\sqrt{9}=3$. The proposed correlation is

\rho_{X,Y}=\frac{7}{(2)(3)}=\frac{7}{6}=1.1667\ldots

This is greater than $1$, so the proposal is impossible. Equivalently, its covariance matrix would have determinant

$$ (4)(9)-7^2=36-49=-13, $$

which is negative and therefore cannot be a covariance matrix for this model. For fixed standard deviations, a valid covariance must satisfy $|\operatorname{Cov}(X,Y)|\leq\sigma_X\sigma_Y=(2)(3)=6$. Thus the largest possible absolute covariance is $6$, attained only at an endpoint correlation.

If the covariance is changed to $5$, then

\rho_{X,Y}=\frac{5}{6}, \qquad \det(\Sigma)=(4)(9)-5^2=36-25=11>0.

The revised parameters are feasible and nondegenerate because the correlation is strictly between $-1$ and $1$ and the determinant is positive. The means do not affect this feasibility check. This method avoids treating covariance itself as if it had to lie between $-1$ and $1$; it is the correlation that has those bounds.

Endpoint Models and Their Supporting Lines

The line equation at an endpoint is not merely an approximate trend. It holds with probability one. For example, if $E[X]=10$, $E[Y]=20$, $\sigma_X=3$, $\sigma_Y=4$, and $\rho_{X,Y}=-1$, the relationship is

Y=20-\frac{4}{3}(X-10) \quad\text{with probability one}.

Every possible pair of values lies on this line. Knowing $X$ determines $Y$ exactly, and knowing $Y$ determines $X$ exactly. At correlation $1$, the same parameters would instead give $Y=20+\frac{4}{3}(X-10)$.

A correlation of $1$ does not mean the two variables have the same units, means, or standard deviations. It means their standardized deviations from their means are identical. A correlation of $-1$ means those standardized deviations have opposite signs and equal magnitudes. The scale factor $\sigma_Y/\sigma_X$ in the line equation accounts for different units and spreads.

Worked Example: Classifying a Covariance Matrix

A bivariate normal pair has means $E[X]=10$ and $E[Y]=20$, with covariance matrix $$ \Sigma= \begin{pmatrix} 9 & 9\\ 9 & 16 \end{pmatrix}. $$ Find the correlation, classify the pair as nondegenerate or degenerate, and write its supporting line if it is degenerate.

Method. Read the variances from the diagonal, calculate the standard deviations and correlation, and verify the determinant.

The variances are $9$ and $16$, so $\sigma_X=3$ and $\sigma_Y=4$. The covariance is $9$, giving

\rho_{X,Y}=\frac{9}{(3)(4)}=\frac{3}{4}=0.75.

The determinant is

\det(\Sigma)=(9)(16)-9^2=144-81=63>0.

Therefore, the model is nondegenerate; there is no single line on which all its probability is concentrated. As a consistency check, $0.75$ is strictly between $-1$ and $1$. If instead the off-diagonal covariance were $12$, the correlation would be $12/((3)(4))=1$, and the determinant would be $144-144=0$. The endpoint relationship would then be

Y=20+\frac{4}{3}(X-10) \quad\text{with probability one}.

The covariance $12$ is the largest possible positive covariance for these standard deviations. This comparison shows why checking the determinant and checking the correlation give the same classification.

Worked Example: A Shared Normal Source at an Endpoint

Let $Z\sim N(0,1)$, and define $X=50+4Z$ and $Y=80-6Z$. Find both means, both variances, the covariance and correlation, and then calculate $P(X>54,\;Y<74)$.

Method. Use the mean and variance of $Z$, and express both event conditions in terms of the same underlying variable. This makes the endpoint dependence explicit.

Because $E[Z]=0$, the means are $E[X]=50$ and $E[Y]=80$. Since $\operatorname{Var}(Z)=1$,

\operatorname{Var}(X)=4^2(1)=16, \qquad \operatorname{Var}(Y)=(-6)^2(1)=36.

The covariance is

\operatorname{Cov}(X,Y)=(4)(-6)\operatorname{Var}(Z)=-24.

The standard deviations are $4$ and $6$, so

\rho_{X,Y}=\frac{-24}{(4)(6)}=-1.

The exact line relationship is $Y=80-6Z$ and $X-50=4Z$, hence

Y=80-\frac{6}{4}(X-50) =80-\frac{3}{2}(X-50).

For the requested event, $X>54$ means $50+4Z>54$, so $Z>1$. Also, $Y<74$ means $80-6Z<74$, which again gives $Z>1$. Their intersection is therefore just $\{Z>1\}$, not a product of two independent events. Thus

P(X>54,\;Y<74)=P(Z>1)\approx 0.1587.

The probability is rounded to four decimal places. Both inequalities describe the same side of the supporting line, which is the key endpoint feature. Treating the two events as independent would give the wrong calculation.

Common Exam Trap: Confusing Covariance With Correlation

A frequent error is to compare a covariance directly with $1$, or to divide covariance by the product of the variances instead of the product of the standard deviations. For example, with variances $4$ and $9$ and covariance $5$, the incorrect calculation

\rho_{\text{wrong}}=\frac{5}{(4)(9)}=\frac{5}{36}

does not use the definition of correlation. The correct calculation is

\rho_{X,Y}=\frac{5}{\sqrt{4}\sqrt{9}}=\frac{5}{(2)(3)}=\frac{5}{6}.

The two numbers differ substantially. Always take the square roots of the variances before forming the denominator. Then check $-1\leq\rho_{X,Y}\leq1$; for a nondegenerate bivariate normal pair, check the stricter condition $-1<\rho_{X,Y}<1$.

A second trap is to call an endpoint correlation “impossible.” It is impossible for a nondegenerate bivariate normal model, but it is valid for a degenerate bivariate normal pair. The correct conclusion at an endpoint is a deterministic line relationship, not a failed model.

Key takeaway: With positive marginal variances, correlation must lie between $-1$ and $1$, inclusive. Strict inequality characterizes the nondegenerate bivariate normal case. Equality forces the standardized variables to be identical or opposites, placing the pair on a straight line.

Check Your Understanding

For each problem, distinguish feasibility from nondegeneracy and use the line relationship when an endpoint occurs.

A proposed pair has standard deviations $5$ and $4$, and covariance $18$. Find the correlation and determine whether the pair can be nondegenerate. (Answer: $\rho=0.9000$; feasible and nondegenerate.)
A pair has means $12$ and $30$, standard deviations $3$ and $5$, and covariance $-15$. State the correlation and its line relationship. (Answer: $\rho=-1$; $Y=30-\frac{5}{3}(X-12)$ with probability one.)
A model specifies standard deviations $2$ and $5$ and correlation $0.6$. Find its covariance and determine whether its covariance matrix is singular. (Answer: $\operatorname{Cov}(X,Y)=6$; it is nonsingular because $1-0.6^2=0.64>0$.)
Let $Z\sim N(0,1)$, $X=7+2Z$, and $Y=10-3Z$. Find the correlation and $P(X>9,\;Y<7)$. (Answer: $\rho=-1$; $P(X>9,\;Y<7)=P(Z>1)\approx0.1587$.)
A covariance matrix has diagonal entries $16$ and $25$, and off-diagonal entries $-20$. Determine whether it is valid and, if so, classify it and give the standardized relationship. (Answer: valid and degenerate; $\rho=-1$ and $(Y-E[Y])/5=-(X-E[X])/4$ with probability one.)

What You'll Learn