Bivariate Normal Density and Correlation

From Marginals to a Joint Density

Two normal variables can each have a familiar bell-shaped marginal distribution while their joint behavior varies substantially. The correlation records how their standardized deviations tend to move together: positive correlation favors deviations with the same sign, while negative correlation favors deviations with opposite signs. To describe that joint behavior precisely, we need the bivariate density and its support.

Let $X$ and $Y$ be jointly normal, with means $\mu_1,\mu_2$, positive standard deviations $\sigma_1,\sigma_2$, and correlation $\rho$. Their covariance is $\rho\sigma_1\sigma_2$, so the covariance matrix is determined by these five parameters. The density formula applies when $-1<\rho<1$; the endpoint cases require a different description.

Definition: For $\mu_1,\mu_2\in\mathbb R$, $\sigma_1,\sigma_2>0$, and $-1<\rho<1$, the nonsingular bivariate normal distribution with means $\mu_1,\mu_2$, standard deviations $\sigma_1,\sigma_2$, and correlation $\rho$ has density $$ f(x,y)=\frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}} \exp\left\{-\frac{1}{2(1-\rho^2)} \left[ \left(\frac{x-\mu_1}{\sigma_1}\right)^2 -2\rho\left(\frac{x-\mu_1}{\sigma_1}\right)\left(\frac{y-\mu_2}{\sigma_2}\right) +\left(\frac{y-\mu_2}{\sigma_2}\right)^2 \right]\right\},\qquad (x,y)\in\mathbb R^2. $$ The corresponding covariance matrix is $$ \begin{pmatrix} \sigma_1^2 & \rho\sigma_1\sigma_2\\ \rho\sigma_1\sigma_2 & \sigma_2^2 \end{pmatrix}. $$

Why the Formula Has This Form

The density is the two-dimensional Gaussian density expressed in terms of its means and covariance matrix. To make its structure and normalization explicit, define the standardized coordinates $u=(x-\mu_1)/\sigma_1$ and $v=(y-\mu_2)/\sigma_2$. The correlation matrix and its inverse are

R=\begin{pmatrix}1&\rho\\\rho&1\end{pmatrix}, \qquad \det(R)=1-\rho^2, \qquad R^{-1}=\frac{1}{1-\rho^2} \begin{pmatrix}1&-\rho\\-\rho&1\end{pmatrix}.

For $-1<\rho<1$, the determinant is positive, and the quadratic form $(u,v)R^{-1}(u,v)^{\mathsf T}$ is exactly the expression in square brackets divided by $1-\rho^2$. The determinant supplies the factor $1/\sqrt{1-\rho^2}$; the change from standardized to original coordinates supplies $1/(\sigma_1\sigma_2)$.

Result: The displayed formula is a probability density: it is nonnegative and integrates to one over $\mathbb R^2$. Its covariance matrix has determinant $\sigma_1^2\sigma_2^2(1-\rho^2)$, which is positive precisely when $-1<\rho<1$.

Proof of normalization. Under the change of variables from $x,y$ to $u,v$, the area element satisfies $dx\,dy=\sigma_1\sigma_2\,du\,dv$. Thus it is enough to integrate the standardized density. Complete the square:

\frac{u^2-2\rho uv+v^2}{1-\rho^2} =u^2+\frac{(v-\rho u)^2}{1-\rho^2}.

For each fixed $u$, substitute $w=(v-\rho u)/\sqrt{1-\rho^2}$, so $dv=\sqrt{1-\rho^2}\,dw$. The integral of the standardized density is therefore

\frac{1}{2\pi\sqrt{1-\rho^2}} \int_{\mathbb R}e^{-u^2/2} \left[\int_{\mathbb R}e^{-(v-\rho u)^2/[2(1-\rho^2)]}\,dv\right]du = \frac{1}{2\pi} \left(\int_{\mathbb R}e^{-u^2/2}\,du\right) \left(\int_{\mathbb R}e^{-w^2/2}\,dw\right)=1.

The final equality uses the standard normal integral $\int_{\mathbb R}e^{-t^2/2}\,dt=\sqrt{2\pi}$. The integrand is nonnegative, so evaluating the iterated integral is justified by Tonelli’s theorem. This verifies normalization as well as the role of the determinant factor. $\square$

Reading Correlation in the Density

The standardized coordinates $u$ and $v$ measure deviations from the two means in units of their respective standard deviations. For a fixed pair of such deviations, the density depends on correlation through both the normalizing factor and the quadratic expression. When $\rho$ is positive, the cross term $-2\rho uv$ lowers that expression for same-sign deviations; when $\rho$ is negative, it lowers the expression for opposite-sign deviations. Since the density decreases exponentially as the quadratic expression increases, these comparisons describe how correlation changes relative density.

Changing $\rho$ while keeping the means and standard deviations fixed does not change either marginal distribution. It changes the covariance from $\rho=0$ upward or downward and changes the joint density. At $\rho=0$, the formula factors into the product of the two univariate normal densities. For a jointly normal pair, this means the variables are independent. Correlation is not a change in the units or marginal scales; it describes dependence in the joint distribution.

Worked Example: Evaluating the Density at a Point

Let $X$ and $Y$ have means zero, standard deviations one, and correlation $0.8$. Find the density at $(1,1)$.

Here $u=v=1$, and $1-\rho^2=1-0.64=0.36$. The quadratic expression in the exponent is

\frac{1-2(0.8)(1)(1)+1}{0.36} =\frac{0.4}{0.36} =\frac{10}{9}.

The exponent is therefore $-5/9$, and the coefficient is $1/(2\pi\sqrt{0.36})=1/(1.2\pi)\approx0.26526$. Consequently,

f(1,1)=\frac{1}{1.2\pi}e^{-5/9} \approx 0.26526(0.57375) \approx 0.1522.

This is a density value, not the probability that $(X,Y)$ equals $(1,1)$. A continuous bivariate distribution assigns probability zero to any single point.

Worked Example: Reversing the Correlation Sign

Keep both means at zero and both standard deviations at one, but compare correlations $0.8$ and $-0.8$. Evaluate the density at $(1,1)$ and at $(1,-1)$.

For either correlation, the coefficient is $1/(1.2\pi)\approx0.26526$. When $\rho=0.8$, at $(1,1)$ the exponent is $-5/9$, as in the previous calculation, giving $f(1,1)\approx0.1522$. At $(1,-1)$, the quadratic expression is

\frac{1+2(0.8)+1}{0.36}=\frac{3.6}{0.36}=10,

so $f(1,-1)=0.26526e^{-5}\approx0.00179$. With $\rho=-0.8$, these two results switch: at $(1,1)$ the exponent is $-5$, giving approximately $0.00179$, while at $(1,-1)$ it is $-5/9$, giving approximately $0.1522$.

The marginal distributions have not changed at all. The comparison shows how the sign of correlation favors same-sign versus opposite-sign standardized deviations.

Support and the Endpoint Cases

The support of a probability distribution is the smallest closed set that has probability one. For a density that is strictly positive throughout the plane, the support is the whole plane. A bivariate normal with $|\rho|<1$ has exactly this property. At $\rho=1$ or $\rho=-1$, the covariance matrix is singular and the distribution instead lies on a line.

Result: If $X,Y$ are jointly normal with positive standard deviations, then for $-1<\rho<1$ their support is $\mathbb R^2$. If $\rho=1$, they satisfy $Y=\mu_2+(\sigma_2/\sigma_1)(X-\mu_1)$ almost surely. If $\rho=-1$, they satisfy $Y=\mu_2-(\sigma_2/\sigma_1)(X-\mu_1)$ almost surely. In either endpoint case, the support is the corresponding line, and there is no density with respect to two-dimensional area.

Proof. For $|\rho|<1$, the density formula is continuous and strictly positive at every $(x,y)\in\mathbb R^2$. Every open ball therefore contains a smaller ball on which the density is bounded below by a positive number. Its probability is positive, so no proper closed subset of the plane can be the support; the support is $\mathbb R^2$.

For the endpoint cases, let $U=(X-\mu_1)/\sigma_1$ and $V=(Y-\mu_2)/\sigma_2$. Both have mean zero and variance one, and $\operatorname{Cov}(U,V)=\rho$. If $\rho=1$, then $E[(U-V)^2]=1+1-2\rho=0$. A nonnegative random variable with expectation zero is zero almost surely, so $U=V$ almost surely. If $\rho=-1$, then $E[(U+V)^2]=1+1+2\rho=0$, so $U=-V$ almost surely. Rewriting these identities gives the two lines stated above.

Because $X$ has a nondegenerate normal marginal, it has positive probability in every open interval. The affine relation at either endpoint therefore gives support equal to the entire indicated line. A line has two-dimensional Lebesgue measure zero, so a distribution assigning probability one to that line cannot have a density with respect to area. The nonsingular density formula cannot be applied by setting $\rho=\pm1$: its denominator vanishes, and the law changes dimension. $\square$

Worked Example: A Singular Pair at Correlation Minus One

Suppose $X$ and $Y$ are jointly normal with $\mu_1=2$, $\sigma_1=3$, $\mu_2=-1$, $\sigma_2=2$, and $\rho=-1$. Determine their support relation.

At correlation $-1$, their standardized deviations obey

\frac{Y-(-1)}{2}=-\frac{X-2}{3}\quad\text{almost surely}.

Solving for $Y$ gives

Y=-1-\frac{2}{3}(X-2) =\frac{1}{3}-\frac{2}{3}X\quad\text{almost surely}.

Thus the support is the line $y=1/3-(2/3)x$. The pair has perfectly negative correlation, but it has no bivariate density with respect to area. Its $X$ and $Y$ marginals are still nondegenerate normal distributions.

What Correlation Does—and Does Not—Change

With the means and standard deviations held fixed, $\rho$ changes the covariance and the relative density assigned to different pairs of standardized deviations. For every $-1<\rho<1$, however, the support remains all of $\mathbb R^2$: any open region has positive probability, even if its density is very small. The support changes to a line only at the endpoint correlations. As $\rho$ approaches an endpoint, the density becomes increasingly concentrated near the corresponding line, but the endpoint law itself is singular.

A frequent mistake is to infer that a high positive correlation rules out pairs with opposite-sign deviations, or that a high negative correlation rules out same-sign deviations. For $|\rho|<1$, the density is positive everywhere, so neither kind of pair is impossible. Correlation changes relative likelihood, not the support. Another mistake is to substitute $\rho=1$ into the density formula and treat the result as an ordinary area density. The vanishing determinant signals that the formula no longer describes a nonsingular distribution; the endpoint law must instead be specified on its line.

Key takeaway: The bivariate normal density is determined by two means, two positive standard deviations, and a correlation strictly between $-1$ and $1$. The sign and magnitude of correlation alter joint behavior without changing the marginals. Only at correlations $1$ and $-1$ does the support collapse from the plane to a line.

Check Your Understanding

Use the density formula and support results to answer the following questions.

For $\sigma_1,\sigma_2>0$, why must a bivariate normal density with respect to area have $-1<\rho<1$?
With both standardized deviations positive, which sign of correlation gives the larger density: $0.6$ or $-0.6$? Explain using the cross term.
If $\rho=0$, what does the bivariate density factor into, and what does this imply for a jointly normal pair?
For $\rho=1$, write the almost-sure relation between $X$ and $Y$ in terms of their means and standard deviations.
Why does positive density throughout the plane imply full-plane support even when some density values are extremely small?