Eccentricity of bivariate normal level sets

Suppose you have a bivariate normal distribution with correlation where

Then the level sets of the density function are ellipses, and the eccentricity e of the ellipses is related to the correlation by

Plots

For example, suppose = 0.8.

Here's a plot of the density function f(x, y).

And here is a plot of the contours, the level sets obtained by slicing the plot of the density horizontally.

Eccentricity and aspect ratio

The equation above says that since = 0.8 the eccentricity e should be (1.6/1.8) = 0.9428.

As I've written about several times, eccentricity is a little hard to interpret. It's a number between 0 and 1, with 0 being a circle and 1 being a degenerate ellipse collapsed to a line segment. Since 0.9428 is closer to 1 than to 0, you might think that an ellipse with such a large value of e is very far from circular. But this is not the case.

Eccentric literally means off-center. Eccentricity measures how far off-center the foci of an ellipse are. This is related to how round or thin an ellipse is, but only indirectly. It is true that such a large value of e means that the foci are close to the ends of the ellipse. In fact, the foci are located at 94.28% of the distance from the center to the end of the semimajor axes. But the aspect ratio is more moderate.

It's easier to imagine the shape of an ellipse from the aspect ratio than from the eccentricity. Aspect ratio r is related to eccentricity e by

Using the expression above for eccentricity e as a function of correlation we have

This says our aspect ratio should be 1/3. So although our ellipse is highly eccentric-the foci are near the ends-it isn't extraordinarily thin. It's three times longer than it is wide, which matches the contour plot above.

Related posts

• Eccentricity, Flattening, and Aspect Ratio
• Example of a highly elliptical orbit
• Length of an elliptical arc

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