1

Bivariate Normal Distribution

and Error Ellipses

Bivariate Normal Distribution

• Joint distribution of two random

variables

• Very useful when dealing with

planimetric (x,y) positions in surveying

• Density function is a bell-shaped surface

centered at x = µx and y = µy

Bivariate Density Function Bivariate Density Function

• The joint density function of two random variables (X

and Y) which have a bivariate normal distribution is:

where:

µx and σx = mean and standard deviation of X

µy and σy = mean and standard deviation of Y

ρ = correlation coefficient of X and Y

22

22

1 1( , ) exp 2

2(1 )2 1

− − − −− = − + − −

y yx x

x x y yx y

y yx xf x y

µ µµ µρ

ρ σ σ σ σπσ σ ρ

= xy

x y

σρ

σ σ

Marginal Density Functions

• Density functions for X and Y

• Components of the bivariate normal distribution at the

X and Y axes

• The same as the usual density functions for individual

normally distributed random variables

2

2

1 1( ) exp

22

1 1( ) exp

22

− = −

− = −

x

xx

y

yy

xf x

yf y

µσσ π

µ

σσ π

Cutting Ellipse/Ellipse of Intersection

• When a plane parallel to the x,y coordinate plane cuts

the bivariate density surface at a height K, an ellipse is

formed

• The equation of this ellipse is: (obtained by making

f(x,y) = K)

where:

22

2 22 (1 ) − − − −

− + = −

y yx x

x x y y

y yx xc

µ µµ µρ ρ

σ σ σ σ

12 2 2 2 2 2ln 4 (1 ) a constant

− = − = x y

c Kπ σ σ ρ

2

Example

The parameters of a bivariate normal distribution

are µx = 4, µy = 5, σx = 1, σy = 0.5, and ρxy = 0.5.

A plane intersects the density function at K = 0.1

above the x,y coordinate plane. Evaluate the

ellipse of intersection.

Solution

The equation of the ellipse is:

Simplifying,

12 2 2 2 2 2

2 2

ln 4 (0.1) (1) (0.5) (1 (0.5) ) 2.60

(1 ) (1 0.25)(2.60) 1.95

− = − =

− = − =

c

c

π

ρ

2 24 4 5 5

2(0.5) 1.951 1 0.5 0.5

− − − − − + =

x x y y

2 2( 4) 2( 4)( 5) 4( 5) 1.95− − − − + − =x x y y

Error Ellipse

• Produced when the bivariate probability

distribution is centered at the origin (µx = µy = 0)

� This equation is used if we want to represent the

random errors only

22

22

1 1( , ) exp 2

2(1 )2 1

− = − + − − x x y yx y

x x y yf x y ρ

ρ σ σ σ σπσ σ ρ

Error Ellipse

• The corresponding equation for the cutting

ellipse in this case would be:

� This equation represents a family of error

ellipses centered on the origin

22

2 22 (1 )

− + = − x x y y

x x y ycρ ρ

σ σ σ σ

Standard Error Ellipse

• When c=1, we get the equation of the standard

error ellipse:

• Represents the area of uncertainty for the location

of a control point

• Size, shape, and orientation of a standard error

ellipse are governed by the parameters σx, σy, and

ρ.

22

22 (1 )

− + = − x x y y

x x y yρ ρ

σ σ σ σ

The Standard Error Ellipse

3

Sample variants of the standard error

ellipse (by varying the parameters)Standard Error Ellipse

• In general, the principal axes (x’ and y’)

do not coincide with the coordinate axes

(x and y)

• The major axis of the ellipse (x’) makes

an angle θ with respect to the x-axis

Positional Errors

• A positional error is expressed in the

x,y coordinate system by the random

vector

• The same positional error is expressed

in the x’,y’ coordinate system by the

random vector

X

Y

′ ′

X

Y

Orthogonal (Rotational) Transformation

• The two vectors can be related by the equation:

• θ is the angle of rotation

• Transformations from one coordinate system to

the other can be made using the above equation

� correlated errors may be transformed to

uncorrelated errors using the equation

cos sin

sin cos

′ = ′ −

X X

Y Y

θ θθ θ

Covariance Matrices

• The covariance matrices for the random

vectors are:

� X’ and Y’ are uncorrelated (they are the

principal axes of the ellipse)

X

Y

′ ′

X

Y

2

2

x xy

xy y

σ σσ σ

2

2

0

0

′

′

x

y

σσ

Covariance Matrices

Recall:

Applying this to the vector relationship, we get:

Solving the matrix:

T

yy xxA AΣ = Σ

2 2

2 2

0 cos sin cos sin

0 sin cos sin cos

x x xy

y xy y

σ σ σθ θ θ θσ σ σθ θ θ θ

′

′

− = −

2 2 2 2 2

2 2 2 2 2

2 2 2 2

cos 2 sin cos sin

sin 2 sin cos cos

0 ( )sin cos (cos sin )

x x xy y

y x xy y

y x xy

σ σ θ σ θ θ σ θ

σ σ θ σ θ θ σ θ

σ σ θ θ σ θ θ

′

′

= + +

= + +

= − + −

Eq. 1

Eq. 2

Eq. 3

4

Orientation of Error Ellipse

But and

The third equation (Eq. 3) therefore becomes:

Which can be further simplified as:

sin 2sin cos

2

θθ θ = 2 2cos sin cos 2θ θ θ− =

2 2

2tan 2

xy

x y

σθ

σ σ=

−

2 21( )sin 2 cos 2 0

2y x xyσ σ θ σ θ− + =

Semimajor and Semiminor Axes

The first two equations (Eq. 1 and Eq. 2) become:

σx’ = semimajor axis

σy’ = semiminor axis

1/ 22 2 2 2 2

2 2

1/ 22 2 2 2 2

2 2

( )

2 4

( )

2 4

x y x y

x xy

x y x y

y xy

σ σ σ σσ σ

σ σ σ σσ σ

′

′

+ −= + +

+ −= − +

Standard Error Ellipse

Example:

The random error in the position of a survey

station is expressed by a bivariate normal

distribution with parameters µx = µy = 0, σx

= 0.22 m, σy = 0.14 m, and ρ = 0.80.

Evaluate the semimajor axis, semiminor axis,

and the orientation of the standard error

ellipse associated with this position error.

Standard Error Ellipse

Solution:

( ) ( ) ( )

( ) ( )

( ) ( ) ( )( )( )

( )

2

2 22 2

2

1 221 2 2 22

2 2

22 2

1 22

2 22 2

2 2

2 2

0 8 0 22 0 14 0 0246 m

0 22 0 140 0340 m

2 2

0 22 0 140 0246 0 0285 m

4 4

0 0340 0 02852 4

0 0625 m

xy x y

x y

//

x y

xy

/

x yx y

x' xy

x'

. . . .

. ..

. .. .

. .

.

σ ρσ σ

σ σ

σ σσ

σ σσ σσ σ

σ

= = =

+ += =

−− + = + =

−+ = + + = +

=

Standard Error Ellipse

( )

( )( ) ( )

1 22

2 22 2

2 2

2 2

2 22 2

0 0340 0 02852 4

0 0055 m

0 0625 0 25 m

0 0055 0 074 m

2 2 0 02462

0 22 0 14

2 1 711

/

x yx y

y' xy

y'

x'

y'

xy

x y

. .

.

. .

. .

.tan

. .

tan .

σ σσ σσ σ

σ

σ

σ

σθ

σ σ

θ

−+ = − + = −

=

= =

= =

= =− −

=

2θ lies in the 1st quadrant since σxy and σx2 - σy

2 are both

positive. Therefore, 2θ = tan-1(1.711) = 59.7° and the

orientation of the error ellipse is θ = 29.8°

Probability of Error Ellipse

Assumption: Independent (Uncorrelated) random errors X

and Y.

Given a point whose position is defined by the random

errors (X and Y), the point will lie within the error

ellipse if:

22

2 22 (1 )

− + = − x x y y

x x y ycρ ρ

σ σ σ σ

22

2

x y

X Yc

σ σ

+ =

22

2

x y

X Yc

σ σ

+ ≤

5

Probability of Error Ellipse

The random variable U such that:

has a chi-square distribution with two degrees of freedom.

(Recall from the past lecture that Y has a chi-square

distribution:

where Zn are independent standard normal random

variables)

22

x y

X YU

σ σ

= +

2 2 2 2

1 2 nY Z Z Z= + + +⋯

Probability of Error Ellipse

For two degrees of freedom, the probability density

function of U can be derived:

( / 2) 1 / 2

/ 2

(2 / 2) 1 / 2

2/ 2

1( )

2 ( / 2)

1( )

2 (2 / 2)

n y

n

u

f y y en

f u y e

− −

− −

=Γ

=Γ

/ 21( ) for u>0

2

uf u e−= Probability Density Function of U

Probability of Error Ellipse

Therefore, the probability that the position given by X and

Y lies on/within the error ellipse is:

2

22

2 2

/ 2

0

1

2

x y

c

u

X YP c P U c

e du

σ σ

−

+ ≤ = ≤

= ∫

2

22

2 / 21 c

x y

X YP c e

σ σ−

+ ≤ = −

Probability

Probability of Error Ellipse

Example:

The random error in the position of a survey station is expressed by a bivariatenormal distribution with parameters µx = µy = 0, σx = 0.22 m, σy = 0.14 m, and ρ = 0.80. Evaluate the semimajor and semiminor axes of the error ellipse within which it is 0.90 probable that the error in position will lie.

Probability of Error Ellipse

Solution:

The semimajor and semiminor

axes are:

[ ]

2

2

2

2

2

2

2

0 90

1 0 90

0 10

2 30262

4 6052

2 146

c /

c /

P U c .

e .

ln . ln e

c.

. c

c .

−

−

≤ =

− =

=

−− =

=

=

( )( )

2 146 0 25 0 54 m

2 146 0 074 0 16 m

x'

y'

c . . .

c . . .

σ

σ

= =

= =