jacobian transformation

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Math 425

Intro to Probability

Lecture 30

Kenneth Harris

[email protected]

Department of MathematicsUniversity of Michigan

April 3, 2009

Kenneth Harris (Math 425) Math 425 Intro to Probability Lecture 30 April 3, 2009 1 / 32

Introduction One Function of Random Variables

Functions of a Random Variable: Density

Let g (x ) = y be a one-to-one function whose derivative is nonzero

on some region A of the real line.

Suppose g maps A onto B , so that there is an inverse map

x = h (y ) from B back to A.

Let X be a continuous random variable with known density f X (x ).

Let Y = G (X ). Then the density of Y is

f Y (y ) = f X

h (y )

· d

dt h (y )

.

Note: Compare to Ross, Theorem 5.7.1, page 243.


Two Functions of Two Random Variables

Problem

Let the continuous random variables (X , Y ) have joint density

f X ,Y (x , y ) and let A = {(x , y ) : f X ,Y (x , y ) > 0}.

(X , Y ) determines a point with xy -coordinates in the region

A.

Consider the continuous random variables (U , V ) given by

U = g 1(X , Y ) V = g 2(X , Y ).

(U , V ) determines a point with uv -coordinates in some region B.

Problem. If the transformation from xy -coordinates to uv -coordinates

given by

u = g 1(x , y ) v = g 2(x , y ).

is nice onA

, then we can produce the joint density f U ,V

(u , v ) for the

random variable (U , V ).



Definition Nice Transformations

Definition

A transformation from xy -coordinates to uv -coordinates (xy ⇒ uv )

given by

u = g 1(x , y ) v = g 2(x , y ).

is nice on A, if

1 The partial derivatives ∂ u ∂ x ,

∂ u ∂ y ,

∂ v ∂ x , and ∂ v

∂ y exist and are continuous

on A.2 The Jacobian of the transformation is nonzero on A:

J (x , y ) =

∂ u ∂ x

∂ u ∂ y

∂ v ∂ x

∂ v ∂ y

= ∂ u

∂ x

∂ v

∂ y − ∂ u

∂ y

∂ v

∂ x = 0

whenever (x , y ) ∈ A.





Change of coordinates

A nice transformation on A (xy ⇒ uv ) amounts to simply a change

of coordinates of the plane from xy -coordinates to uv -coordinates.

We can recover the original xy -coordinates from the new

uv -coordinates.

Suppose (xy ⇒ uv ) is nice transformation on Au = g 1(x , y ) v = g 2(x , y )

to uv -coordinates on a region B.

There is a reverse transformation (uv ⇒ xy ) from uv -coordinates to

xy -coordinates

x = h 1(u , v ) y = h 2(u , v ).

which maps B onto A and which are also nice on B.



Jacobians

The Jacobian of the original transformation (xy ⇒ uv ) is thedeterminant

J (x , y ) =

∂ u ∂ x

∂ u ∂ y

∂ v ∂ x

∂ v ∂ y

= ∂ u

∂ x

∂ v

∂ y − ∂ u

∂ y

∂ v

∂ x

The Jacobian of the inverse transformation (uv

⇒xy ) is the

determinantJ (u , v ) =

∂ x ∂ u

∂ x ∂ v

∂ y ∂ u

∂ y ∂ v

= ∂ x

∂ u

∂ y

∂ v − ∂ x

∂ v

∂ y

∂ u

Since (xy ⇒ uv ) is nice on A,

J (x , y ) = 0 whenever (x , y ) ∈ A and

J (u , v ) = 0 whenever (u , v ) ∈ B.

Furthermore, the two Jacobian determinants are inverses

J (x , y ) = J (u , v )−1



Main Theorem

Theorem

Let (X , Y ) be continuous random variables with joint density f X ,Y (x , y ), and (U , V ) be random variables given by

U = g 1(X , Y ) V = g 2(X , Y ).

Suppose the (xy ⇒ uv ) transformation

u = g 1(x , y ) v = g 2(x , y ).

is nice on A = {(x , y ) : f X ,Y (x , y ) = 0}.Let the inverse (uv ⇒ xy ) from B to A be

x = h 1(u , v ) y = h 2(u , v ).

The joint density of (U , V ) is given for (u , v ) ∈ B by either equation f U ,V (u , v ) = f X ,Y

h 1(u , v ), h 2(u , v )

·J (u , v )

f U ,V (u , v ) = f X ,Y

h 1(u , v ), h 2(u , v )

· J (x , y )−1

whichever is more convenient to compute.



Picture of Theorem

f U ,V (u , v ) · du · dv ≈ P{(U , V ) ∈ ∆B }

P {(X , Y ) ∈ ∆A} ≈ f X ,Y (x , y ) · J (u , v )

· du · dv

x,yu,v

du

dv

UV XY

B

A

PU,VB PX,YA

Area of B dudv Area of A Ju,vdudv





Sketch of Proof of Theorem

Let B ⊆ B and suppose (uv ⇒ xy ) maps B to A ⊆ A.

P {(U , V ) ∈ B } = P {(X , Y ) ∈ A}=

(x ,y )∈A

f X ,Y (x , y ) dx dy

= (u ,v )∈B

f X ,Y h 1(u , v ), h 2(u , v ) · J (u , v ) dudv

using the Change of Variables Theorem of analysis.

Intuitively, we can break B into small regions ∆B which (uv ⇒ xy )transforms to small regions ∆A of A where for any (u , v ) ∈ ∆B :

f U ,V (u , v ) · Area

∆B ≈ f X ,Y (h 1(u , v ), h 2(u , v )) · Area

∆A

where Area

∆A

≈ Area

∆B · J (u , v )

. Differentiate the integrals to get the transformation rule:

f U ,V (u , v ) =

f X ,Y

h 1(u , v ), h 2(u , v )

· J (u , v )

if (u , v ) ∈ B0 otherwise.


Example


Example. Let X and Y be continuous random variables with joint

density f X ,Y (x , y ) and where X = 0. Consider

U = XY V = X .

The transformation (xy ⇒ uv ) is given by

u = xy v = x .

The inverse transformation (uv ⇒ xy ) is given by

x = v y = u

v .

The Jacobian for transformation for (uv ⇒ xy ) is

J (u , v ) =

0 1

v 1 − u

v 2

= −1

v


Example


So, the joint density is

f U ,V (u , v ) = f X ,Y

h 1(u , v ), h 2(u , v ) · J (u , v )

= f X ,Y

v ,

u

v

· 1

|v |

We can compute the marginal f U (u ) = f XY (u ) by

f XY (u ) =

∞

−∞

f X ,Y

v ,

u

v

· 1

|v | dv


Examples: Polar Coordinates

Rectangle to Polar coordinates

It is often convenient to change from rectangular coordinates xy topolar coordinates r θ. The transformation (xy ⇒ r θ) is

r = x 2 + y 2 θ = arctan y x

.

where r > 0 and −π < θ ≤ π.

The inverse transformation (r θ ⇒ xy ) from polar back torectangular is

x = r cos θ y = r sin θ.

The transformation is (xy

⇒r θ) nice in the punctured plane

R2 − {(0, 0)}. Verified in three slides.





Converting Rectangle to Polar

Rectangle xy -coordinates to polar r θ-coordinates:

r =

x 2 + y 2, r > 0 θ = arctan y

x , −π < θ ≤ π,

Polar r θ-coordinates to rectangle xy -coordinates

x = r cos θ y = r sin θ −∞ < x , y < ∞.

x,yr,Θ

yrsinΘ

xrcosΘ

r x2 y

2

Θarctan y

x




Plot of tan y x

on [−π, π]. The four quadrants of the plane are

I : x , y > 0 II : x < 0, y > 0 III : x , y < 0 IV : x > 0, y < 0

II III I IV

Π Π

2

Π

2 Π

6

4

2

2

4

6



Problem: Rectangle to Polar

Problem. Let (X , Y ) be randomly chosen in some region R of the

xy -plane with joint density f X ,Y (x , y ).

Consider the random variables giving the polar coordinates

R =

X 2 + Y 2 Θ = arctan Y

X

where R > 0 and −π < Θ ≤ π.

The Jacobian is easiest to compute on the r θ-plane:

J (r , θ) =

cos θ sin θ

−r sin θ r cos θ

= r cos2 θ + r sin2 θ = r

The joint distribution of R , Θ is

f R ,Θ(r , θ) = r · f X ,Y (r cos θ, r sin θ) r > 0,−π < θ ≤ π.



Example

Example. Let (X , Y ) be uniformly distributed in R = unit circle. So,

f X ,Y (x , y ) = 1

π when x 2 + y 2 ≤ 1.

So,

f R ,Θ(r , θ) = r · f X ,Y (r cos θ, r sin θ) = r

π 0 < r ≤ 1,−π < θ ≤ π

The marginals are

f R (r ) =

π−π

r

π d θ = 2r 0 < r ≤ 1,

f Θ(θ) = 1

0

r

π

dr = 1

2π −π < θ

≤π.

Thus, Θ is uniformly distributed on (−π, π].





Example

Example. Let (X , Y ) be independent and normally distributed in the

plane with (µ = 0, σ2). So,

f X ,Y (x , y ) = 1

2πσ2e −(x 2+y 2)/2σ2

Since f R ,Θ(r , θ) = r

·f X ,Y (r cos θ, r sin θ),

f R ,Θ(r , θ) = r

2πσ2e −(r 2 cos2 θ+r 2 sin2 θ)/2σ2

= r

2πσ2e −r 2/2σ2

0 < r ,−π < θ ≤ π.

The marginals are

f R (r ) =

π−π

r

2πσ2e −r 2/2σ2

d θ = r

σ2e −r 2/2σ2

0 < r ,

f Θ(θ) =

∞0

r

2πσ2e −r 2/2σ2

dr = 1

2π − π < θ ≤ π.

Thus, Θ is uniformly distributed on (−π, π] and R is the Rayleigh

distribution (the distance of (X , Y ) from the origin).Kenneth Harris (Math 425) Math 425 Intro to Probability Lecture 30 April 3, 2009 21 / 32


Example

Example. Let R be exponentially distributed with mean 2 and Θ beuniformly distributed in (−π, π], both independent. The joint distributionis

f R ,Θ(r , θ) = 1

2π · 1

2e −r /2 0 < r ,−π < θ ≤ π

Let X and Y be random variables determined by

X =√

R cos Θ Y =√

R sin Θ

Solve for r , θ in the transformation x =√

2r cos θ and y =√

2r sin θ:

r = x 2 + y 2 θ = arctan y

x .

The Jacobian determinant is easiest to compute using r θ-coordinates:

J (r , θ) =

cos θ2√

r sin θ2√

r

−√ r sin θ√

r cos θ

= cos2 θ

2 +

sin2θ

2 =

1

2.



Example – continued

f R ,Θ(r , θ) = 1

2π · 1

2e −r /2 0 < r ,−π < θ ≤ π

So,f X ,Y (x , y ) = f R ,θ(x 2 + y 2, arctan

y

x ) · 2

= 1

2πe −(x 2+y 2)/2

X and Y are independent and normally distributed randomvariables with (µ = 0, σ2 = 1). The marginals are obtained byintegrating f X ,Y (x , y ):

f X (x ) = 1√

2πe −x 2/2

f Y (y ) = 1√ 2π

e −y 2/2



Example–continued

Let U and V be uniformly distributed on (0, 1).

Consider the random variable Θ:

Θ = 2πV − π

So, Θ is uniformly distributed on (−π, π).

Consider the random variable R :

R = 2 ln 1

U solving, u = e −r /2.

By Proposition 5.7.1 (Ross, page 243),

f R (r ) = f U (e −r /2) · u = 1

2e −r /2

So, R is exponentially distributed with mean 2.






Let X and Y be random variables determined by

X =√

R cos Θ Y =√

R sin Θ

Then X and Y are independent standard normal random variables!!

We can simulate a standard normal random variable X by using two

independent uniform random variables U and V on (0, 1):

X =

2 ln

1

U cos

2πV − π

.




Simulating a standard normal random variable with a pair of independent

uniform random variables on (0, 1).

3 2 1 0 1 2 30.0

0.1

0.2

0.3

0.4

0.5 1000 data points

3 2 1 0 1 2 30.0

0.1

0.2

0.3

0.4

0.5 10,000 data points


Example of Joint Distribution

Example

Example. Let X and Y be independent and uniformly distributed on(0, 1]. Find the joint probability density function for the randomvariables

U =

X

Y V = XY .

Individually, the distribution of X and Y are

f X (x ) =

1 if 0 ≤ x ≤ 1

0 otherwisef Y (y ) =

1 if 0 ≤ y ≤ 1

0 otherwise

So, the joint distribution f X ,Y (x , y ) is

f X ,Y (x , y ) = f X (x ) · f Y (y ) = 1 if 0 ≤ x , y ≤ 1

0 otherwise.




The transformation into uv -coordinates

u = x

y v = xy ,

is one-to-one and has an inverse

x =√

uv y =

v

u .

The Jacobian determinant is easiest when computed in xy coordinates:

J (x , y ) =

1y − x

y 2

y x

= 2x

y = 2u

So, u , v > 0 and

f U ,V (u , v ) = f X ,Y √ uv ,

v

u · 1

2u

= 1

2u if 0 <

√ uv , v

u ≤1

0 otherwise.






It remains to compute the bounds on u and v .

0 <√

uv , v

u ≤ 1 =⇒ 0 < v ≤ 1

u , 0 < v ≤ u .

Only one of these ranges need be retained, depending upon whetheru ∈ (0, 1] or u ∈ [1,∞):

f U ,V (u , v ) =

12u

if 0 < u < 1, 0 < v ≤ u ,

or, if u ≥ 1, 0 < v ≤ 1u

,

0 otherwise.




f U ,V (u , v ) =

12u

if 0 < u < 1, 0 < v ≤ u ,

or, if u ≥ 1, 0 < v ≤ 1u

,

0 otherwise.

We compute the marginals.

f U (u ) =

u

01

2u dv if 0 < u < 1 1

u

01

2u dv if u ≥ 1

0 otherwise

=

12

dv if 0 < u < 11

2u 2 dv if u ≥ 1

0 otherwise

f v (v ) =

1v

v 1

2u du if 0 < v ≤ 1

0 otherwise=

ln 1

v if 0 < v ≤ 1

0 otherwise




Plot of area determined by

0 < u < 1 =⇒ 0 < v ≤ u and u ≥ 1 =⇒ 0 < v ≤ 1

u .

v

1

v

1 2 3 4u

0.2

0.4

0.6

0.8

1.0

v


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