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Stat 310Bivariate Change of Variables

Garrett Grolemund

1. Review of distribution function

techniques

2. Change of variables technique

3. Determining independence

4. Simulating transformations

Lakers' final scores

Laker's

Mean

87.2

Opponent's

Mean

81.2

Let

X = The Lakers' Score

Y = The opponent's score

U = X -Y

Then the Lakers will win if U is positive,

and they will lose if U is negative.

How can we model U? (i.e, How can we

find the CDF and PDF of U?)

Recall from Tuesday

U = X - Y is a bivariate transformation

The Distribution function technique gives

us two ways to model X - Y:

1. Begin with FX,Y(a):

Compute FU(a) in terms of FX,Y(a) by

equating probabilities

1. Begin with FX,Y(a):

Compute FU(a) in terms of FX,Y(a) by

equating probabilities

FU(a) = P(U < a)

= P(X - Y < a)

= P(X < Y + a)

= ?

2. Begin with fX,Y(a) :

Compute FU(a) by integrating fX,Y(a)

over the region where U < a

X

Y

f(x,y)

Set A

P(Set A)

2. Begin with fX,Y(a) :

Compute FU(a) by integrating fX,Y(a)

over the region where U < a

X

Y

f(x,y)

Set A

P(Set A)

FU(a) = ∫∞

-∞∫Y + a

-∞fX,Y(a) dxdy

Change of variables

X U

If

U = g(X) X = h(U)

Where h is the inverse of g, then

fU(u) = fx(h(u)) |h'(u)|

Method works for bivariate case, once we make the appropriate modifications.

Univariate change of variables

(X,Y) (U,V)

if

U = g1(X, Y) X = h1(U, V)

V = g2(X, Y) Y = h2(U, V)

Where h is the inverse of g, then

fU,V(u, v) = fx,y(h1(U, V) , h2(U, V) ) | J |

Bivariate change of variables

fU(u) = fx(h(u)) |h'(u)|

Since U depends on both X and Y, we replace fx(h(u)) with the joint density fx,y(h(u), * )

fU(u) = fx,y(h(u), * ) |h'(u)|

A joint density must be a function of two random variables

Let X = h1(u) and Y = h2(u)

fU(u) = fx,y(h1(u), h2(u)) |h'(u)|

But for equality to hold, we must have a function of two variables on the left side as well

Define V = g2(X, Y) however you like.

fU,V(u, v) = fx,y(h1(u, v), h2(u, v)) |h'(u)|

Now we have two equations to take derivatives of (h1, h2) and two variables to take the derivative with respect to, (U,V)

The multivariate equivalent of h'(u) is the Jacobian, J

fU,V(u, v) = fx,y(h1(u, v), h2(u, v)) | J |

And we're finished

Jacobian

δx δxδu δv

J = δy δyδu δv

δx δxδu δv

J = δy δyδu δv

a b

c d= ad - bc

= δx δy - δx δyδu δv δv δu

fU,V(u, v) = fx,y(h1(u, v), h2(u, v)) | J |

U = X - Y

What should V be?

fU,V(u, v) = fx,y(h1(u, v), h2(u, v)) | J |

U = X - Y

What should V be?

• Sometimes we want V to be something specific

• Otherwise keep V simple or helpful

e.g., V = Y

fU,V(u, v) = fx,y(h1(u, v), h2(u, v)) | J |

What should fx,y(*, *) be?

fU,V(u, v) = fx,y(h1(u, v), h2(u, v)) | J |

What should fx,y(*, *) be?

First consider: what should fx(*) and fy(*) be?

How would we model the Lakers score

distribution?

How would we model the Lakers score

distribution?

• Discrete Data

How would we model the Lakers score

distribution?

• Discrete Data

• Sum of many

bernoulli trials

How would we model the Lakers score

distribution?

• Discrete Data

• Sum of many

bernoulli trials

Poisson?

Lakers' scores vs. simulated Poisson (87.2) distributions

Opponent's scores vs. simulated Poisson (81.2)

fU,V(u, v) = fx,y(h1(u, v), h2(u, v)) | J |

X ~ Poisson (87.2)

Y ~ Poisson (81.2)

fx,y(h1(u, v), h2(u, v)) = fx(h1(u, v)) fy(h2(u, v))

fx(h1(u, v)) = e-87.2 (87.2)h1(u,v)

h1(u, v)!

fy(h2(u, v)) = e-81.2 (81.2)h2(u,v)

h2(u, v)!

?

ρ = 0.412

fU,V(u, v) = fx,y(h1(u, v), h2(u, v)) | J |

Your Turn

Calculate the Jacobian of our transformation. Let U = X - Y and V = Y.

δx δxδu δv

J = δy δyδu δv

= δx δy - δx δyδu δv δv δu

Your Turn

Calculate fU,V(u, v) and express fU(u) as an

integral (you do not need to solve that

integral).

Let U = X - Y and V = Y. Let X ~ Poisson(87.2) and Y ~ Poisson(81.2)

fU,V(u, v) = e-(87.2 + 81.2) (87.2)h1(u,v)(81.2)h2(u,v)

h1(u, v)! h2(u, v)!

Skellam Distribution

http://en.wikipedia.org/wiki/Skellam_distribution

U values

U values vs. Skellam Distribution

Testing Independence

Recall:

1. U and V are independent if

fU,V(u, v) = fU(u) fV(v)

2. By change of variables:

fU,V(u, v) = fx,y(h1(u, v), h2(u, v)) | J |

Complete the proof

Complete the handout to show that U = X + Y and V = X - Y are independent when X, Y ~ N(0, 1).

Simulation

We can also learn a lot about the distribution of a random variable by simulating it.

Let Xi ~ Uniform (0,1)

Let U = (X1 + X2 ) / 2

If we generate a 100 pairs of X1 and X2 and plot (X1 + X2 ) / 2 for each pair, we will have a simulation of the distribution of U

For comparison, V1 = X1 ~ uniform(0,1)

10,000 samples

V1 = (X1 + X2) / 2

10,000 samples

V1 = (X1 + X2 + X3) / 3

10,000 samples

V1 = (X1 + X2 + … + X10) / 10

10,000 samples

V1 = Σ11000 Xi / 1000

10,000 samples