15 Bivariate Change Of Variables
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Transcript of 15 Bivariate Change Of Variables
Quiz• Pick up quiz and handout on your way in.
• Start at 1pm
• Finish at 1:10pm
• The quiz is not for a grade, but I will be
collecting it.
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