- The quiz will be handed out and start at 1pm, finishing at 1:10pm.
- Though it is not for a grade, the instructor will collect the completed quizzes.
6. 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?)
7. Recall from Tuesday
U = X - Y is a bivariate transformation
The Distribution function technique gives
us two ways to model X - Y:
8. 1. Begin with FX,Y(a):
Compute FU(a) in terms of FX,Y(a) by
equating probabilities
9. 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)
=?
10. 2. Begin with fX,Y(a) :
Compute FU(a) by integrating fX,Y(a)
over the region where U < a
f(x,y)
P(Set A)
X
Set A
Y
11. 2. Begin with fX,Y(a) :
Compute FU(a) by integrating fX,Y(a)
over the region where U < a
f(x,y)
∞ Y+a P(Set A)
FU(a) = ∫ ∫ fX,Y(a) dxdy
-∞ -∞
X
Set A
Y
13. Univariate 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.
14. Bivariate 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 |
15. 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), * )
16. 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)
17. 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.
18. 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
21. a b
= ad - bc
c d
δx δx
δu δv
J= = δx δy - δx δy
δy δy δu δv δv δu
δu δv
22. fU,V(u, v) = fx,y(h1(u, v), h2(u, v)) | J |
U=X-Y
What should V be?
23. 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
24. fU,V(u, v) = fx,y(h1(u, v), h2(u, v)) | J |
What should fx,y(*, *) be?
25. 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?
26. How would we model the Lakers score
distribution?
27. How would we model the Lakers score
distribution?
• Discrete Data
28. How would we model the Lakers score
distribution?
• Discrete Data
• Sum of many
bernoulli trials
29. How would we model the Lakers score
distribution?
• Discrete Data
• Sum of many
bernoulli trials
Poisson?
35. Your Turn
Calculate the Jacobian of our transformation. Let
U = X - Y and V = Y.
δx δx
δu δv
J= = δx δy - δx δy
δy δy δu δv δv δu
δu δv
36. 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)
44. 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