This document discusses bivariate transformations and calculating probabilities of transformed random variables. It introduces calculating the probability that the maximum and minimum of iid random variables fall in a given range using the distribution function technique. This involves finding the probability that all variables are individually greater than some value a to find the probability the minimum is greater than a. Similarly, the probability the maximum is less than a involves finding the probability all variables are individually less than a. It also discusses finding the joint distribution of the maximum and minimum using a double integral over the joint density function.

1. Stat 310
Bivariate Transformations
Garrett Grolemund

2. Pick up handout

3. 1. Example
2. Bivariate transformations
3. Calculating probabilities
4. Distribution function technique

11. Question
Suppose the basket is at (25, 0). Devise a
way to calculate each shot’s distance from
the basket using X and Y.

12. Polar Coordinates
r = √((x -
25)2 2
+ )
y
Ө = tan-1 (y/x)

13. Polar Coordinates
r = √((x -25)2 2
+ )
y
Ө = tan-1 (y/(x – 25))

14. Bivariate Transformations
(Transformations that involve two
random variables at a time)

15. Transformed Data

19. Your Turn
Suppose you own a portfolio of stocks. Let X1 be
the amount of money your portfolio earns
today, X2 be the amount of money it earns
tomorrow, and so on…
How would you calculate U and V, where U is
the amount of money you’ll make on your
best day during the next week, and V is the
amount you’ll make on your worst day?

20. Calculating Probabilities

21. What is the probability that
max(X1, X2 , X3 , X4 , X5 , X6 , X7) ≤ $100 ?
min(X1, X2 , X3 , X4 , X5 , X6 , X7) ≤ $ -100 ?

22. Recall from the univariate case, we have
two methods of calculating probabilities of
transformed variables
Distribution Change of
function variable
technique technique

23. Distribution function technique

24. Suppose the Xi are iid. Is this a reasonable
assumption?
Then, we can calculate Fv(a) by
P(V ≤ a) = P(min(Xi) ≤ a)

25. Suppose the Xi are iid. Is this a reasonable
assumption?
Then, we can calculate Fv(a) by
P(V ≤ a) = P(min(Xi) ≤ a)
= 1 – P(min(Xi) > a)

26. Suppose the Xi are iid. Is this a reasonable
assumption?
Then, we can calculate Fv(a) by
P(V ≤ a) = P(min(Xi) ≤ a)
= 1 – P(min(Xi) > a)
= 1 – P(all Xi > a)

27. = 1 – [P(X1 > a, X2 > a, … X7 > a)]

28. = 1 – [P(X1 > a, X2 > a, … X7 > a)]
= 1 – [P(X1 > a) P(X2 > a) … P(X7 > a) ]

29. = 1 – [P(X1 > a, X2 > a, … X7 > a)]
= 1 – [P(X1 > a) P(X2 > a) … P(X7 > a) ]
(Because the Xi are independent)

30. = 1 – [P(X1 > a, X2 > a, … X7 > a)]
= 1 – [P(X1 > a) P(X2 > a) … P(X7 > a) ]
(Because the Xi are independent)
= 1 – [P(X1 > a) P(X1 > a) … P(X1 > a) ]

31. = 1 – [P(X1 > a, X2 > a, … X7 > a)]
= 1 – [P(X1 > a) P(X2 > a) … P(X7 > a) ]
(Because the Xi are independent)
= 1 – [P(X1 > a) P(X1 > a) … P(X1 > a) ]
(because the Xi are identically distributed)

32. = 1 – [P(X1 > a, X2 > a, … X7 > a)]
= 1 – [P(X1 > a) P(X2 > a) … P(X7 > a) ]
(Because the Xi are independent)
= 1 – [P(X1 > a) P(X1 > a) … P(X1 > a) ]
(because the Xi are identically distributed)
= 1 – [P(X1 > a) 7 ]

33. = 1 – [P(X1 > a, X2 > a, … X7 > a)]
= 1 – [P(X1 > a) P(X2 > a) … P(X7 > a) ]
(Because the Xi are independent)
= 1 – [P(X1 > a) P(X1 > a) … P(X1 > a) ]
(because the Xi are identically distributed)
= 1 – [P(X1 > a) 7 ]
= 1 – [ (1 – P(X1 ≤ a) )7 ]

34. = 1 – [P(X1 > a, X2 > a, … X7 > a)]
= 1 – [P(X1 > a) P(X2 > a) … P(X7 > a) ]
(Because the Xi are independent)
= 1 – [P(X1 > a) P(X1 > a) … P(X1 > a) ]
(because the Xi are identically distributed)
= 1 – [P(X1 > a) 7 ]
= 1 – [ (1 – P(X1 ≤ a) )7 ]
= 1 – [ (1 – Fx(a) ) 7 ]

35. So P(V ≤ -100) = Fv(-100) = 1 – [ (1 – Fx(-100) ) 7 ]
We can find the density of V by differentiating:
fv(a) = Fv(a)

36. So P(V ≤ -100) = Fv(-100) = 1 – [ (1 – Fx(-100) ) 7 ]
We can find the density of V by differentiating:
fv(a) = Fv(a)
= {1 – [ (1 – Fx(a) ) 7 ]}

37. So P(V ≤ -100) = Fv(-100) = 1 – [ (1 – Fx(-100) ) 7 ]
We can find the density of V by differentiating:
fv(a) = Fv(a)
= {1 – [ (1 – Fx(a) ) 7 ]}
= -7(1 – Fx(a) ) 6 (1 - Fx(a))

38. So P(V ≤ -100) = Fv(-100) = 1 – [ (1 – Fx(-100) ) 7 ]
We can find the density of V by differentiating:
fv(a) = Fv(a)
= {1 – [ (1 – Fx(a) ) 7 ]}
= -7(1 – Fx(a) ) 6 (1 - Fx(a))
= 7(1 – Fx(a) ) 6 fx(a)

39. Your Turn
Work through the handout to find FU(a) and
fU(a).

40. What if we wish to find the joint distribution
FU,V(a,b)?
U = max(X, Y)
V = min(X, Y)
P(U < 2, V < 5) = ?

41. Probability as volume under a surface
f(x,y)
P(Set A)
X
Set A
Y

42. P(U < 2, V < 5) = P( max(X, Y) < 5 min(X, Y) > 2)
f(x,y)
P(Set A)
X
Set A
Y
5 5
P(U < 2, V < 5) = ∫ ∫ fx,y (x,y) dx dy
2 2

43. But…
•Computing double integrals can
be hard
•Finding correct bounds can be
hard
r = √((x - 25)2 + y2 )
Ө = tan-1 (y/(x – 25))

44. Next time: Change of Variables

45. Read Section 3.4