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Stat 310Bivariate Transformations
Garrett Grolemund
Pick up handout
1. Example
2. Bivariate transformations
3. Calculating probabilities
4. Distribution function technique
Suppose the basket is at (25, 0). Devise a
way to calculate each shot’s distance from
the basket using X and Y.
Question
Polar Coordinates
r = √((x - 25)2 + y2 )
Ө = tan-1 (y/x)
Polar Coordinates
r = √((x - 25)2 + y2 )
Ө = tan-1 (y/(x – 25))
(Transformations that involve two
random variables at a time)
Bivariate Transformations
Transformed Data
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?
Calculating Probabilities
What is the probability that
max(X1, X2 , X3 , X4 , X5 , X6 , X7) ≤ $100 ?
min(X1, X2 , X3 , X4 , X5 , X6 , X7) ≤ $ -100 ?
Recall from the univariate case, we have
two methods of calculating probabilities of
transformed variables
Distribution
function
technique
Change of
variable
technique
Distribution function technique
Suppose the Xi are iid. Is this a reasonable assumption?
Then, we can calculate Fv(a) by
P(V ≤ a) = P(min(Xi) ≤ a)
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)
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)
= 1 – [P(X1 > a, X2 > a, … X7 > a)]
= 1 – [P(X1 > a, X2 > a, … X7 > a)]
= 1 – [P(X1 > a) P(X2 > a) … P(X7 > a) ]
= 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, 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) ]
= 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, 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 – [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 – [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 ]
So P(V ≤ -100) = Fv(-100) = 1 – [ (1 – Fx(-100) ) 7 ]
We can find the density of V by differentiating:
fv(a) = Fv(a)
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 ]}
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))
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)
Your Turn
Work through the handout to find FU(a) and
fU(a).
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) = ?
Probability as volume under a surface
X
Y
f(x,y)
Set A
P(Set A)
X
Y
f(x,y)
Set A
P(Set A)
P(U < 2, V < 5) = P( max(X, Y) < 5 min(X, Y) > 2)
P(U < 2, V < 5) = ∫5
2∫5
2fx,y (x,y) dx dy
But…
•Computing double integrals can
be hard
•Finding correct bounds can be
hard
r = √((x - 25)2 + y2 )
Ө = tan-1 (y/(x – 25))
Next time: Change of Variables
Read Section 3.4
