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# 14 Bivariate Transformations

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### 14 Bivariate Transformations

1. 1. Stat 310 Bivariate Transformations Garrett Grolemund
2. 2. Pick up handout
3. 3. 1. Example 2. Bivariate transformations 3. Calculating probabilities 4. Distribution function technique
4. 4. Question Suppose the basket is at (25, 0). Devise a way to calculate each shot’s distance from the basket using X and Y.
5. 5. Polar Coordinates r = √((x - 25)2 2 + ) y Ө = tan-1 (y/x)
6. 6. Polar Coordinates r = √((x -25)2 2 + ) y Ө = tan-1 (y/(x – 25))
7. 7. Bivariate Transformations (Transformations that involve two random variables at a time)
8. 8. Transformed Data
9. 9. 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?
10. 10. Calculating Probabilities
11. 11. What is the probability that max(X1, X2 , X3 , X4 , X5 , X6 , X7) ≤ \$100 ? min(X1, X2 , X3 , X4 , X5 , X6 , X7) ≤ \$ -100 ?
12. 12. Recall from the univariate case, we have two methods of calculating probabilities of transformed variables Distribution Change of function variable technique technique
13. 13. Distribution function technique
14. 14. Suppose the Xi are iid. Is this a reasonable assumption? Then, we can calculate Fv(a) by P(V ≤ a) = P(min(Xi) ≤ a)
15. 15. 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)
16. 16. 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)
17. 17. = 1 – [P(X1 > a, X2 > a, … X7 > a)]
18. 18. = 1 – [P(X1 > a, X2 > a, … X7 > a)] = 1 – [P(X1 > a) P(X2 > a) … P(X7 > a) ]
19. 19. = 1 – [P(X1 > a, X2 > a, … X7 > a)] = 1 – [P(X1 > a) P(X2 > a) … P(X7 > a) ] (Because the Xi are independent)
20. 20. = 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) ]
21. 21. = 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)
22. 22. = 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 ]
23. 23. = 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 ]
24. 24. = 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 ]
25. 25. So P(V ≤ -100) = Fv(-100) = 1 – [ (1 – Fx(-100) ) 7 ] We can find the density of V by differentiating: fv(a) = Fv(a)
26. 26. 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 ]}
27. 27. 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))
28. 28. 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)
29. 29. Your Turn Work through the handout to find FU(a) and fU(a).
30. 30. 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) = ?
31. 31. Probability as volume under a surface f(x,y) P(Set A) X Set A Y
32. 32. 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
33. 33. But… •Computing double integrals can be hard •Finding correct bounds can be hard r = √((x - 25)2 + y2 ) Ө = tan-1 (y/(x – 25))
34. 34. Next time: Change of Variables