1. Random Variables and Probability Distributions Schaum’s Outlines of Probability and Statistics Chapter 2 Presented by Carol Dahl Examples by Tyler Hodge

2. 2-2 Outline of Topics Topics Covered:  Random Variables  Discrete Probability Distributions  Continuous Probability Distributions  Discrete Joint Distributions  Continuous Joint Distributions  Joint Distributions Example  Independent Random Variables  Changing of Variables  Convolutions

3. 2-3 Random Variables Variables events or values given probabilities Examples Drill for oil - may hit oil or a dry well Quantity oil found

4. 2-4 Random Variables Notation: P (X=x k )= P(x k ) probability random variable, X, takes value x k = P(x k ). Example Drilling oil well in Saudi Arabia outcome is either (Dry, Wet) 5% chance of being dry P (X = Dry well) = 0.05 or 5% P (X = Wet well) = 0.95 or 95%

5. 2-5 Discrete Probability Distributions Discrete probability distribution takes on discrete, not continuous values. Examples: Mineral exploration discrete - finds deposit or not Amount of deposit found Amount of deposit found continuous - any amount may be found continuous - any amount may be found

6. 2-6 Discrete Probability Distributions If random variable X defined by: P(X=x k )= P(x k ), k=1, 2, …. P(X=x k )= P(x k ), k=1, 2, ….Then, 0<P (x k )<1 ∑ k P(x) = 1

7. 2-7 Discrete Probability Distributions Example: Saudi oil well drilling Example: Saudi oil well drilling So, ∑ f (well) = 0.05 +0.95 = 1.0 Drilled wellDryWet f (well)0.050.95

8. 2-8 Discrete Probability Distributions Random variable X cumulative distribution function probability X  x

9. 2-9 Discrete Probability Distributions Example: Find cumulative distribution function wind speed at a location (miles per hour) Wind Speed 45678910 P(x)4/365/366/36 5/364/36 F(x)4/369/3615/3621/3627/3632/3636/36

10. 2-10 Continuous Probability Distributions Definition: Continuous probability distribution takes any value over a defined range Similar to discrete BUT replace summation signs with integrals Examples: amount of oil from a well student heights & weights

11. 2-11 Continuous Probability Distributions Continuous probability distribution describe probabilities in ranges f (x)  0 Equation for probability X lies between a and b:

12. 2-12 Continuous Probability Distributions Example: Engineer wants to know probability gas well pressure in economical range between 300 and 350 psi

13. 2-13 Continuous Probability Distributions Example cont: let f(x) = x/180,000 for 0<X<600 = 0 everywhere else verify that f(x) is bonafied probability distribution if not fix it

14. 2-14 Continuous Probability Distributions Example cont: let f(x) = x/180,000 for 0<X<600 = 0 everywhere else

15. 2-15 Discrete Joint Distributions X and Y - two discrete random variables joint probability distribution of X and Y P(X=x,Y=y) = f(x,y) Where following should be satisfied: f(x,y)  0 and

16. 2-16 Discrete Joint Distributions joint distribution of two discrete random table cells make up joint probabilities column and row summations marginal distributions. X\Yy1y1 y2y2...ynyn P(X) x1x1 f(x 1,y 1 )f(x 1,y 2 )...f(x 1,y n )f 1 (x 1 ) x2x2 f(x 2,y 1 )...f(x 1,y n )f 1 (x 2 )... xmxm f(x m,y 1 )f(x m,y 2 )...f(x m,y n )f 1 (x m ) P(Y)f 2 (y 1 )f 2 (y 2 )...f 2 (y n )1 (grand total)

17. 2-17 Continuous Joint Distributions X and Y two continuous random variables joint probability distribution of X and Y f(x,y) with f(x,y)  0 and

18. 2-18 Discrete Joint Distribution (Example) Example: Yield from two forests has distribution f(x,y) = c(2x+y) where x and y are integers such that: 0  X  2 and 0  Y  3 and f(x,y)=0 otherwise

19. 2-19 Joint Distributions (Example) cont. For previous slide, find value of “c” P (X=2, Y=1) Find P(X  1, Y  2) Find marginal probability functions of X & Y

20. 2-20 Joint Distributions (Example) cont. P (X=0, Y=0) = c(2X+Y) = c(2*0+0) = 0 P (X=1, Y=0) = c(2X+Y) = c(2*1+0) = 2c Fill in all possible probabilities in table

21. 2-21 Since, 42c=1 which implies that c=1/42. Joint Distributions (Example) cont. X\Y0123totals 00c2c3c6c 12c3c4c5c14c 24c5c6c7c22c totals6c9c12c15c42c

22. 2-22 Joint Distributions (Example) cont. find P( X=2,Y=1) from table: X\Y0123totals 001/422/423/426/42 12/423/424/425/4214/42 24/425/426/427/4222/42 totals6/429/4212/4215/4242/42 P(X=2,Y=1)=5/42

23. 2-23 Joint Distributions (Example) cont. Evaluate P(X  1,1<Y  2) sum cells of shaded region below: X\Y0123totals 001/422/423/426/42 12/423/424/425/4214/42 24/425/426/427/4222/42 totals6/429/4212/4215/4242/42 P(X  1,1<Y  2) = (3 + 4 + 5 + 6)/42 = 18/42=0.429

24. 2-24 Joint Distributions (Example) cont. Functions: Marginal X = P(X=x i ) =  j (X=x i, Y=y j ) Marginal Y = P(Y=y j ) =  i (X=x i, Y=y j )

25. 2-25 Joint Distributions (Example) cont. Marginal probability functions read off totals across bottom and side of table: X\Y0123P(X) 001/422/423/426/42 12/423/424/425/4214/42 24/425/426/427/4222/42 P(Y)6/429/4212/4215/4242/42

26. 2-26 Joint Distributions Continuous Case X, Y ~ f(X,Y) f x (X) =  f(X,Y)dx f y (Y) =  f(X,Y)dy Example: ore grade for copper and zinc f(x,y) = ce -x-y 0<x<0.4 0<Y<0.3 = 0 elsewhere

27. 2-27 Exp -x-y

28. 2-28 Joint Distributions Continuous Case Example: ore grade for copper and zinc f(x,y) = ce -x-y 0<X<0.4 0<Y<0.3 what is c ∫ ∫ cf(x,y)dxdy = 1 f x (x) =  0 0.3 ce -x-y dy = - ce -x-y | 0 0.3 = - ce -x-0.3 + ce -x-0. =  0 0.4 (-ce -x-0.3 + ce -x-0 ) = ce -x-0.3 - ce --x | 0 0.4 = ce -0.4 -0.3 - ce –0.4 - [ce -0.3 - ce -0 ] = 1 = c[e -0.7 - e –0.4 - e -0.3 + 1] = 1 c = 1/[e -0.7 - e –0.4 - e -0.3 + 1]= 1/3.084 = 0.324

29. 2-29 Independent Random Variables Random variables X and Y are independent if occurrence of one ->no affect other’s probability 3 tests 1. iff P(X=x|Y=y) = P(X=x) for all X and Y P(X|Y) = P(X) for all values X and Y otherwise dependent

30. 2-30 Independent Random Variables Random variables X and Y are independent if occurrence of one ->no affect other’s probability 2. iff P(X=x,Y=y) = P(X=x)*P(Y=y) P(X,Y) = P(X)*P(Y) otherwise X and Y dependent

31. 2-31 Independent Random Variables Random variables X and Y are independent if occurrence of one ->no affect other’s probability 3. iff P(X  x,Y  y) = P(X  x)*P(Y  y) F(X,Y) = F(X)*F(Y) otherwise X and Y dependent

32. 2-32 State of Pennsylvania problems with abandoned coal mines Government officials-reclamation bonding requirements new coal mines size of mine influence probability reclamation Independent Random Variables (Reclamation Example)

33. 2-33 Office of Mineral Resources Management compiled joint probability distribution Size of Mine (000 st) Abandons Mine Reclaims Mine P(X) 0- 1001/182/183/18 100 – 5002/184/186/18 500 – 1,0002/184/186/18 > 1,0001/182/183/18 P(Y)6/1812/18 Independent Random Variables (Reclamation Example cont.)

34. 2-34 Mine Size and Reclamation Independent? 1. Does P(X|Y) = P(X) for all values X and Y? Does P(Mine 0-100|Reclaim) = Does P(Mine 0-100|Reclaim) = P(Mine 0-100 and reclaim)/P(Reclaim) P(Mine 0-100 and reclaim)/P(Reclaim) (2/18)/(12/18) = 2/12 = 1/6 (2/18)/(12/18) = 2/12 = 1/6 P(mine 0 - 100 ) = 3/18 = 1/6 P(mine 0 - 100 ) = 3/18 = 1/6 Holds for these values Holds for these values Check if holds for all values X and Y Check if holds for all values X and Y If not hold for any values dependent If not hold for any values dependent Independent Random Variables – Check 1

35. 2-35 2. Does P(X,Y) = P(X)*P(Y) for all values X and Y? P(Mine 0-100 and reclaim) = 2/18 = 1/9 P(mine 0-100)* P(reclaim) = 3/18*12/18 = 36/324 = 1/9 = 36/324 = 1/9 Holds for these values Checking it holds for all values X and Y Checking it holds for all values X and Y If not hold for any values – dependent If not hold for any values – dependent Independent Random Variables – Check 2

36. 2-36 3. Does F(X,Y) = F(X)*F(Y) for all values X and Y? P(Mine<500 and reclaim) = 2/18 + 4/18 = 1/3 P(Mine < 500)* P(reclaim) =6/18+3/18 = 9/18*12/18 = 9/18*12/18 =108/324 = 1/3 =108/324 = 1/3 Holds for these values Checking if holds for all values X and Y Checking if holds for all values X and Y If not hold for any values – dependent If not hold for any values – dependent Independent Random Variables – Check 3

37. 2-37 Are Copper and Zinc Ore Grades Independent: Continuous Case f(x,y) = 0.324e -x-y Discrete: P(X|Y) = P(X) Continuous f(x|y) = f(x) f(x|y) = f(x,y)/f(y) = 0.324e -x-y /(-0.324e -0.3-y+ 0.324e -y ) f(x|y) = f(x,y)/f(y) = 0.324e -x-y /(-0.324e -0.3-y+ 0.324e -y ) ?= (-0.324e -0.4-x+ 0.324e -x ) ?= (-0.324e -0.4-x+ 0.324e -x ) Discrete: P(X,Y) = P(X)*P(Y) Continuous: f(x,y) = f(x)*f(y) 0.324e -x-y? = (-0.324e -0.4-x+ 0.324e -x )*(-0.324e -0.3-y+ 0.324e -y ) Discrete: P(X  x,Y  y) = P(X  x)*P(Y  y)

38. 2-38 Convolutions - Example density function of their sum U=X+Y : X, Y ~ e -X-Y Y = U-X

39. 2-39 Integrate Exponential Functions Integrate Exponential functions Let u = x 2 du/dx = 2x dx = du/2x From oil well example :

40. 2-40 discrete probability distribution size Pennsylvania coal mine P(X = 1) = 2 -1 = 1/2 P(X = 2) = 2 -2 = 1/4 P(X = 3) = 2 -3 = 1/8, etc. Changing of Variables Example

41. 2-41 MRM wants probability distribution of reclamation bond amounts where reclamation = U U = g(X) = X 4 + 1 What is pdf of U P(X = 1) = 2 -1 = 1/2 =>P(U =X 4 + 1= 1 4 + 1=2)= 1/2 P(X = 2) = 2 -2 = 1/4 =>P(U =X 4 + 1= 2 4 + 1=17)= 1/4 P(X = 3) = 2 -3 = => P(U =X 4 + 1= ) = P(X = x) = 2 -x => P(U =X 4 + 1) Changing of Variables

42. 2-42 P(X = x) = 2 -x = 1/2 x => P(U =X 4 + 1) = 1/2 x Want in terms of U U = X 4 + 1 Solve for X = (U – 1) (1/4) P(U = x) = 2 -x = 2 (U – 1)^(1/4) for U = (1 4 +1, 2 4 +1, 3 4 +1,.. Changing of Variables Discrete Example

43. 2-43 Changing Variables General Case Discrete X ~ Px(X) for X = a,b,c,.. U = g(X) => X = g -1 (U) U = g(X) => X = g -1 (U) U ~ Pu(U) = Px(g -1 (U)) for g(a), g(b), g(c),... Continuous X ~ fx(X) where fx(X) = f(X) for a < X < b fx(X) = 0 elsewhere U = g(X) => X = g -1 (U) U = g(X) => X = g -1 (U) U ~ fu(U) = fx(g -1 (U)) for g(a) < U < g(b)

44. 2-44 Normal Practice Variable Change Problem X ~ N( ,  ) –  < X <  f(X) = 1 exp(-(X-  ) 2 /(2  2 )) dX  (2  ) 0.5  (2  ) 0.5 Show that (X –  ) /  ~ N (0,1) –  < Z < 

45. 2-45 Convolutions Extending Variable Change to Joint Distributions X, Y ~ f(X,Y) Want density function of sum U = X + Y

46. 2-46Convolutions special case X and Y are independent, f (x,y) = f 1 (x) f 2 (y) and previous equation is reduced to following:

47. 2-47 Example: X (Oil Production) and Y (Gas Production) independent random variables f(x,y) = e -2x* 3e -3y Find density function of their sum U=X+Y? g(u) =  0  2e -2x* 3e -3(u-x) dx = complete this example = complete this example Convolutions - Example

48. 2-48 Sum Up Random Variables and Probability Distributions (PS 2) Discrete Probability Distributions values with probabilities attached Cumulative Discrete Probability Functions Continuous Probability Distributions

49. 2-49 Chapter 2 Sum up Discrete Joint Distributions P(X=x,Y=y) = f(x,y) Independent Random Variables P(X|Y) = P(X) for all values X and Y P(X,Y) = P(X)*P(Y) F(X,Y) = F x (X)*F y (Y) f(x,y) = f 1 (x)*f 2 (y)

50. 2-50 Changing Variables Discrete X ~ Px(X) for X = a,b,c,.. U = g(X) => X = g -1 (U) U = g(X) => X = g -1 (U) U ~ Pu(U) = Px(g -1 (U)) for g(a), g(b), g(c),... Continuous X ~ fx(X) where fx(X) = f(X) for a < X < b fx(X) = 0 elsewhere U = g(X) => X = g -1 (U) U ~ fu(U) = fx(g -1 (U)) for g(a) < U < g(b)

51. 2-51 Convolutions: Extending Variable Change to Joint Distributions Convolutions: Extending Variable Change to Joint Distributions X, Y ~ f(X,Y) Want density function of sum U = X + Y

52. 2-52Convolutions Special case X and Y are independent, f (X,Y) = f1(X) f2(Y) and previous equation is reduced to following: