1. 1 Introduction to Stochastic Models GSLM 54100

2. 2 Outline  conditional probability & Binomial  recursive relationships  examples of similar random phenomena

3. 3 Conditional Probability

4. 4 A former Mid-Term Question (Compared to Example 3.4 in Ross ) Example 3.4 in Ross Example 3.4 in Ross  three types of cakes, chocolate, mango, and strawberry in a bakery  each customer choosing chocolate, mango, and strawberry w.p. 1/2, 1/3, and 1/6, respectively, independent of everything else  profit from each piece of chocolate, mango, and strawberry cake ~ $3, $2, and $1, respectively  4 cream cakes sold on a particular day  (a).Let X c be the number of chocolate cream cakes sold on that day. Find the distribution of X c.  (b).Find the expected total profit of the day from the 4 cream cakes.  (c).Given that no chocolate cream cake is sold on that day, find the variance of the total profit of the day.

5. 5 A former Mid-Term Question (Compared to Example 3.4 in Ross )  (a). X c = the number of chocolate cream cakes sold on that day  X c ~ Bin(4, 1/2)  (b). E(total profit of the day) = E(3X c + 2X m + X s ) = 3E(X c ) + 2E(X m ) + E(X s ) = 6+(8/3)+(2/3) = 28/3.

6. 6 A former Mid-Term Question (Compared to Example 3.4 in Ross )  (c). Given that no chocolate cream cake is sold on that day, find the variance of the total profit of the day.  given X c = 0, each cake is of mango of probability 2/3 and of strawberry of probability 1/3.  (X m |X c = 0) ~ Bin(4, 2/3) and (X s |X c = 0) ~ Bin(4, 1/3).  V(X m |X c = 0) = V(X s |X c = 0) = 8/9  the total profit = (Y|X c = 0) = 2(X m |X c = 0) + (X s |X c = 0)  (X m |X c = 0) + (X s |X c = 0) = 4  (Y|X c = 0) = 4 + (X m |X c = 0)  V(Y) = V(4+X m |X c = 0) = V(X m |X c = 0) = 8/9

7. 7 Recursive Relationships

8. 8 Two Innocent Equations  A and B : two events  P(A) = P(A|B)P(B) + P(A|B c )P(B c )  generalization:  j B j =  and B i  B j =  for i  j  P(A) =  useful recursive equations finding probabilities and expectations by these equations

9. 9 Recursive Relationship  a special property in some random phenomena: changing back to oneself, or to something related  flipping a coin until getting the first head flipping a coin until getting the first head first flip = T first flip = H THE END flipping a coin until getting the first head one flip +

10. 10 Recursive Relationship random phenomenon type 1 outcome simple type 1 problem........ type k outcome simple type k problem type k+1 outcome difficult type k+1 problem random phenomenon type 1 outcome simple type 1 analysis........ type k outcome simple type k analysis type k+1 outcome type k+1 problem being related to the original random phenomenon the problem may become easy if the type k+1 problem is related to the original random phenomenon

11. 11 More Recursive Relationships random phenomenon A type 1 outcome simple type 1 problem type 2 outcome difficult problem related to random phenomenon B random phenomenon B type 1’ outcome simple type 1’ problem type 2’ outcome difficult problem related to random phenomenon A

12. 12 About Recursive Relationships  more forms, possibly involving more than 2 random phenomena  identifying the relationships among random phenomena being an art, not necessarily science

13. 13 Examples of Recursive Relationships (The identification of a similar structure is an art.)

14. 14 Exercise 3.1.2 of Notes  n contractors bidding for m projects (n  m)  one project for each contractor  all projects being equally profitable  random independent bids by contractors  A i = project i, i  m, is bid (by at least one contractor)  (a).Find  (b).Find P(A 1 )  (c).Find  (d).Find P(A 2 | A 1 )

15. 15 Exercise 3.1.2 of Notes  A i = the project i, i  m is bid (by at least one contractor)  (a).  (b). P( A 1 ) =  (c).  (d). to find P(A 2 | A 1 ), note that random bids by n contractors on m projects (n  m), one project for each contractor

16. 16 E(X) = E[E(X|Y)] ..  suppose that B y = {Y = y} ..  define indicator variable  E(1 A ) = P(A)

17. 17 E(X) = E[E(X|Y)]  let X = 1 A ..  the expressions being true in general  E(X) = E[E(X|Y)]  P(A) = E[P(A|Y)]

18. 18 E(X) = E[E(X|Y)]  discrete X and Y  E(X|Y = y) =  x xP(X = x|Y = y)  E[E(X|Y)] =  y E(X|Y = y)P(Y = y)

19. 19 c.f. Example 3.11 of Ross (Expectation of a Random Sum) Example 3.11 of RossExample 3.11 of Ross  a tailor-shop equally likely to sell 0, 1, or 2 suits per day  net profit from a sold suit $300 to $800 dollars  customer arrivals and prices all being independent  mean profit per day = ?  P(net profit in a day  7) = ?

20. 20 c.f. Example 3.11 of Ross (Expectation of a Random Sum) Example 3.11 of RossExample 3.11 of Ross  N = number of suits sold in a day  equally likely to be 0, 1, 2  X j = profit from the jth sold suit, j = 1, 2, 3  uniform 3 to 8 (hundreds)  all random variables being independent with each other  S = total profit in a day,

21. 21 c.f. Example 3.11 of Ross (Expectation of a Random Sum) Example 3.11 of RossExample 3.11 of Ross ..  hard to find E(S) from the distribution of S  finding E(S) without using its distribution

22. 22 c.f. Example 3.11 of Ross (Expectation of a Random Sum) Example 3.11 of RossExample 3.11 of Ross  E(S) = (0+5.5+11)/3 = 5.5  another way

23. 23 c.f. Example 3.11 of Ross (Expectation of a Random Sum) Example 3.11 of RossExample 3.11 of Ross  to find P(S > 7)  P(S > 7) = E[P(S > 7|N)] 4 4 3 8 38

24. 24 Example 3.12 of Ross Example 3.12 of Ross  X ~ geo (p)  E(X) = E(X|X=1)P(X=1) + E(X|X>1)P(X>1)  E(X|X = 1) = 1  E(X|X > 1) = 1 + E(X)  E(X) = (1)p + (1 + E(X))(1  p), i.e. E(X) = 1/p flipping a coin until getting the first head first flip = T first flip = H THE END flipping a coin until getting the first head one flip +

25. 25 Example 3.4.3 of Notes  find E(X) for X = max{Y, Z}, where Y & Z ~ i.i.d. unif[0,1]  E(X) = E[E(X|Y)]

26. 26 Example 3.13 of Ross Example 3.13 of Ross  A miner is trapped in a mine containing three doors. The first door leads to a tunnel that takes him to safety after two hours of travel. The second door leads to a tunnel that returns him to the mine after three hours of travel. The third door leads to a tunnel that returns him to his mine after five hours. Assuming that the miner is at all times equally likely to choose any one of the doors, what is the expected length of time until the miner reaches safety?

27. 27 Example 3.13 of Ross Example 3.13 of Ross every time equally likely to choose one of the three doors 2 hr 3 hr 5 hr

28.  X = time taken  Y = door initially chosen  E[X|Y = 1] = 2  E[X|Y = 2] = 3 + E[X]  E[X|Y = 3] = 5 + E[X]  E[X] = (2+3 +E[X]+5+E[X])/3  E[X] = 10 28 Example 3.13 of Ross Example 3.13 of Ross

29. 29 Example 3.5.4 of Notes  Comparing Two Exp Random Variables  X ~ exp(  ) and Y ~ exp( ), independent  P(X > Y) = E[P(X > Y|Y)]  P(X > Y|Y = y) = P(X > y) = e  y  P(X > Y) = E[P(X > Y|Y)] = E[e  Y ] =

30. 30 Example 3.5.3 of Notes  Random Partition of Poisson  Z ~ Poisson( )  each item being type 1 w.p. p, 0 < p < 1, and type 2 o.w., independent of everything else  X = number of type 1 items in Z  P(X = k) = E[P(X = k|Z)] =

31. 31 Ex. #4 of WS#10  #1.(Solve Exercise #4 of Worksheet #10 by conditioning, not direct computation.) Let X and Y be two independent random variables ~ geo(p).  (a).Find P(X = Y).  (b).Find P(X > Y).  (c).Find P(min(X, Y) > k) for k  {1, 2, …}.  (d).From (c) or otherwise, Find E[min(X, Y)].  (e).Show that max(X, Y) + min(X, Y) = X + Y. Hence find E[max(X, Y)].

32. 32 Ex. #4 of WS#10  (a). different ways to solve the problem  by computation:  P(X = Y) = = = = = p/(2-p)

33. 33 Ex. #4 of WS#10  by conditioning:  P(X = Y|X = 1, Y = 1) = 1  P(X = Y|X = 1, Y > 1) = 0  P(X = Y|X > 1, Y = 1) = 0  P(X = Y|X > 1, Y > 1) = P(X = Y)  P(X = Y) = P(X = 1, Y = 1)(1) + P(X > 1, Y > 1)P(X = Y)  P(X = Y) = p 2 + (1  p) 2 P(X = Y) i.e., P(X = Y) = p/(2  p)

34. 34 Ex. #4 of WS#10  (b) P(X > Y)  by symmetry  P(X > Y) + P(X = Y) + P(X < Y) = 1  P(X > Y) = P(X < Y)  P(X > Y) = =

35. 35 Ex. #4 of WS#10  by direct computation  P(X > Y) =

36. 36 Ex. #4 of WS#10  by conditioning  P(X > Y) = E[P(X > Y|Y)]  P(X > Y|Y = y) = P(X > y) = (1-p) y  E[P(X > Y|Y)] = E[(1-p) Y ]

37. 37 Ex. #4 of WS#10  yet another way of conditioning  P(X > Y|X = 1, Y = 1) = 0  P(X > Y|X = 1, Y > 1) = 0  P(X > Y|X > 1, Y = 1) = 1  P(X > Y|X > 1, Y > 1) = P(X > Y)  P(X > Y) = P(X > 1, Y = 1) + P(X > 1, Y > 1)P(X > Y) = (1-p)p + (1-p) 2 P(X > Y)  P(X > Y) = (1  p)/(2  p)

38. 38 Ex. #5 of WS#10  In the sea battle of Cape of No Return, two cruisers of country Landpower (unluckily) ran into two battleships of country Seapower. With artilleries of shorter range, the two cruisers had no choice other than receiving rounds of bombardment by the two battleships. Suppose that in each round of bombardment, a battleship only aimed at one cruiser, and it sank the cruiser with probability p in a round, 0 < p < 1, independent of everything else. The two battleships fired simultaneously in each round.

39. 39 Ex. #5 of WS#10  (b)Now suppose that initially the two battleships aimed at a different cruiser. They helped the other battleship only if its targeted cruiser was sunk before the other one.  (i)What is the probability that the two cruisers were sunk at the same time (with the same number of rounds of bombardment).

40. 40 Ex. #5 of WS#10  (b). (i). Let N i be the number of rounds taken to sink the ith cruiser. N i ~ Geo (p); N 1 and N 2 are independent.  p s = P(2 cruisers sunk at the same round) = P(N 1 = N 2 )  discussed before

41. 41 Stochastic Modeling  given a problem statement  formulate the problem by defining events, random variables, etc.  understand the stochastic mechanism  deduce means (including probabilities, variances, etc.)  identifying special structure and properties of the stochastic mechanism In the sea battle of Cape of No Return, two cruisers of country Landpower (unluckily) ran into two battleships of country Seapower. With artilleries of shorter range, the two cruisers had no choice other than receiving rounds of bombardment by the two battleships. Suppose that in each round of bombardment, a battleship only aimed at one cruiser, and it sank the cruiser with probability p in a round, 0 < p < 1, independent of everything else. The two battleships fired simultaneously in each round. (b) Now suppose that initially the two battleships aimed at a different cruiser. They helped the other battleship only if its targeted cruiser was sunk before the other one. (i) What is the probability that the two cruisers were sunk at the same time (with the same number of rounds of bombardment).

42. 42 Examples of Ross in Chapter 3 Examples of Ross in Chapter 3  Examples 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.11, 2.12, 3.13

43. 43 Exercises of Ross in Chapter 3 Exercises of Ross in Chapter 3  Exercises 3.1, 3.3, 3.5, 3.7, 3.8, 3.14, 3.21, 3.23, 3.24, 3.25, 3.27, 3.29, 3.30, 3.34, 3.37, 3.40, 3.41, 3.44, 3.49, 3.51, 3.54, 3.61, 3.62, 3.63, 3.64, 3.66