1 Introduction to Stochastic Models GSLM 54100. 2 Outline conditional probability & Binomial ...

1

Introduction to Stochastic ModelsIntroduction to Stochastic ModelsGSLM 54100GSLM 54100

2

OutlineOutline

conditional probability & Binomial

recursive relationships

examples of similar random phenomena

3

Conditional ProbabilityConditional Probability

4

A former Mid-Term Question A former Mid-Term Question (Compared to (Compared to 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 Xc be the number of chocolate cream cakes sold on

that day. Find the distribution of Xc. (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

A former Mid-Term Question A former Mid-Term Question (Compared to Example 3.4 in Ross )(Compared to Example 3.4 in Ross )

(a). Xc = the number of chocolate cream cakes sold on that day

Xc ~ Bin(4, 1/2)

(b). E(total profit of the day)

= E(3Xc+ 2Xm + Xs)

= 3E(Xc) + 2E(Xm) + E(Xs)

= 6+(8/3)+(2/3) = 28/3.

7

Recursive RelationshipsRecursive Relationships

8

Two Innocent EquationsTwo Innocent Equations

A and B : two events

P(A) = P(A|B)P(B) + P(A|Bc)P(Bc)

generalization: jBj = and BiBj = for i j

P(A) =

useful recursive equations finding probabilities and expectations by these equations

( | ) ( )j j jP A B P B

9

Recursive RelationshipRecursive Relationship

a special property in some random phenomena: changing back to oneself, or to something related

flipping a coin until getting the first head


first flip = T

first flip = H

THE END


one flip +

10

Recursive RelationshipRecursive Relationship

random phenomenon

type 1 outcome

simple type 1 problem.

...

type k outcome simple type

k problemtype 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

More Recursive RelationshipsMore 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

About Recursive RelationshipsAbout Recursive Relationships

more forms, possibly involving more than 2 random phenomena

identifying the relationships among random phenomena being an art, not necessarily science

13

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

14

Exercise 3.1.2 of NotesExercise 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 Ai = project i, i m, is bid (by at least one contractor) (a). Find (b). Find P(A1) (c). Find (d). Find P(A2| A1)

1( )cP A

2 1( | )cP A A

15

Exercise 3.1.2 of NotesExercise 3.1.2 of Notes

Ai = the project i, i m is bid (by at least one contractor)

(a).

(b). P( A1) =

(c).

(d). to find P(A2| A1), note that

1( )cP A

random bids by n contractors on m projects (n m),

one project for each contractor

11 ( )cP A

2 1( | )cP A A

(none of the contractors bids on project 1)P n1(1 )nm

=

2 11 ( | )c cP A A 11

1 (1 )nm

2 2 1 1 2 1 1( ) ( | ) ( ) ( | ) ( )c cP A P A A P A P A A P A

16

EE((XX) = ) = EE[[EE((XX||YY)])]

.

suppose that By = {Y = y}

.

define indicator variable

E(1A) = P(A)

( ) ( | ) ( )y y yP A P A B P B

( | ) ( )( )

( | ) ( )

y

Y

P A Y y P Y yP A

P A Y y f y dy

1, if occurs,

10, . .A

A

o w

17

EE((XX) = ) = EE[[EE((XX||YY)])]( | ) ( )

( )( | ) ( )

y

Y

P A Y y P Y yP A

P A Y y f y dy

let X = 1A

.

the expressions being true in general E(X) = E[E(X|Y)]

P(A) = E[P(A|Y)]

( | ) ( )( )

( | ) ( )

y

Y

E X Y y P Y yE X

E X Y y f y dy

18

EE((XX) = ) = EE[[EE((XX||YY)] )]

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)

( , )

( )xxP X x Y y

P Y y

( , )( )

( )y xxP X x Y y

P Y yP Y y

( , )y x xP X x Y y

( , )x y xP X x Y y ( )x xP X x ( )E X

19

c.f. c.f. Example 3.11 of Ross(Expectation of a Random Sum)(Expectation of a Random Sum)

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

c.f. Example 3.11 of Rossc.f. Example 3.11 of Ross(Expectation of a Random Sum)(Expectation of a Random Sum)

N = number of suits sold in a day equally likely to be 0, 1, 2

Xj = 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, 1

N

jj

S X

21


.

hard to find E(S) from the distribution of S

finding E(S) without using its distribution

1

N

jj

S X

1 1( ) ( ) |

N N

j jj j

E S E X E E X N

1| 0 0

N

jj

E X N

1

1| 1 ( ) 5.5

N

jj

E X N E X

1 21

| 2 ( ) 11N

jj

E X N E X X

22


E(S) = (0+5.5+11)/3 = 5.5

another way

( ) [ ( | )] [5.5 ] 5.5 ( ) 5.5E S E E S N E N E N

1 1| ( ) 5.5

N n

j j jj j

E X N n E X nE X n

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))(1p), i.e. E(X) = 1/p


first flip = T

first flip = H

THE END


one flip

+

25

Example 3.4.3 of NotesExample 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)]

21122

( | )] [max( , )] yyE X Y y E y Z y zdz

21 22 3

( ) [ ( | )] ( )YE X E E X Y E

( | ) ( )( )

( | ) ( )

y

Y

E X Y y P Y yE X

E X Y y f y dy

26

Example 3.13 of RossExample 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


every time equally likely to choose one of the three doors 2 hr

3 hr5 hr

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


29

Example 3.5.4 of NotesExample 3.5.4 of Notes Comparing Two Exp Random Variables 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) = ey

P(X > Y) = E[P(X > Y|Y)] = E[eY]

= = 0y ye e dy

30

Example 3.5.3 of NotesExample 3.5.3 of Notes Random Partition of Poisson 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)] = ( | ) ( )

m kP X k Z m P Z m

!(1 )

!( )! !

mk m k

m k

m ep p

k m k m

(1 )

! ( )!

k mm k

m k

e pp

k m k

0

( )(1 )

! !

k nn

n k

e pp

k n

(1 )( )

!

kpe p

ek

( )

!

p ke p

k

31

Ex. #4 of WS#10Ex. #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


(a). different ways to solve the problem

by computation:

P(X = Y) = =

= =

= p/(2-p)

1( )

kP X Y k

1( , )

kP X k Y k

1( ) ( )

kP X k P Y k

1 1

1(1 ) (1 )k k

kp p p p

33


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) = p2 + (1p)2 P(X = Y)

i.e., P(X = Y) = p/(2p)

34


(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) = =1 ( )

2

P X Y 1

2

p

p

35


by direct computation

P(X > Y) = 1

2 1( , )

i

i jP X i Y j

1

2 1( ) ( )

i

i jP X i P Y j

1 1 1

2 1(1 ) (1 )

i i j

i jp p p p

12 1 1

2 1(1 ) (1 )

ii j

i jp p p

12 1

2

1 (1 )(1 )

ii

i

pp p

p

1 2 2

2(1 ) (1 )i i

ip p p

2

2

1 (1 )

(1 ) 1 (1 )

p pp

p p

1

2

p

p

36


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] 1

1(1 ) (1 )y y

yp p p p

2

(1 )

1 (1 )

p p

p

1

2

p

p

37

Ex. #4 of WS#10Ex. #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)2P(X > Y) P(X > Y) = (1p)/(2p)

38


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


(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


(b). (i). Let Ni be the number of rounds taken to sink the ith cruiser. Ni ~ Geo (p); N1 and N2 are independent.

ps = P(2 cruisers sunk at the same round) = P(N1 = N2)

discussed before

41

Stochastic ModelingStochastic 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

Examples of Ross in Chapter 3Examples 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

Exercises of Ross in Chapter 3Exercises 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

1 Introduction to Stochastic Models GSLM 54100. 2 Outline conditional probability & Binomial ...

Documents

Transcript of 1 Introduction to Stochastic Models GSLM 54100. 2 Outline conditional probability & Binomial ...