11/15/16 Conditional distributions Let X and Y be discrete r.v.s. - PDF document

11/15/16 Conditional distributions Let X and Y be discrete r.v.s. Conditional probability mass function of X given that Y=y p X | Y ( x | y ) = Pr ( X = x | Y = y ) Conditional distributions, conditional expectation = Pr ( X = x, Y = y ) = p ( x, y ) and the law of total expectation. Pr ( Y = y ) p Y ( y ) X p X | Y ( x | y ) =? x p ( x, y ) p Y ( y ) = p Y ( y ) X X p X | Y ( x | y ) = p Y ( y ) = 1 x x 2 3 Conditional distributions Sum of Poisson random variables is Poisson p X | Y ( x | y ) = Pr ( X = x | Y = y ) X ∼ Poi ( λ 1 ) Y ∼ Poi ( λ 2 ) If X and Y are independent Poisson random variables with Show that X + Y ∼ Poi ( λ 1 + λ 2 ) respective parameters and , calculate the conditional λ 1 λ 2 n Proof: distribution of X, given that X + Y = n. X P ( X + Y = n ) = P ( X = k, Y = n − k ) k =0 n X = P ( X = k ) P ( Y = n − k ) k =0 P ( X = k | X + Y = n ) = P ( X = k, X + Y = n ) n e − λ 1 λ k λ n − k X k ! · e − λ 2 1 2 = ( n − k )! P ( X + Y = n ) k =0 n λ k 1 λ n − k = P ( X = k, Y = n − k ) = e − ( λ 1 + λ 2 ) X 2 k !( n − k )! k =0 P ( X + Y = n ) n = e − ( λ 1 + λ 2 ) n ! X k !( n − k )! λ k 1 λ n − k = P ( X = k ) P ( Y = n − k ) 2 n ! k =0 = e − ( λ 1 + λ 2 ) P ( X + Y = n ) ( λ 1 + λ 2 ) n 4 n ! 5 Conditional distributions Conditional Expectation If X and Y are independent Poisson random variables with Expected value of random variable X given event A respective parameters and , calculate the conditional λ 1 λ 2 distribution of X, given that X + Y = n. X E ( X | A ) = xPr ( X = x | A ) e − λ 1 λ k k ! e − λ 2 λ n − k 1 2 P ( X = k ) P ( Y = n − k ) ( n − k )! x ∈ Range ( X ) = e − ( λ 1 + λ 2 ) ( λ 1 + λ 2 ) n P ( X + Y = n ) n ! Law of Total Expectation (example) λ n − k ✓ n ◆ λ k 49.8% of population male 1 2 = ( λ 1 + λ 2 ) k · ( λ 1 + λ 2 ) n − k k Average height 5’11’’ (men) 5’5’’ (female) The conditional distribution of X, given that X+Y=n is: E ( H ) = E ( H | M ) Pr ( M ) + E ( H | F ) Pr ( F ) Binomial (n, ) λ 1 = 511 12 · 0 . 498 + 5 5 12 · 0 . 502 ( λ 1 + λ 2 ) 6 7 1

11/15/16 Law of Total Expectation Law of Total Expectation X random variable on a sample space S X random variable on a sample space S partition of S partition of S A 1 , A 2 , . . . , A k A 1 , A 2 , . . . , A k X E ( X ) = E ( X | A i ) Pr ( A i ) X E ( X ) = E ( X | A i ) Pr ( A i ) i Version with conditional distributions i X X = xPr ( X = x | A i ) Pr ( A i ) X E ( X ) = E ( X | Y = y ) P ( Y = y ) i x X X = xPr ( X = x | A i ) Pr ( A i ) y x i X X = x Pr ( X = x | A i ) Pr ( A i ) x i X = xPr ( X = x ) X E ( X | Y = y ) = xp X | Y ( x | y ) = X xPr ( X = x | Y = y ) i x x 8 9 Linearity of expectation applies Law of Total Expectation : Application To conditional expectation too!! System that fails in step i independently with probability p X # steps to fail E(X+ Y | A) = E(X | A) + E(Y |A) E(X) ? E(aX + b | A)= a E(X | A) + b Let A be the event that system fails in first step. E ( X ) = E ( X | A ) Pr ( A ) + E ( X | A ) Pr ( A ) = p + (1 + E ( X ))(1 − p ) = 1 + (1 − p ) E ( X ) E ( X ) = 1 p 10 11 Law of Total Expectation : Example Law of Total Expectation : Example A miner is trapped in a mine containing 3 doors. A miner is trapped in a mine containing 3 doors. • D 1 : The 1 st door leads to a tunnel that will take him to safety after 3 • D 1 : The 1 st door leads to a tunnel that will take him to safety after 3 hours. hours. • D 2 : The 2 nd door leads to a tunnel that returns him to the mine after 5 • D 2 : The 2 nd door leads to a tunnel that returns him to the mine after 5 hours. hours. • D 3 : The 3 rd door leads to a tunnel that returns him to the mine after a • D 3 : The 3 rd door leads to a tunnel that returns him to the mine after a number of hours that is Binomial with parameters (12, 1/3). number of hours that is Binomial with parameters (12, 1/3). At all times, he is equally likely to choose any one of the doors. At all times, he is equally likely to choose any one of the doors. E(time to reach safety) ? E(time to reach safety) ? E ( T ) = E ( T | D 1 )1 3 + E ( T | D 2 )1 3 + E ( T | D 3 )1 3 = 3 · 1 3 + 5 · 1 3 + (4 + E ( T )) · 1 3 E ( T ) = 6 12 13 2

11/15/16 Problem Problem The number of people who enter an elevator on the ground floor The number of people who enter an elevator on the ground floor is a Poisson random variable with mean 10. If there are N floors is a Poisson random variable with mean 10. If there are N floors above the ground floor, and if each person is equally likely to get above the ground floor, and if each person is equally likely to get off at any one of the N floors, independently of where the others off at any one of the N floors, independently of where the others get off, compute the expected number of stops that the elevator get off, compute the expected number of stops that the elevator will make before discharging all the passengers. will make before discharging all the passengers. X number of people who enter Pr ( X = k ) = e − 10 10 k Y number of stops k ! ∞ X E ( Y ) = E ( Y | X = k ) P ( X = k ) k =0 E ( Y | X = k ) = E ( Y 1 + . . . + Y N | X = k ) Y i indicates a stop on floor i � 1 − (1 − 1 /N ) k � E ( Y i | X = k ) = 14 15 Game of Craps • Begin by rolling an ordinary pair of dice • If the sum of dice is 2, 3 or 12, the player loses • If the sum of dice is 7 or 11, the player wins • If it is any other number, say k, the player continues to roll the dice until the sum is either 7 or k. • If it is 7, the player loses. • If it is k, the player wins. Let R denote the number of rolls of the dice in a game of craps. • What is E(R)? 16 3