Tel Hai Academic College Department of Computer Science Prof. Reuven Aviv Probability and Random variables

Random Experiments and Events Do Random Experiment, again and again – get outcomes examples? Same experiment - Outcomes might be different Event: an outcome or subset of outcomes Example: outcomes: Num of packets in router, N – Event : N = 20 – set A {N any of 20, 21, 22, …30} an event? – set E: {N any even number} an event? – set O: {N any odd number}, an event? …..

Probability Define Probability of event A Pr[A] = NumOutcomes(A)/TotalNumExperiments  When Total Number of experiments  ∞  What is Pr[A or B] (Pr [ A U B]) Pr[A or B] = Pr[A] + Pr[B] – Pr[A and B] What if A, B cannot happen in same experiment  E.g. Events E, O Pr[A and B] = 0  Pr[A or B] ≡ Pr[A + B] = Pr[A] + Pr[B]

Q0: The Birthday Paradox n people in a room. What’s the probability that 2 of them has same birthday (day, month)? Experiment: All persons “choose” birthdays at random  Outcome: a series of n numbers from range {1..365} Number of all possible outcomes 365 n Event E (of interest): Set of all outcomes (series) in which 2 of the numbers are the same Other event O: Set of all outcomes with distinct numbers All outcomes are in either E or O.

Q0: The Birthday Paradox Assume birthdays equally probable Pr[O] = (364/365)*(363/365)***([365-(n-1)]/365) let x = 1/365 Pr[O] = (1- x)(1-2x)***(1-(n-1)x) Pr[E] = 1- Pr[O] n = 22  Pr[E] = ; n = 23  Pr[E] = Approximation: use e -x ≈ 1-x small x  Pr[O] ≈ e -n 2 x/2 Pr[E] ≈ 1- e -n 2 x/2

Q1 – Birthday paradox (cont’d) n people in a room. What’s the probability that one of them has same birthday as mine? For what value of n it will be 0.5? Event E: set of all outcomes (series) in which one of the numbers is equal to B (my birthday) Event O: set of all outcomes in which all numbers are different from B Pr[O] = (364/365) n Pr[E] = 1 – (364/365) n (364/365) n = 0.5  n *log 2 (364/365) = -1 n = -1/ log 2 (364/365) =

Random Variables (1) R.V: a variable X that on each experiment gets a value x from a set, with certain probability: Pr[X=x)]≡ Pr[x] Example: LAN States: number of frames transmitted in unit time Define X: number of transmissions in unit time  What are the possible values of X X = 1 if one station has a frame to transmit. Pr[1] = a X = 0 if no station has a frame to transmit. X = k, number of transmitting stations, from N  Pr[k] = u k = [N!/(k!(N-k)!] a k (1-a) N-k Pr[X=0] = ?

Random Variables (2) Example: X is the collision state of the LAN channel during a time unit What could be possible states? X = idle (0) if no frame is transmitted.  Pr[idle] = ? X = transmitting (1) if 1 frame is transmitted.  Pr[tran]=? X = collided (2) if more than 1 frame is transmitted.  Pr[col] Here X is a discrete random variable

Poisson Process N(t) a R.V., number of arrivals during time interval [0..t]  Probability of n arrivals during time 0..t P n (t) ≡ Pr[N(t) = n] = ( t) n e - t /n!  Poisson process with parameter Average number of arrivals = E[N(t)] =  n nPr[N(t)=n] = t Variance also t  is the average rate of arrivals (packets/time unit) P n (1): probability that n packets arrived during 1’st hour t = 0 can be arbitrary!

Probability Functions Probability Mass Function (PMF) Pr[X=x] Cumulative Distribution Function (CDF): ???  F (x) ≡ Pr[X ≤ x]  F X (-∞) = 0, F X (+∞) = 1  F ( x 1 < X ≤ x 2 ) = F X (x 2 ) – F X (x 1 ) Tail Function (Complementary Cumulative Distribution Function) T(x) ≡ Pr[X > x] = 1 – F(x) Poisson: F(x) = e- t  k ( t) k /k! (sum over k from 0 to x)

Q 3 Event Q3: the fourth packet arrived during the first two hours? What is Pr[Q3] Event Q3 occurs if and only if during [0..2] 4, OR 5, OR 6, OR 7, OR any number k>= 4 packets arrived, …  the probability that at least 4 packets arrived during [0..2] Pr[Q3] = P 4 (2) + P 5 (2) + P 6 (2) + … ≡ T 4 (2) Pr[Q3] = T 4 (2) ≡ 1 - P 0 (2) – P 1 (2) –P 2 (2) - P 3 (2) =1 – [ (4/3) 3 ]e -2 Note:  has to be given in units packets/hour

Q4 Event Q4: n th packet arrived during t hours What is Pr[Q4]? Pr[Q4] = the sum of the probabilities that at least n packets arrived during [0..t]  The Tail (complementary Cumulative) function T n (t) Pr[Q4] = T n (t) ≡ P n (t) + P n+1 (t) + P n+2 (t) +…= = 1- P 0 (t) – P 1 (t) –P 2 (t).. = 1-  k P k (t) k = 0..n

Q5 Event Q5: during the 2 nd hour 4 packets arrived What is Pr[Q5] ? A: Since time 0 is arbitrary, we can choose it to be at the end of first hour Pr[Q5] is equal to the probability that during the first hour 4 packets arrived, Pr[Q5] = P 4 (1) = 4 e  /4! Probability depends on the length of the time interval (here 1), not on its location

Q6 Event Q6: the 4 th packet arrived during the 2 nd hour What is Pr[Q6] Event Q6 occurs if and only if: 0 arrived 1 st hour, AND 4 or more arrived 2 nd hour, OR 1 arrived 1 st hour, AND 3 or more arrived 2 nd hour, OR 2 arrived 1 st hour, AND 2 or more arrived 2 nd hour, OR 3 arrived 1 st hour, AND 1 or more arrived 2 nd hour Pr[Q6] = P 0 (1)T 4 (1) + P 1 (1)T 3 (1) + P 2 (1)T 2 (1) + T 3 (1)F 0 (1)

Q6 (alternative method) Consider the probability that the 4 th packet arrived during two hours, (which is T 4 (2)). This is equal to: Pr [4 th packet arrived during 1 st Hr] + (this is T 4 (1)) + Pr[4 th packet arrived during 2 nd Hr] (what we want)  T 4 (2) = T 4 (1) + Pr[Q6]  Pr[Q6] = T 4 (2) – T 4 (1) Pr[Q6], the probability that the 4 th packet arrived during the 2 nd hour is the probability that the 4 th packet arrived during two hours minus the probability that the 4 th packet arrived during one hour

Q7 Event Q7: time until 1 st packet arrived is larger than 1 hour; What is Pr[Q7] ? Define T k arrival time of packet k, k = 1, 2, 3, …  Continuous Random variable, can have any non-zero value We are looking for Pr[Q7] ≡ Pr(T 1 >1) Event Q7 occurs if and only if zero packets arrived during 1 st hour  Pr[Q7] ≡ Pr(T 1 >1) = P 0 (1) = e -

Review: Conditional Probability Consider many experiments, observing two RVs, S 1, S 2 e.g. number of packets in buffer at time steps 1, 2 Count number of events with S 1 = n 1  Then count the fraction of these with S 2 = n 2  Define: Pr[S 2 = n 2 | S 1 = n 1 ] ≡ NumEvents[S 1 = n 1 & S 2 = n 2 ]/NumEvents[S 1 =n 1 ] When TotalNumExperiments  ∞ Pr[S 2 = n 2 | S1 = n 1 ] is the probability of S 2 equal n 2, conditioned on S 1 equal n 1

Conditional Probability (2) Pr[S 2 = n 2 | S 1 = n 1 ] ≡ NumEvents[S 1 = n 1 & S 2 = n 2 ]/NumEvents[S 1 =n 1 ] Divide numerator and denominator on RHS by TotalNumEvents (no change in LHS). Result?? Pr[S 2 =n 2 | S 1 =n 1 ] = Pr[S 2 = n 2, S 1 = n 1 ]/Pr[S 1 = n 1 ]  Pr(S 2 = n 2, S 1 =n 1 ] = Pr[S 2 = n 2 |S 1 = n 1 ]*Pr[S 1 = n 1 ]  Low of Total Probability: Pr(S 2 = n 2 ) =  n 1 Pr(S 2 = n 2 |S 1 = n 1 )*Pr(S 1 = n 1 ) How did we derive this?

Q8 Time interval [0..t+s] consists of 2 non-overlapping intervals [0..t], [t..t+s] Event E1: during full interval [0..t+s] n Poisson arrivals Event E2: during sub-interval [0..t] k arrivals (k < n) What is the conditional probability Pr[E2|E1]? Pr[E2|E1] ≡ Pr[E2 and E1]/Pr[E1] = = Pr [(N(t) = k) and (N(t+s) = n] / Pr[N(t+s) = n] = Pr [(N(t) = k) and (N(s) = n-k] / Pr[N(t+s) = n] Note: arrivals during 0…t are independent of arrivals during t..t+s, so the numerator is a product of probabilities

Q8 (cont’d) Numerator: Pr [(N(t) = k) and (N(s) = n-k] = = Pr[N(t) = k]*Pr[N(s) = n-k] = P k (t)*P n-k (s) = =( t) k e - t ( s) n-k e - s /(k!(n-k)!) Denominator: Pr[N(t+s) = n] = (  t+s)) n e -  t+s) /n! Pr[E2|E1] = [n!/(k!(n-k)!)][t/(t+s)] k [s/(t+s)] n-k  Pr[E2|E1] is independent

Q9 Jar has 5 white balls, 8 red balls. 2 balls are taken out, one after the other, (balls are not returned). X k denotes the color of the k th ball, k = 1, 2  X k = 0 is red ball. X k = 1 is white ball Calculate the conditional probabilities: Given that 2 nd ball was white, the 1 st ball was red  Pr[X 1 = 0| X 2 = 1] Given that 2 nd ball was red, the 1 st ball was red  Pr[X 1 =0|X 2 =0]  We need to calculate the joint and total probabilities

Q9 (Cont’d) Joint Probabilities (sum of these is 1): Red Red: Pr[X 1 =0,X 2 = 0] = (8/13)(7/12) = 14/39 Red White: Pr[X 1 = 0, X 2 = 1] = (8/13)(5/12) = 10/39 White Red: Pr[X 1 = 1, X 2 = 0] = (5/13)(8/12) = 10/39 White White: Pr[X 1 = 1, X 2 =1] = (5/13)(4/12) = 5/39 Total probabilities for first ball (sum of these is 1): 1 st Red: Pr[X 1 = 0] = Pr[X 1 =0, X 2 = 0] + Pr[X 1 =0, X 2 =1] = 24/39 1 st White: Pr[X 1 = 1] = Pr[X 1 =1, X 2 =0] + Pr[X 1 =1, X 2 = 1] = 15/39

Q9 (Cont’d) Total probabilities fro second ball (sum of these is 1) 2 nd Red:  Pr[X 2 = 0] = Pr[X 1 =0, X 2 =0] + Pr[X 1 =1, X 2 =0] = 24/39 2 nd White:  Pr[X 2 = 1] = Pr[X 1 =0, X 2 =1] + Pr[X 1 =1, X 2 =1] = 15/39 Pr[X 1 =0|X 2 =1] = Pr[X 1 =0, X 2 =1]/Pr[X 2 =1] = 2/3 Pr[X 1 =0|X 2 =0] = Pr[X 1 =0, X 2 =0]/Pr[X 2 =0] = 7/12

Q10 Engine  If Engine is faulty, the probability of a flash is 0.99  If Engine is OK, the probability of a flash is 0.01  The engine is faulty only one day out of 100 Given a flash, what’s the probability that engine is faulty? E engine state: E = 1 OK, E = 0 faulty F: Flash state: F = 1 flash, F = 0 no flash Pr[F = 1|E= 0] = 0.99; Pr[F = 1|E = 1] = 0.01 Pr[E = 0] = 0.01; Pr[E = 1] = 0.99 What is Pr[E = 0 |F = 1]

Q10 (Cont’d) By definition of conditional probability Pr[E = 0, F = 1] = Pr[E = 0 | F = 1]*Pr[F=1] Pr[E = 0, F = 1] = Pr[F = 1 | E = 0]*Pr[E = 0] Pr[E = 0 | F = 1]*Pr[F=1] = Pr[F = 1 | E = 0]*Pr[E = 0]  Bayes Formula Pr[E = 0 | F = 1] = Pr[F =1 | E =0]*Pr[E=0] / Pr[F=1] Numerator = 0.99* 0.01 Denominator: What is Pr[F = 1]?

Q10 (Cont’d) The probability that there will be a flash consists of the probability for a flash when the engine is OK (E=1) and the probability for a flash when the engine is faulty (E=0) The Complete Probability Rule: Pr[F=1] = Pr[F=1|E=0]*Pr[E= 0] + Pr[F=1|E=1]*Pr[E=1] = 0.99* *0.99 = 2*0.99*0.01 Pr[E=0 | F = 1] = 0.99*0.01/(2*0.99*0.01) = 0.5 If there is a flash, only 50% chance that engine is faulty!

Q11 1% of the population has a sickness If a person is sick, probability that the test is + is 0.99 If a person is not sick, probability that the test is + is 0.01 Given that the test is +, what is the probability that the person is sick? Bayes: Pr[sick |Test+] = Pr[Test+|sick]*Pr[sick]/Pr[Test+] Complete Probability formula  Pr[Test+] = Pr[Test+|sick]*Pr[sick] +Pr[Test+| not sick] * Pr[not sick] Pr[sick] = 0.01; Pr[not sick = 0.99]; Pr[Test+|sick] = 0.99; Pr[Test+| not sick] = 0.01; Pr[sick| Test +] = 0.05

Q12 Three payphones 1 st, 2 nd, 3 rd. One phone is Bad - never works, another is Reliable - always work, third phone Unstable - work with probability 0.5; We do not know which is which 3 phone tests: 1 st didn’t work, 2 nd worked, 2 nd worked again What’s the probability that 2 nd is the Reliable phone? Event A: 1 st didn’t work, 2 nd worked, 2 nd workd Use Bayes Theorem to find Pr[2 nd Reliable | A]

Q12 (cont’d) Bayes Theorem: Pr[2 nd Reliable | A] = Pr[A | 2 nd Reliable] * (Pr[2 nd Reliable] / Pr[A]) Complete Probability Rule: Pr[A] =  Pr[A|1 st Reliable]*Pr[1 st Reliable]+  Pr[A| 2 nd Reliable]*Pr[2 nd Reliable] +  Pr[A | 3 rd Reliable]* Pr[3 rd Reliable] Pr[1 st Reliable] = Pr[2 nd Reliable] = Pr[3 rd Reliable] = 1/3

Q12 (cont’d) If 2 nd is the Reliable, A will happen when : Either 1 st is Bad, probability ½, definitely wouldn’t work  OR 1 st is Unstable, Pr = ½, probability 0.5 it works Pr[A | 2 nd Reliable] = Pr[1 st didn’t work| 1 st Bad] *Pr[1 st Bad|2 nd Reliable]+ Pr[1 st didn’t work|1 st Unstable]*Pr[1 st Unstable | 2 nd Reliable] =1* Pr[1 st Bad|2 nd Reliable]+0.5Pr[1 st Unstable|2 nd Reliable] = *0.5 = 0.75

Q12 (cont’d) If 1 st is reliable A will never occur  Pr[A|1 st is Reliable] = 0 Calculate Pr[A| 3 rd Reliable] by using chain rule with all states of 2 nd Pr[A | 3 rd Reliable] = Pr[A| 2 nd Unstable]) *Pr[2 nd Unstable| 3 rd Reliable] + Pr[A| 2 nd Bad]*Pr[2 nd Bad| 3 rd Reliable] Id 3 rd is Reliable, 2 nd can be Unstable or Bad with Pr = 1/2 If 2 nd is Unstable event A occur with probability 1/4 If 2 nd is Bad, the event A does not occur Pr[A | 3 rd Reliable] = 0.25* = 1/8

Q12 (cont’d) To sum up: Pr[A] = (1/3( 0 + ¾ +1/8) Pr[2 nd Reliable] = 1/3 Pr[A|2 nd Reliable] = 3/4 Use Bayes Theorem: Pr[2 nd Reliable| A] = Pr[A | 2 nd Reliable] * Pr[2 nd Reliable]/Pr[A] = (3/4)*(1/3)/[(1/3)*(0 + ¾ + 1/8)] = (3/4)/(3/4 + 1/8) = 6/7

Bernoulli Random Variable Bernoulli Experiment: outcomes Success or Failure  Described by Bernoulli Variable X; values ? X gets one of two values: 1 (Success ), 0 (Failure )  Example: a packet arrive/doesn’t arrive to a router Bernoulli PMF: Pr[X=1] ≡ a Pr[X=0] ≡ b = 1 – a Bernoulli CDF: F X (x) ≡ Pr[X ≤ x] = b for all x < 1 How much is  PMF CD F  a + 0*b = a  2 = (1-  ) 2 a + (0-  ) 2 *b   2 = ab

A series of Bernoulli Random Variables Packets arrive to a Router: a series of Bernoulli experiments Described by a series of Bernoulli variables  X j : = 1 (arrival); X j = 0 (no arrival) at a time step j  Assume all X j have same distribution Pr[X j = 1] ≡ a; Pr[X j = 0] ≡ b = 1-a; Other examples: Rain or No-Rain in a certain day Packet with/out error in a stream of arriving packets

First Success Random variable– Geometric Distribution Packets arrive to a Router Random Var T: number of time units up to (inc) first arrival What is Event T = n n-1 failures (no arrivals), then success (arrival) X j = 0 for j = 1, 2,.. n-1. X n = 1  Pr[T = n] = a*b n-1 = a*(1-a) n--1 n ≠ 0 why?  =  n nPr(T = n) =  n na*(1-a) n—1  = 1/a what’s the meaning of   2 =  n (n-  ) 2 Pr(T = n) =  n (n-  ) 2 a*(1-a) n--1  2 = (1-a)/a 2

Counter Random Variable: Binomial PMF Same N Bernoulli Variables: X j = 1 (success),X j = 0 (failure)  Pr[X j = 1] ≡ a Pr[X j = 0] ≡ b = 1- a Bernoulli Success Counter K: counts number of successes  E.g. number of packets arriving during N time units  Pr[K = k] = [N!/(k!(N-k)!]*a k b N-k 0 ≤ k ≤ N   = Na  2 = Nab what’s the meaning of  ? Poisson Approximation: N  ∞ with finite  =Na  Pr[K=k] ≈ [(Na) k /k!] *e -Na = (  k /k!)*e - 

Usage Example: A file is downloaded from a remote site by 500 packets Assume:1 percent of received packets through the channel are in error. What’s the probability that 5 received packets are in error Which PMF should be used? These are 500 Bernoulli experiments, with a = 0.99 ?? Need Probability of 495 successes, 5 failures Pr[K=495] = [500!/(495!*5!)](0.99) 495 (0.01) 5 =  How many packets, on the average, arrived with error?