Probability Distributions Lecture Notes PDF

Modelling and Simulation. 403 ‫ﺣﺴﺐ‬ Lect-06: Some of the known Probability Distributions..‫ﻋزاﻟدﯾن ﻛﺎﻣل أﻣﯾن ﻣﺻطﻔﻰ‬.‫ﻛﻠﯾﺔ اﻟﻌﻠوم اﻟرﯾﺎﺿﯾﺔ‬ ‫ﺟﺎﻣﻌﺔ اﻟﺧرطوم‬  Content. l Bernoulli Distributions. l Poisson Process. l Genera...

Modelling and Simulation. 403 ‫ﺣﺴﺐ‬ Lect-06: Some of the known Probability Distributions..‫ﻋزاﻟدﯾن ﻛﺎﻣل أﻣﯾن ﻣﺻطﻔﻰ‬.‫ﻛﻠﯾﺔ اﻟﻌﻠوم اﻟرﯾﺎﺿﯾﺔ‬ ‫ﺟﺎﻣﻌﺔ اﻟﺧرطوم‬  Content. l Bernoulli Distributions. l Poisson Process. l Generalization …. l Single Server Queue. l Random Variables. l Normal Distribution. l Probability Distribution l Standard Normal Curve Funstions (Cumulative Tables. Probability Distribution Functions) l Probability Density Functions (Probability Mass Functions) l Memoryless Property (Markov Property) l Poisson Distribution. 2/ 2 BERNOULLI TRIALS …. PROBABILITY MASS FUNCTION VS PROBABILITY DENSITY FUNCTION.  Bernoulli Trials. l A random experiment that has two outcomes, ”success” (p) or “failure” (q). Now consider a compound sequence of n indpendent repetitions of this experiment. This is known as sequence of Bernoulli trial. l Probability of exactly k success after n trials: " ! 𝑃 𝑘 = 𝐶! 𝑝 (𝑞)"#! Verification: ∑"!$% 𝑃 𝑘 = 1? 4/ 4 Binomial Theorem and Beroulli Trials. $ ! 𝐶!$ 𝑝 ! 𝑞 $%! = (𝑝 + 𝑞)$ 𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 !"# = 1  Characteristics of Binomial Distribution l Mean: ! "#$% l &'()'$*+, -! "#$%. l /0'$1'(1#2+3)'0)4$, - "5$%. 6/  Example. l A coin is tossed 6 times, with head as success, Then n=6, p=q=1/2. l The probability that exactly two heads occur is: l B(k; n, p) = B(2; 6, ½) l =C(6,2).(1/2)2(1/2)4 l =6!/(2!4!) ¼. 1/16 =15/64 7/  Example. l A die is tossed 180 times. Then the expected number of 6 is l !###"##$6% "#78967:;"+#?0'$1'(1#2+3)'0)4$#)?, l -###"##5@78967:;#6#A:;B#"#A 8/  Example. l A communication channel with p being the proability of succesful transmission of a bit. Assume we design a code that can tolerate up to e bit error with n bit word code. l Find the probability of transmitting a word succesfully,Pn? l Pn = P(e or fewer errors in n trials) ß (What assumption made?) l = ∑$!"# 𝐶!% (1 − 𝑝)! 𝑞%&! l If no parity of checking, e=0, single error- correction Hamming code e= 1. 9/ 9  Generalization. l Sequence of n independent trials, each trial the result is EXACTLY one of k possibilities, b1, b2, …, bk with probability p1, p2, …, pk such that: pi ≧. 0. and ! - 𝑝& = 1 &$' 10/ 10  Generalization...2 l Let s ∊ S be an element in a sample space: l 𝑠 = {𝑏' 𝑏' 𝑏' … 𝑏' , 𝑏( 𝑏( … 𝑏( , …., 𝑏! 𝑏! … 𝑏! } l 𝑛' , 𝑛(. …. 𝑛! "! "" "# "$ l Such that: 𝑝' 𝑝( 𝑝) … 𝑝! 𝑎𝑛𝑑 ∑!&$' 𝑛& = 𝑛 "! "! "" "# "$ l P(𝑛' , 𝑛( , …. , 𝑛! ) = "! !"" !…."$ !. 𝑝! 𝑝# 𝑝$ … 𝑝% 11/ 11  Generalization...3 l Example: l A diode can function properly with a probability (p1); i.e pass a signal; short circuit (p2) or open circuit (as if it does not exist!) (p3); Assume that I have a series of diods connected in series: B A l What is the probability that it will work properly? 12/ 12  Example …1 lThe circuit will work properly if it has type 1 and type 2 diods but not type 3 (Open circuit: does not pass any signal): ∑"!-',""-',"!/""$" 𝑝(𝑛' , 𝑛( , 0)= 𝑛 "! "" % ∑"!-',""-',"!/""$" 𝑛 , 𝑛 , 0 𝑝' 𝑝( 𝑝) ' ( "! "! "#"! =∑"!-' " ! "#" ! 𝑝' 𝑝( ! ! " "! "#"! "! % " " =∑"!$% 𝐶"! 𝑝' 𝑝( − %!"! 𝑝' 𝑝( = 𝑝' + 𝑝( " − 𝑝(" = =(1 − 𝑝) )" −𝑝(" 13/ 13  Example … l Let use assume that we have the following code: If B then repeat S1 until B1 else repeat S2 until B2. l and assume that S1 = S2 =Print (”hello”) l If we have Pr[B=TRUE]=p; Pr[B1=TRUE)= 3/5 l Pr[B2=TRUE]=2/5 l After a long experiment we found that: l Pr[EXACTLY 3 ”Hello”]=3/25; What is p? 14/ 14  Solution …. l In order to get 3 Hello printed, we can have it through two ways: (B=T, B1 =F, B1=F, B1=T) or (B=F, B2=F, B2=F, B2=T) l P[A = Having Three ”Hello” Printed]=3/25 =P[B=T].P[B1=F]2P[B1=T] + P[B=F]P[B2=F]2P[B2=T] =p.(2/5)2.3/5 + (1-p)(3/5)2(2/5) =3/25 =12p/125+18/125-18p/125=3/25 6p/125 =3/125 15/ 15 P = 3/6=0.5  Random Variables (R.V). We have the probability system: (S, ∑, P) Random Variable (RV) is a variable whose value depends upon outcome of a random experiment. The outcome of our random experiment is an w∊S. We associate a real number X(w) to w. Thus, our rv X(w) is nothing more than a function defined on the sample Space S. X: S à R, Throwing a dice to start a football Game. 16/  Sample Point. P(s). X(s) 111 0.125 3 110 0.125 2 101 0.125 2 100 0.125 1 011 0.125. 2 001 0.125 1 010 0.125 1 000 0.125 0 l Random experiment of three Bernoulli trials. The sample consists of eight triples of 0’s and 1’s. The random variable X(s) is the number of ones 17/ 17 appears in the experiment (See table above)  Probability Distribution Function (PDF) or Cumulative Distribution. 𝑋 ≤ 𝑥 = {𝑤: 𝑋(𝑤) ≤ 𝑥} 1 l PDF is Defined as: 5/8 𝐹& 𝑥 = 𝑃 𝑋 ≤ 𝑥 3/8 Properties: 1. 𝐹& ≥ 0 2. 𝐹& ∞ = 1 -5 5 3. 𝐹& −∞ = 0 4. 𝐹& 𝑏 − 𝐹& 𝑎 = 𝑃 𝑎 < 𝑋 ≤ 𝑏 𝑓𝑜𝑟 𝑎 < 𝑏 5. 𝐹& 𝑏 ≥ 𝐹& 𝑎 𝑓𝑜𝑟. 𝑎 ≤ 𝑏 18/ 18 Probability Density Function (pdf). Or Probability Mass Function (pmf). 19/ 19 Special Discrete Distribution. Geometric pmf. 21/ Memoryless or Markov Property. 22/..... ‫ﺗوزﯾﻊ ﺑوﯾﺳون‬ POISSON DISTRIBUTION. 23  Poisson random variable. l a random variable x, taking on one of the values 0,1,2,3,…. Is said to be a Poisson random variable with parameter C, if for some 𝜆>0 𝜆& #0 𝑝 𝑖 =𝑝 𝑥=𝑖 = 𝑒 𝑖! 24/ Poisson Distribution. l Poisson Distribution of a discrete random variable N which can take value n = 0, 1, 2,... l The probability mass function is: 𝛼 % 𝑒 &; 𝑝 𝑛; 𝛼 = ' 𝑛! , 𝑛 = 0, 1, 2, … 0, 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒 𝑤ℎ𝑒𝑟𝑒 𝛼 > 0 𝑖𝑠 𝑎 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟. 25/ Probability Distribtion Function. Value of N 0 1 2 …. Probaility e-𝛂 𝛂. e-𝛂 (𝛂2. e-𝛂 )/2 … The Characteristics of this distribution is: !"""#""$ %! #"$ % #"& $ 26/  Poisson Process l The counting process { N(t), t>=0) is said to be a Poisson process having rate l , l > 0 if 1) N(0)=0 2) the process has independent increments 3) the number of events in any interval of length t is Poisson distributed with mean 𝛌t. that is, for all s,t >=0 P{N (t + s ) - N ( s ) = n} = e (lt ) / n! - lt n n = 0,1... 27/ l Note: from condition 3) it follows that a Poisson process has stationary increments and also that E[N(t)] =𝛌t b P{a t} = òt f (T1 ) dt1 ¥ = ò le -lT1 dt1 t 28/  " = 𝜆# #$! 𝑒 1 𝑑𝑡% ! 1 #$! " = 𝜆[(− )𝑒 1 ]! 𝜆 !"# $ = [𝑒 < ]# !"# =𝑒  Interarrival Time Distribution consider a Poisson process T1 ⟹ Time of the first event Tn ⟹ Elapsed time between the (n-1)th and the nth event {Tn, n= 1,2,…} ⟹ sequence of the interarrival times T1 T2 T3. Tn t=0 t=T1 (n-1)th nth event event Distribution of poisson Tn - lt p{T1>t} = P{N(t)=0} = e T1 has an exponential distribution with the mean 1/𝛌. P{T2 > t | T1=s} = P{0 events in (s,s+t)|T1=s} = P{0 events in (s,s+t)} - lt = e T2 is also an exponential random variable with the mean 1/ 𝛌, and T2 is independent of T1 Obs: Tn, n=1,2,… are independent identically distributed exponential random variables having mean 1/ l.  Poisson Process. l Let N(t) be a random number of some events in the interval [0, t]. l Assumptions: 1. Events can occur one at a time. 2. The number of events between t and t+s depends on s. 3. The number of events in non-overlapping time intervals are independent random variables. Then: ! !"# (#$)$ Prob[N(t) =n] = 𝑓𝑜𝑟 𝑡 ≥ 0, 𝑛 = 0, 1, 2,.. &! 1. 𝛌 – mean number of events in a unit of time. 33/ 33  Single Server Queue. l For a single server queueing system: l Suppose that the number of arriving customers per hour is a Poisson random variable with a mean 𝛼 =2 customers/hour. l What is the probability of arriving 3 customers during 1 hour? l We have n=3, 𝛼=2, so that: & !".(# p(3; 2) = = 0.18 34/ 34 l )!...... ‫اﻟﺗوزﯾﻊ اﻟطﺑﯾﻌﻲ‬ NORMAL DISTRIBUTION. 35  Use of Normal Distribution Tables. l The normal Distribution with mean ! '$1# 3'()'$*+#-! 1 2#3 " # ". #56265 𝑓 𝑥 = 𝑒 (4 𝜎 2𝜋 l If we substitute Z= (X- !B: -D#E+#F+0, 1 #7 "/( 𝑓 𝑧 = 𝑒 2𝜋 36/  Standard Normal Curve. l From the following graph: l P(-1 ≤ z ≤ +1) = 0.6827; P(-2 ≤ z ≤ +2) = 0.9545; P(-3 ≤ z ≤ +3) = 0.9973; 37/  Example. l Find the area under the standrd normal curve between z= -0.46 and z=2.21? 38/ ‫ب اْﻟَﻌﺎَﻟِﻣﯾَن‬ ِ ّ ‫ي َوَﻣَﻣﺎﺗِﻲ ِﻟﻠﱠـِﮫ َر‬ َ ‫ﺳِﻛﻲ َوَﻣْﺣ َٰﯾﺎ‬ َ ‫ﻗُْل ِإﱠن‬ ُ ُ‫ﺻَﻼِﺗﻲ َوﻧ‬ ‫ﺳِﻠِﻣﯾَن‬ْ ‫ت َوأََﻧﺎ أَﱠوُل اْﻟُﻣ‬ ُ ‫ﺷِرﯾَك َﻟﮫُ ۖ◌ َوِﺑذَِﻟَك أ ُِﻣْر‬َ ‫﴾ َﻻ‬١٦٢﴿ (‫﴾ )اﻷﻧﻌﺎم‬١٦٣﴿ ( 162 ) Say, "Indeed, my prayer, my rites of sacrifice, my living and my dying are for Allah, Lord of the worlds. ( 163 ) No partner has He. And this I have been commanded, and I am the first [among you] of the Muslims."