Geometric Random Variables N ~ Geometric(p) # Bernoulli trials until the first success pmf: f(k) = (1-p) k-1 p memoryless: P(N=n+k | N>n) = P(N=k) –probability.

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Presentation transcript:

Geometric Random Variables N ~ Geometric(p) # Bernoulli trials until the first success pmf: f(k) = (1-p) k-1 p memoryless: P(N=n+k | N>n) = P(N=k) –probability that we must wait k more coin flips for the first success is independent of n, the number of trials that have occurred so far

Previously… Conditional Probability Independence Probability Trees Discrete Random Variables –Bernoulli –Binomial –Geometric

Agenda Poisson Continuous random variables: –Uniform, Exponential E, Var Central Limit Theorem, Normal

Poisson N ~ Poisson( ) N = # events in a certain time period average rate is Ex. cars arrivals at a stop sign –average rate is 20/hr –Poisson(5) = #arrivals in a 15 min period

Poisson pmf: P(N=k) = e - k /k! Excel: POISSON(k,,TRUE/FALSE) =12.5 =3

Poisson N 1 ~Poisson( 1 ), N 2 ~Poisson( 2 ) N 1 +N 2 ~ Poisson( ) Splitting: –Poisson( ) people arrive at L-stop –probability p person is south bound –Poisson(p ) people arrive at L-stop south bound

other slides… from Prof. Daskin’s slides

E and Var X random variable E[g(X)]=∑ k g(k) P(X=k) E[a X+b] = aE[X] +b Var[a X + b] = a 2 Var[X] –always X 1,…,X n random variables E[X 1 +…+ X n ] = E[X 1 ]+…+E[X n ] –always Var[X 1 +…+ X n ] = Var[X 1 ]+…+Var[X n ] –when independent E[X 1 ·X 2 ·…· X n ] = E[X 1 ]·E[X 2 ] ·…·E[X n ] –when independent

E, Var X~Bernoulli(p) E[X]=p, Var[X]=p(1-p) X~Binomial(N,p) E[X]=Np, Var[X]=Np(1-p) N~Geometric(p) E[N]=1/p, Var[N]=(1-p)/p 2 N~Poisson( ) E[N]=, Var[N]= X~U[a,b] E[X]=(a+b)/2, Var[X]=(b-a) 2 /12 X~Exponential( ) E[X]=1/, Var[X]=1/ 2

Central Limit Theorem X 1,…,X n i.i.d, µ=E[X 1 ],  2 =Var[X 1 ] independent, identically distributed S n = X 1,…,X n E[S n ]=nµ, Var[S n ] = n  2 distribution approaches shape of Normal –Normal(nµ,n  2 )

Normal Distribution mean=0  =1  =2  =4

Normal Distribution X 1 ~ N(µ 1,  1 2 ), X 2 ~ N(µ 2,  2 2 ) X 1 +X 2 ~ N(µ 1 +µ 2,   2 2 ) pdf, cdf NORMALDIST(x,µ, , TRUE/FALSE ) fractile / inverse cdf –p=P(X≤z) –NORMINV(p,µ,  )

Newsvendor Problem must decide how many newspapers to buy before you know the day’s demand q = #of newspapers to buy b = contribution per newspaper sold c = loss per unsold newspaper random variable D demand