ETM 607 – Random-Variate Generation Inverse-transform technique - Exponential Distribution - Empirical Continuous Distributions - Normal Distribution?

ETM 607 – Random-Variate Generation

• Inverse-transform technique- Exponential Distribution- Empirical Continuous Distributions- Normal Distribution?

• Discrete Distributions- Empirical- Discrete Uniform- Geometric

• Acceptance-Rejection Technique- Poisson Process- Non-stationary Poisson Process

Random-Variate Generation – Chapter 8:

Random-Variate generation is converting from a random number (Ri) to a Random Variable, Xi ~ some distribution.

Inverse transform method:Step 1 – compute cdf of the desired random variable XStep 2 – Set F(X) = R where R is a random number ~U[0,1)Step 3 – Solve F(X) = R for X in terms of R. X = F-1(R).Step 4 – Generate random numbers Ri and compute desired random variates:

Xi = F-1(Ri)

ETM 607 – Random Number and Random Variates

Recall Exponential distribution: is often thought of as the arrival rate, and 1/ as the time between arrivals

Insert figure 5.9

2

1][

1][

0,1

0,0)(

,0

0,)(

XV

XE

xe

xxF

otherwise

xexf

x

x

ETM 607 – Random-Variate Generation

Inverse transform method – Exponential Distribution:Step 1 – compute cdf of the desired random variable X

Step 2 – Set F(X) = R where R is a random number ~U[0,1)

Step 3 – Solve F(X) = R for X in terms of R. X = F-1(R).

Step 4 – Generate random numbers Ri and compute desired random variates:


xeRxF 1)(

)ln(1

),1ln(1

,1 RXorRXeR x

0,1

0,0)(

xe

xxF x

)ln(1

ii RX

Empirical Continuous Distribution:

2 distinct methods:

• Few observations• Many observations


Empirical Continuous Distribution – Few Observations:

Given a random variable X which has an empirical continuous distribution, and the following observed values:

0.7 1.6 2.8 1.9 First sorting the values smallest to largest, a cdf can be generated, assuming X can take on values between 0 and the largest observed value.



Considering the inverse-transform method, one needs to find the slope of each line, ai:


n

xx

nini

xxa iiii

i /1/)1(/)1()1(

And knowing that each line segment begins at x(i-1) , then:

n

iRaxX

and

a

xX

n

iRXF

iii

i

iii

)1(

1)(

)1(

)1(


In-class Exercise:

Create a random-variate generator for the variable X which has an empirical continuous distribution, but assume there are technological reasons the value can never be below 2.5, or above 5.0.

Also, the following values were observed: 2.7 3.6 4.8 3.9 4.2

What are the values of X, given random numbers: .05, .27, .75, and .98


Empirical Continuous Distribution – Many Observations:

Major difference from “few observations” approach – don’t use 1/n to construct cdf, use n/N, where N is the total number of observations, and n is the observed number of observations for a given interval of the random variable X.

For the slope, ai:

(length of the interval) / (relative frequency of observations within the interval)

Where ci is the value of the cumulative distribution at xi.

See example next slide.


iiii cRaxX )1(

Empirical Continuous Distribution – Many Observations:

Insert example 8.3


Distributions with no closed form cdf:

What if a distribution has no closed form cdf (e.g. normal, gamma, beta)?

An inverse transform approximation for the Normal Dist.:

Insert Table 8.4


1975.

)1()(

135.0135.01 RR

RFX

Discrete Distributions – Empirical:

Recall discrete distribution means the random variable X, takes on countable values (integers).

For example, observations for the number of shipments per day were:

Then, insert fig 8.6


x p(x) F(x)

0 0.5 0.5

1 0.3 0.8

2 0.2 1.0

x

x

x

x

xF

20.1

218.0

105.0

00

)(

Discrete Distributions – Empirical:

Since, insert fig 8.6

Then,


x

x

x

x

xF

20.1

218.0

105.0

00

)(

0.18.0,2

8.05.0,1

5.0,0

R

R

R

X

Discrete Uniform Distribution:

Inclass Exercise

Consider a random variable X which follows a uniform discrete distribution between and including the numbers [5,10]. Develop an inverse transform for X with respect to R.


Geometric Distribution:Probability mass function (pmf):

Cumulative distribution function (cdf):


1

0

11

)1()(

x

x

j

j

p

ppxF

...2,1,0,1

...3,2,1,)1()( 1

xpp

xppxpx

x

Geometric Distribution:Knowing the cdf:

Using the inverse transform method:


x

x

pxF

pxF

11)1(

11)( 1

1)1ln(

)1ln(

)1ln(

)1ln(1

)1ln(

)1ln(

)1(1)1(

)1(111

)()1(

1

1

p

RX

p

Rx

p

R

pRp

pRp

xFRxF

xx

xx

Random Variate generation for a discrete random variable X with a Poisson Distribution:

Recall the pmf:

– referred to as the rate parameter (or mean number of occurrences per unit time)

Poisson process is the number of arrivals from an exponential inter-arrival stream with mean time (1/).

Then over one unit of time, x arrivals occur if and only if:

Where Ai is the inter-arrival time for the ith unit.


otherwisexp

xx

exp

x

,0)(

...2,1,0,!

)(

12121 ....1.... xxx AAAAAAA

Random Variate for a Poisson Distribution:Recall inverse transform method for exponential distribution:

Then,

Becomes:

This leads to a acceptance-rejection algorithm.


12121 ....1.... xxx AAAAAAA

)ln(1

ii RX

1

11

1

11

1

11

1

11

lnln

)ln()ln(

)ln(1

1)ln(1

x

ii

x

ii

x

ii

x

ii

x

ii

x

ii

x

ii

x

ii

ReR

RR

RR

RR

Using:

Step 1: Set n = 0, P = 1.

Step 2: Generate a random number Rn+1, and replace P by P*Rn+1.

Step 3: If P < e, then accept N = n. Otherwise, reject the current n, increase n by one, and return to step 2.

In Class Exercise, Given X~Poisson( = 6 arrivals per unit time), use table A.1 (94737, 08225, 35614, 24826, 88319, 05595, 58701, 57365, 74759, etc…) to generate the first value for X for unit time.


1

11

x

ii

x

ii ReR

Nonstationary Poisson Process (NSPP):

A poisson arrival process whose arrival rate (i) changes over time.Think “fast food”. Arrival rates at the lunch and dinner hour much greater than arrival rates during “off hours”.

Thinning Process:Generates Poisson arrivals at the fastest rate, but “accept” only a portion of the arrivals, in effect thinning out just enough to get the desired time-varying rate.


Nonstationary Poisson Process (NSPP):

Suppose arrival rates of 10, 5 and 15 for the first three hours. Thinning (in effect) generates arrivals at a rate of 15 for each of the three hours, but accepts approximately 10/15th’s of the arrivals in the first hour and 5/15th’s of the arrivals in the second hour.

Original

Thinned


0 1 2 3

0 1 2 3

Nonstationary Poisson Process (NSPP) – Thinning Algorithm:

To generate successive arrival time (Ti) when rates vary:

Step 1 – Let max{t} be the maximum arrival rate, and set t = 0and i = 1.

Step 2 – Generate E from the exponential distribution with rate and let t = t+E (the arrival time of the next arrival using max rate).

Step 3Generate random number R from U[0,1). If R < (t)then Ti = t and i = i + 1.

Step 4 – Go to step 2.


)ln(1

RE

Documents

ETM 607 – Random-Variate Generation Inverse-transform technique - Exponential Distribution - Empirical Continuous Distributions - Normal Distribution?