04-random-variate generation.ppt
TRANSCRIPT
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Random-Variate Generation
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Inverse Transform Technique --
Exponential Distribution
To generate samples from exponential distribution we use the inverse
transform technique
Step 1. Compute the cdf of the desired random variable X, F(x).
Step 2. Find the inverse of F(x) function
Step 4. Generate uniform random variables R 1, R 2, R 3, … and compute
the desired random variates by
0,0
0,)(
x
xe x f
x
0,0
0,1)(
x
xe x F
x
)1ln(1
)(1)( 1 y y F xe y x F y x
)1ln(1
)(1 iii R R F X
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Inverse Transform Technique --
Exponential Distribution
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Inverse Transform Technique --
Exponential Distribution
Example: Generate 200 variates Xi with distribution
exp( = 1)
Check: Does the random variable X 1 have the desired
distribution?
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Proof?
Can you prove that the numbers you have
generated are indeed samples from an
exponential distribution?
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Other Distributions
Uniform Distribution [ UN(a,b)] (X = a + (b-a)R) Does it really work?
Weibull Distribution
Derive the transformation
Triangular …. The moral is if we can find a closed-form inverse of thecdf for a distribution we can use this method to getsamples from that distribution
0,0
0,)(
1
x
xe x x f
x
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Continuous Functions without a
Closed-Form Inverse
Some distributions do not have a closed form expression
for their cdf or its inverse (normal, gamma, beta, …)
What can be done then?
Approximate the inverse cdf
For the standard normal distribution:
This approximation gives at least one-decimal place
accuracy in the range [0.0012499, 0.9986501]
1975.0
)1()(
135.0135.01 R R R F X
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Discrete Distributions
An Empirical Discrete Distribution
p(0) = P(X=0) = 0.50
p(1) = P(X=1) = 0.30
p(2) = P(X=2) = 0.20
Can we apply the inverse transform technique?
x
x
x
x
x F
2,0.1
21,8.0
10,5.0
0,0
)(
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Discrete Distributions
Let x0 = -, and x1, x2, …, xn, be the ordered probability
mass points for the random variable X
Let R be a random number
iii x X x F R x F if )()( 1
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Discrete Distributions
A Discrete Uniform Distribution
]1,1[
,1
,1,
1,0
)(
.,..,2,1,1
)(
k
xk
i for i xik
i
x
x F
k x
k
x p
)(
1
kRroundup X or
i X k
i R
k
iif
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Discrete Distributions
The Geometric Distribution
Some algebraic manipulation and …
...,2,1,0,11)(
...,2,1,0,1)(
1
x p x F
x p p x p
x
x
1
1ln
1ln
p
Rroundup X
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Acceptance-Rejection Technique
Useful particularly when inverse cdf does not exist inclosed form, a.k.a. thinning
Illustration: To generate random variates, X ~ U(1/4, 1)
R does not have the desired distribution, but R
conditioned ( R’ ) on the event { R ¼} does.
Efficiency: Depends heavily on the ability to minimizethe number of re ections.
Procedures:
Step 1. Generate R ~ U[0,1]
Step 2a. If R >= ¼, accept X=R.
Step 2b. If R < ¼, reject R, return
to Step 1
Generate R
Condition
Output R’
yes
no
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Acceptance-Rejection Technique
Poisson Distribution
N can be interpreted as number of arrivals from a
Poisson arrival process during one unit of time
Then time between the arrivals in the process are
exponentially distributed with rate
...,2,1,0,
!
)()(
n
n
en N P n p
n
1
11
1n
i
i
n
i
i A An N
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Acceptance-Rejection Technique
Step 1. Set n = 0, and P = 1
Step 2. Generate a random number R n+1 and let P = P. R n+1
Step 3. If P < e-, then accept N = n. Otherwise, reject
current n, increase n by one, and return to step 2
How many random numbers will be used on the average to
generate one Poisson variate?
1
11
1
11
1
11
ln1
1ln1
1
n
i i
n
i i
n
i
i
n
i
i
n
i
i
n
i
i
Re R
R R A A
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Direct Transformations
dxe x t
x
2
2
2
1)(
Consider two normal variables Z1 and Z
2
2 2 2
1 2
1 1
2
1 1 2
2 1 2
~ with twodegrees of freedom
(Exponential with parameter 2)
tan ~ [0, 2 ]
2 ln
2 ln cos 2
2ln sin 2
B Z Z Chi square
Z Uniform
Z
B R
Z R R
Z R R
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Direct Transformation
Approach for normal( m ,s 2 ):
Generate Z i ~ N(0,1)
Approach for lognormal( m ,s 2 ):
Generate X ~ N( m ,s 2
)
Y i = e X i
X i = m + s Z
i
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Convolution Method
Erlang Distribution
An Erlang-K random variable X with parameters (K, ) (1/ is the
mean, K is the stage number) can be obtained by summing K
independent exponential random variables each having mean1/(K )
K
i
ii
K
i
K
i
i
R K
R K
X
X X
11
1
ln1
ln1