LECTURE 25 SIMULATION AND MODELING Md. Tanvir Al Amin, Lecturer, Dept. of CSE, BUET CSE 411.
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Transcript of LECTURE 25 SIMULATION AND MODELING Md. Tanvir Al Amin, Lecturer, Dept. of CSE, BUET CSE 411.
LECTURE 25
SIMULATION AND MODELINGMd. Tanvir Al Amin, Lecturer, Dept. of CSE, BUET
CSE 411
Recap : Inverse Transform Method Find cumulative distribution function Let it be F Find F-1
X = F-1(R) is the desired variate where R ~ U(0,1)
Other methods are Accept Reject technique Composition method Convolution technique Special observations
Random variate for Continuous Distribution We already covered for continuous case :
Proof of inverse transform method(Kelton 8.2.1 pg440)
Exponential distribution (Banks 8.1.1) Uniform distribution (Banks 8.1.2) Weibull distibution (Banks 8.1.3) Triangular distribution (Banks 8.1.4) Empirical continuous distribution (Banks 8.1.5) Continuous distributions without a closed form
inverse (Self study – Banks 8.1.6)
Banks Chapter 8.1.7 + Kelton 8.2.1
Inverse Transform method for Discrete Distributions
Things to remember
If R ~ U(0,1), A Random variable X will assume value x whenever,
Also we should know the identities :
)(
)(1
)()1(
1
1
RFx
xRFx
xFRxF
1
1
1
1
xnxnx
xnxnx
nxnnx
nxnnx
We could use this identity also, It results in F-1(R) = floor(x) which is not useful because x is already an integer for discrete distribution.
Discrete Distributions
0 1 2 3
R1 = 0.65
F(x) = F(1) = 0.8
F(x-1) = F(0) = 0.4
X = 1
Empirical Discrete Distribution Say we want to find random variate for
the following discrete distribution.
At first find the F
x p(x)
0 0.50
1 0.30
2 0.20
x p(x) F(x)
0 0.50 0.50
1 0.30 0.80
2 0.20 1.00
2 ,1
21 ,8.0
10 ,5.0
0 ,0
)(
x
x
x
x
xF
Empirical Discrete Distribution
0 1 2 3
0.5
1.0
R1 = 0.73
0.1.80 ,2
8.05.0 ,1
5.0 ,0
R
R
R
X
Discrete Uniform Distribution
Consider kxk
xp ... 3 ,2 ,1 ,1
)(
xk
kxkk
k
xk
xk
x
xF
,1
1,1
32,2
21,1
1,0
)(
k
1
k
2
k
3
k
k 1
1
1 2 3 … K-1 k
Discrete Uniform Distribution Generate R~U(0,1)
k
1
k
2
k
3
k
k 1
1
1 2 3 … K-1 k
, output X= 1k
R1
0
output X= 2k
Rk
21
output X= 3k
Rk
32
, output X= ik
iR
k
i
1
output X= 4k
Rk
43
If generated
Discrete Uniform Distribution
, output X=1k
R1
0
, output X = i
, output 4
, output X=2k
Rk
21
, output X=3k
Rk
32
k
iR
k
i
1
kR
k
43
If generated
RkX
XRki
RkiRk
RkiiRk
iRkRki
iRkRki
iRkik
iR
k
i
output
1
1 and
and 1
and 1
1
1
Imp : Expand this method for a general discrete uniform distribution DU(a,b)
Discrete Uniform Distribution Algorithm to generate random variate
for p(x)=1/k where x = 1, 2, 3, … k
Generate R~U(0,1) uniform random number
Return ceil(R*k)
Geometric Distribution
1)1ln(
)1ln()(
1)1ln(
)1ln(
)1ln()1ln()1(
1)1(
)1(1
)1(1)1()(
...... 2, 1, 0, x where,)1()(
1
1
1
1
0
p
RRF
p
Rx
RPx
Rp
pR
pppxF
ppxp
x
x
xx
j
x
x
)()1( xFRxFRX
Recall :
1)1ln(
)1ln()(1
p
RRFX
Geometric Distribution
Algorithm to generate random variate for
Generate R~U(0,1) uniform random number
Return ceil(ln(1-R)/ln(1-p)-1)
...... 2, 1, 0, x where,)1()( xppxp
Proof of inverse transform method for Discrete case Kelton 8.2.1 Page 444 We need to show that ixpxXP ii allfor )()(
)()()()]()([)(
1)()(0
)(such that i chooses algorithm thesince,
),()( iff ,2 When,
)()( )1,0(~ since,
order) increasingin are the(since
)()( iff ,1for
11
1
1
11
111
iiiiii
ii
i
iii
i
xpxFxFxFUxFPxXP
xFxF
xFU
xFUxFxXi
xpxXPUU
x
xpxFUxXi
Disadvantage of Inverse Transform Method Kelton page 445-448 Disadvantage
Need to evaluate F-1. We may not have a closed form. In that case numerical methods necessary. Might be hard to find stopping conditions.
For a given distribution, Inverse transform method may not be the fastest way
Advantage ofInverse Transform Method Straightforward method, easy to use Variance reduction techniques has an
advantage if inverse transform method is used
Facilitates generation of order statistics.
Kelton 8.2.2
Composition Technique
Composition Technique
Applies when the distribution function F from which we wish to generate can be expressed as a convex combination of other distributions F1, F2, F3, …
i.e., for all x, F(x) can be written as
1
)()(j
jj xFpxF
Composition Algorithm
To generate random variate for Step 1 : Generate a positive random integer
j such that
Step 2 : Return X with distribution function Fj
1
)()(j
jj xFpxF
jpjJP )(
Geometric Interpretation
For a continuous random variable X with density f, we might be able to divide the area under f into regions of areas p1,p2,…(corresponding to the decomposition of f into its convex combination representation)
Then we can think of step 1 as choosing a region Step 2 as generating from the distribution
corresponding to the chosen region
Example
1 ,0
11,1
1
1 ,1
1
0,1
0 ,0
)(
x
xaa)a(
x
axaa
axa)a(
xx
xf
0 a 1-a 1
Example
a
aarea
a)a(
xxF
aa
xxf
1
2/
12)(
)1()(
2
a
aarea
a)(
xxF
axf
1
21
1)(
)1(
1)(
a
aarea
a)(
xaxF
aa
axf
1
2/
1
)1()(
)1(
1)(
Proof of composition technique Kelton page 449