2G1503 Simulation and Modeling Exercise #3
Ali Ghodsi [email protected]
1
Statistics (chapter 5) PDF 1/3
• Probability Density Function (PDF)
– For continous domains, function is usually called f
• Ex: range Rx = ℝ any real number
– Swedish: täthetsfunktion
• Probability Mass Funktion (PMF)
– For discrete domains, function is usually called p
• Ex: range Rx = { 1, 2, 3, 4, 5, 6 } for a dice
– Swedish: frekvenskfunktion
• Cumulative Distribution Function (CDF) derived from the PDF/PMF
– Swedish: fördelningsfunktion
– For the PDF:
x
–
F ( x) =
∫ f (t )dt
−∞
–
– For the CDF:
– F ( x) =
p( y )
∑
y≤ x
2
1
Statistics (chapter 5) PDF 2/3
• Probability Density Function (PDF)
– Satisfies:
a ) f ( x ) ≥ 0 for all x
b)
∫ f ( x )dx = 1
Rx
∞
=
∫ f ( x )dx = 1
−∞
c ) f ( x ) = 0 if x not in Rx
• Probability Mass Funktion (PMF)
– Satisfies:
a ) p ( x ) ≥ 0 for all x in Rx
b)
∑
p(x) = 1
x ∈ Rx
3
Statistics (chapter 5) PDF 3/3
• Cumulative Distribution Function (CDF) derived
from the PDF/PMF
– Satisfies:
a) F is non − decreasing, a ≤ b ⇒ F (a ) ≤ F (b)
b) lim x → ∞ F ( x) = 1
c) lim x → −∞ F ( x) = 0
4
2
Inverse Transform Technique
• To generate variates (sampling from
a distribution) according to a pdf f
1.
2.
3.
4.
Calculate the CDF for the PDF: F(X)
Set F(X) = R, and define the Range
Calculate the inverse, F-1, of F
Generate random numbers [0,1) and
use the inverse F-1, to generate variates
5
Random Variates: exponential pdf
• Create random variates according
to the pdf f(x)=λe-λx
1. F(x) =
1 - e-λx
for x≥0
0
for x<0
2. F(x)=R, the range is x≥0
3. Calculate F-1, -ln(1-R)/ λ
4. Use random numbers on [0,1) to sample.
Ex: 0.3 : -ln(1-0.3)/ λ
6
3
Random Variates: triangular 1/4
• Generate random variates
accoring to the triangular
distribution:
f(x) = x,
0≤x≤1
2-x,
1<x≤2
7
Random Variates: triangular 2/4
Calculate CDF:
 x2
+ C1 ,
0 ≤ x ≤1

0 ≤ x ≤1
 x,
CDF : F ( x) =  2 2
f ( x) = 
 x − 2, 1 < x ≤ 2
2 x − x + C 2 , 1 < x ≤ 2

2
C1 and C2 need to be calculated :
C1 = 0, since we know that F (0) = 0
We know that F (1) =
2 *1 −
12
1
x2
1
+ 0 = . So when x = 1 we have, 2 x − + C2 =
2
2
2
2
12
1
+ C2 =
⇒ C2 = − 1
2
2
8
4
Random Variates: triangular 3/4
Calculate the inverse function F-1 :
x2
, when 0 ≤ x ≤ 1
2
x2 = 2R
x = ± 2R
⇒
R =
F −1 ( r ) =
R = 2x −
2r
when
1
(x − 4 x + 2 )
2
2 − 2 R = ( x − 2) 2
F
F ( 0 ) ≤ r ≤ F (1)
complete square
(kvadratkomplettering)
x2
− 1, when 0 ≤ x ≤ 1
2
R =−
−1
Choose plus, as F-1
shouldn’t be negative
(r ) = 2 −
2 − 2r
(
)
⇒
1
( x − 2)2 − 4 + 2
2
Choose minus,
x = 2 ± 2 − 2R
when
F (1) ≤ r ≤ F ( 2 )
⇒
R =−
as F-1(0.5)=1
9
Random Variates: triangular 4/4
Calculate the inverse function F-1 :
1

2r ,
0≤r ≤

2
F −1 (r ) = 
1
2 − 2 − 2 r
< r ≤1
2

Generate random numbers, R1, ..., Rn,
uniformly distributed on [0,1)
F-1(R1),..., F-1(Rn), will be the sample from the
specified triangular distribution
10
5