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Proof

1

Simulation Lecture Notes
Wanlin Juan
September 2019

1

Simulation From a Known Distribution

Assumptions

We know how to simulate from U(0,1)
...
, xn ∼ U (0, 1), i
...
d
...
rk ,
...

M
yi
E
...
M=50000
...
Simulate from an exponential distribution x ∼ exp(λ)
Density function: f (x) = λe−λx 1{x>0}

Algorithm
1
...
Compute x = − λ1 log(u)
Claim: x ∼ f (x)

Reason

∀t > 0,
1
P (x ≤ t) = P (− log(u) ≤ t)
λ
= P (u ≥ e−λt )
= 1 − P (u < e−λt )
= 1 − e−λt)
Z t
=
λe−λs ds
0
Z t
=
fexp (s)ds
0

2

B
...
Simulate u ∼ U (0, 1)
2
...
g
...
Then x = − λ1 log(1 − t) = F −1 (t)

C
...
2
...
Simulate u1 , u2 ,
...
Compute x = u1 +
...
Yn ∼ g(•),

√ ¯
n(Y − EY )
p
→ N (0, 1), n → ∞
V ar(Y )

E(u) = 21
R1
V ar(u) = E(u2 ) − (E(u))2 = 0 u2 du − ( 12 )2 =

1
12

Q: n=12 large enough?
A: It is enough for u ∼ U (0, 1)
...
Simulate u1 ∼ U (0, 1) and calculate θ = 2πu1 ∼ U (0, 2π)
√
2
...
Calculate x = rcosθ, y = rsinθ
Claim:
x ∼
N (0, 1),
1), x is independent with y

 y∼ N (0, 
x
0
1 0
∼N
,
y
0
0 1
Joint density function:
2
2
1
1 − x2 +y2
1
2
f (x, y) = φ(x)φ(y) = √ e−x /2 √ e−y /2 =
e
2π
2π
2π

Take the Box-Muller transformation,
r=

p

x2 + y 2

x
cosθ = p
2
x + y2
Cartesian coordinate (x, y) ↔ Polar coordinate (r, θ)
f (r, θ) = f (x, y)|J|
(Jacobian Matrix)

D1
...
Call Algorithm for normal distributions to get x1 ,
...
Calculate y = x21 +
...
Simulate x1 ,
...
Calculate y = x21 +
...
Simulate from F-distribution
F-distribution: X =

Y1 /d1
Y2 /d2 ,

where Y1 ∼ χ2d1 , Y2 ∼ χ2d2 , Y1 and Y2 are independent
...
Simulate from multivariate normal distribution
 

µ1
σ12
ρσ1 σ2
x=
∼N
,
µ2
ρσ1 σ2
σ22





x1
µ1
 x2 



 ∼ N  µ2  , Pp×p 
More generally, x=

...


xp
µp


x1
x2





Algorithm:
1
...
Calculate x1 = µ1 + ρσ1 z1 , x2 = µ2 + ρσ2 z1 + 1 − ρ2 σ22 z2



 

x1
µ1
σ12
ρσ1 σ2
Claim: x=
∼N
,
x2
µ2
ρσ1 σ2
σ22
Why: Linear combinations of normal ⇒ still normal
Thus, we only need to check
E(x1 ) = µ1 + ρσ1 z1 = µ1
q
E(x2 ) = µ2 + ρσ2 z1 + 1 − ρ2 σ22 z2 = µ2
V ar(x1 ) = σ12 V ar(z1 ) = σ12
V ar(x2 ) = ρ2 σ22 V ar(z1 ) + (1 − ρ2 )σ22 V ar(z2 ) = σ22
q

(1 − ρ2 )σ22 z2 )
q
= Cov(σ1 z1 , ρσ2 z1 ) + Cov(σ1 z1 , (1 − ρ2 )σ22 z2 )
q
= ρσ1 σ2 V ar(z1 ) + σ1 (1 − rho2 )σ22 Cov(z1 , z2 )

Cov(x1 , x2 ) = Cov(σ1 z1 , ρσ2 z1 +

= ρσ1 σ2

5

Check 

σ1 p
0
σ1
AAT =
2 )σ 2
ρσ
(1
−
ρ
0
2
2


P
σ12
ρσ1 σ2
=
=
ρσ1 σ2
σ22
More generally,
suppose

 
x1
 x2  
 
Then x=

...

µp



p ρσ2
(1 − ρ2 )σ22

p×p

1
T
can
s
...
AA
 = 4 
 find A 
z1
 z2 


p×p



+A

...

µp

Scalar version:
z ∼ N (0, 1)
x = c + az ∼ N (c, a2 )

Pp×p
Call Algorithm to solve for AAT =
Algorithm:
1
...
, zp ∼ N (0, 1), i
...
d
2
...
, zp )T
Claim:
x ∼ N (µ,

X

)

This leads to:
for i = 1, 2,
...

For j = 1, 2,
...
,p
Set vi = σij ;
For k = 1, 2,
...
Simulate from double exponential distribution (Laplace Distribution)
x ∼ fDE (x)
|x−µ|

1 − θ
Density: fDE (x) = 2θ
e
, for x ∈ (−∞, +∞)
Rx
|x−µ|
CDF: FDE (x) = −∞ fDE (t)dt = 12 + 12 sign(x − µ) • {1 − e θ }
−1
FDE
(t) = µ − θ • sign(t − 12 ) • log{1 − 2|t − 21 |}, for t ∈ (0, 1)

Algorithm1: use algorithm in Part B
...
Simulate u1 , u2 ∼ U (0, 1)
2
...
If u2 < 0
...
Return x
Claim: x ∼ fDE (x; µ, θ)

G1
...
Simulate u ∼ U (0, 1)
2
...
Simulate from Binomial Distribution
Algorithm
1
...
un ∼ U (0, 1), i
...
d
2
...
+ 1{un ≤p}
Claim: x ∼ Binomial(n, p)
Why: x =

Pn

i=1 {Independent

Bernoulli}

G3
...
))
Multinomial (1, (p1 , p2 ,
...


Algorithm
1
...
Set (x1 , x2 , x3 ) = (1{0Claim: x ∼ M ultinomial(1,(p1 , p2 ,
...
Simulate from Multinomial Distribution
Algorithm
1
...
, un P
∼ U (0, 1)
Pn
Pn
n
2
...
))

H
...


e−λ λk
k!

Connections between Poisson(λ) and exp(λ):
Poisson x is the number of events in [0,1], when the time between consecutive events are i
...
d
exponential(λ)
...
, Tx , Tx+1 ∼ exp(λ) ⇒ x ∼ P oisson(λ)

Algorithm
1
...
While(s ≤ 1)

8

Simulate u ∼ U (0, 1)
...
Return x = k − 1
Alternatively, recall U ∼ U (0, 1) ⇒ X = F −1 (u) ∼ F (•)

Q: In discrete case, how to define F −1 (u)?
A: x = F −1 (u) =the largest integer such that F (x) ≤ u

Algorithm(numerical)
1
...

Generate u ∼ U (0, 1)
2
...
Return x

9

2

Simulation Techniques and Methods

A
...

Assumption
1
...

2
...


R
R
Remark C has to be more than 1, since f (x)dx = g(x)dx = 1
...


Algorithm
1
...

f (z)
, set x = z (accept z), and then go to 3;
2
...

3
...

Claim: x ∼ f (x)

Reason(intuitively)

10

Reason(in math)
Define
Aj = {Accept z as x in the j − th loop}
= {Reject loop 1} ∩ {Reject loop 2} ∩
...

C
C

We can easily verify that
+∞
X

P (Aj ) =

(1 −

j=1

j=1

Algorithm converges!
Rejection rate at each loop is 1 −

+∞
X

1
C , if

P (x ≤ t) =

=

1 j−1 1
)
=
...
So we can only
use Laplace distribution to generate Normal distribution, but cannot use Normal distribution to
get Laplace
...
Importance Sampling Method
Goal

To simulate x ∼ f (x), which is difficult
...
We know f(x) (up to a constant)
...
g
...
Exists a candidate distribution (helper) g(z) close to f(z) and easy to simulate from
...


The assumptions are similar to the rejection sampling but weaker than rejection sampling
...
g
...

Algorithm
1
...
, zN ∼ g(z)
2
...
zN according to the probability
P (x = zj |z1 ,
...
, zN ) = PN
g(zj )
j=1 wj

3
...
, zN ) = PN

k=1

P (X ≤ t|z1 ,
...
zN )P (X = zk |z1 ,
...
zN ) •

−∞

Remark
For computing efficiency, try to find g(z) close to f(z), so that wj ’s are not too close to 0 or 1
...

Z
E[a(x)] = a(x)f (x)dx
Z
a(x)f (x)
g(x)dx
=
g(x)


a(z)f (z)
=E
g(z)
≈

N
1 X a(zi )f (zi )
N i=1 g(zi )

=

N
1 X
a(zi )wi
N i=1

13

C
...


Assumption We know how to simulate from f (x|y, z), f (y|x, z) and f (z|x, y)
...

Algorithm
1
...
For k=0, 1,
...
Stop after a large number N iterations,
set (x, y, z) = (x(N ) , y (N ) , z (N ) ) and return (x,y,z)
Claim: When N is large, (x, y, z) ∼ f (x, y, z)
Remarks
1
...

2
...

Markov Chain:

 (1) 
 (2) 
 (N ) 
x(0)
x
x
x
 y (0)  →  y (1)  →  y (2)  →
...
Junker from CMU)
Need to simulalte


x
y


∼ f (x, y) =

2x + 3y + 2
, f or 0 < x < 2, 0 < y < 2
...
We can use Rejection
Sampling and Importance Sampling to simulate from f(x,y), using uniform distribution as the
helper
...

f (x|y) =

f (x, y)
f (X, y)
2x + 3y + 2 3y + 4
2x + 3y + 2
=R
=
/
=
f (y)
28
14
6y + 8
f (x, y)dx

f (y|x) =

f (x, y)
f (X, y)
2x + 3y + 2 2x + 5
2x + 3y + 2
=R
=
/
=
f (x)
28
14
4x + 10
f (x, y)dy
14

Suppose we know how to simulate from f (x|y) for a fixed y and f (y|x) for a fixed x
...
Set k = 0 and starting point (x(0) , y (0) ) = (1, 1)
2
...

x(k+1) ∼ f (x|y (k)
y (k+1) ∼ f (y|x(k)
3
...
5y 2 + 2(x + 1)y
2x + 3y + 2
dt =
F (y|x) =
f (t|x)dt ==
4x + 10
4x + 10
0
0
Solve u = F (y|x) for y
...
Simulate u ∼ U (0,
q1)
2
...

F −1 (v|y) =

p

v(6y + 8) + (1
...
5y + 1)

1
...
Given y, set x = v(6y + 8) + (1
...
5y + 1)

15

D
...
−→ xk −→
...
, xk ,
...

Theorem 1
...
−→ xk −→
...
Key Condition: There exists a density function f(x) such that (*) holds
...
The chain is also π-irreducible(the set of possible values for xk not change for all k =
0, 1, 2,
...
The chain is aperiodic(not going to the same value periodically)
...

Recall Gibbs,
x(k+1) ∼ f (x|y (k) , z (k) )
y (k+1) ∼ f (y|x(k+1) , z (k) )
z (k+1) ∼ f (z|x(k+1) , y (k+1) )
⇒p(x(k+1) , y (k+1) , z (k+1) |x(k) , y (k) , z (k) )
=f (x(k+1) |y (k) , z (k) )f (y (k+1) |x(k+1) , z (k) )f (z (k+1) |x(k+1) , y (k+1) )

16

Z Z Z

p(x(k+1) , y (k+1) , z (k+1) |x(k) , y (k) , z (k) )f (x(k) , y (k) , z (k) )dx(k) dy (k) dz (k)

Z Z Z

f (x(k+1) |y (k) , z (k) )f (y (k+1) |x(k+1) , z (k) )f (z (k+1) |x(k+1) , y (k+1) )f (x(k) , y (k) , z (k) )dx(k) dy (k) dz (k)
Z

Z Z
f (x(k) , y (k) , z (k) )dx(k) dy (k) dz (k)
=
f (x(k+1) |y (k) , z (k) )f (y (k+1) |x(k+1) , z (k) )f (z (k+1) |x(k+1) , y (k+1) )
Z Z
=
f (x(k+1) |y (k) , z (k) )f (y (k+1) |x(k+1) , z (k) )f (y (k) , z (k) )dy (k) dz (k) f (z (k+1) |x(k+1) , y (k+1) )
Z Z
=
f (x(k+1) , y (k) , z (k) )f (y (k+1) |x(k+1) , z (k) )dy (k) dz (k) f (z (k+1) |x(k+1) , y (k+1) )
Z
= f (x(k+1) , z (k) )f (y (k+1) |x(k+1) , z (k) )dz (k) f (z (k+1) |x(k+1) , y (k+1) )
Z
= f (x(k+1) , y (k+1) , z (k) )dz (k) f (z (k+1) |x(k+1) , y (k+1) )

=

=f (x(k+1) , y (k+1) )f (z (k+1) |x(k+1) , y (k+1) )
=f (x(k+1) , y (k+1) , z (k+1) )
Verify (*)!

E
...

Goal
from
...

M-H Algorithm
1
...

2
...
, choose a new xk+1 as follows
a
...
Accept xk+1 = z by the following probability
α(xk , z) = min{1,

f (z)/f (xk )
}
g(z|xk )/g(xk |z)

Otherwise set xk+1 = xk
3
...
Stop!
Claim: When N is large, x∗ ∼ f (x)
Reason: verify (*)
Special cases (Algorithms)
1
...
Independent Chain M-H Algorithm: g(z|x) = g(z)
(z)g(xk )
⇒ In M-H, change α(xk , z) = min{1, ff (x
}
k )g(z)
(MCMC always depends on the previous one, even if x and z are independent
...
Initial value: x0 = 0
2
...
Simulate z ∼ g(z) = ft3 (z)
...
Simulate u ∼ Bernoulli(α(xk , z)), where




ft (z)[1 − sin(20z)]1|z|≤3 ft3 (xk )
[1 − sin(20z)]1|z|≤3
α(xk , z) = min 1, 3
= min 1,
ft3 (xk )[1 − sin(20xk )]ft3 (z)
[1 − sin(20xk )]
Set


xk+1 =

z, if u = 1
xk , if u = 0

3
...
Pick {x5000 , x5050 ,
...

Claim: {x5000 , x5050 ,
...
Pearson Correlation Coefficient
(Continuous x and y
...
)
Population Version
ρ(X, Y ) = p

EXY − EXEY
=p
2
V ar(X)V ar(Y )
[EX − (EX)2 ][EY 2 − (EY )2 ]
Cov(X, Y )

Sample Version
ρˆ =
where s1 =

1
n−1

Pn

i=1 (Xi

¯ 2 , s2 =
− X)

1/n

1
n−1

P

Pn

¯ Y¯
Xi Yi − X
s1 s2
(Yi − Y¯ )2

i=1

1
...

2
...

3
...
Spearman’s Correlation
(Good for ordinal data and skew data
...
Then
E[F1 (X)F2 (Y )] − E[F1 (X)]E[F2 (Y )]
q
1
1
12 × 12
Z Z
=12E[F1 (X)F2 (Y )] − 3 = 12
F1 (x)F2 (y)f (x, y)dxdy − 3

ρs (X, Y ) =ρ(F1 (X), F2 (Y )) =

Sample Version Sort the samples within X’s and Y’s separately
...
, Xn ⇒ r1

(Y )

(X)

, r2

,
...
, Yn ⇒ r1 , r2 ,
...
ρˆs → ρs when n → ∞ (consistent estimator)
2
...
X and Y are not linearly related, but F1 (X) and F2 (Y ) are linearly related
...
Kendall’s τ
(Also related to ranks
...
)
19

Population Version
Suppose (X, Y ) ∼ F (X, Y ), (X 0 , Y 0 ) ∼ F (X, Y )
τk = P {(X − X 0 )(Y − Y 0 ) > 0} − P {(X − X 0 )(Y − Y 0 ) < 0}
Concordance: (X − X 0 )(Y − Y 0 ) > 0
Discordance: (X − X 0 )(Y − Y 0 ) < 0
X and Y positively correlated ⇒ 0 < τk ≤ 1
X and Y negatively correlated ⇒ −1 ≤ τk < 0
X and Y independent ⇒ τk = 0

Sample Version
Suppose
dataset (x1 , y1 ),
...
For each pair, we say (xi , yi ), (xj , yj )
...

τˆk =

# of positive pairs # of negative pairs




−
n
n
Pn

=

i=1

Pn 2

2

j=1 [1{(xi −xj )(yi −yj )>0} − 1{(xi −xj )(yi −yj )<0} ]


n
2

Remark
1
...
When n is large, computation can be a little slow
...
Further/Alternative expression:
Z Z
τk = 4
F (X, Y )f (X, Y )dXdY − 1
4
...


D
...

Independence ⇒ ρ = 0, ρs = 0, τk = 0
Let’s start with a random vector of length = 2 (general case length > 2)
...
It is a way to describe the dependence
between X and Y
...
F1 (x) is the marginal distribution of X, F2 (y) is the
marginal distribution of Y
...
Vice versa, since F1 (x) = F (x, y)dy, F2 (y) = F (x, y)dx
...

C(t, s) =

21

X, Y are independent⇒ C(t, s) = ts ⇒ τk = 4

RR

ts(tds + sdt) − 1 = 0

More generally, for random vector (x1 , x2 ,
...
, xn )
...
, 0 ≤ tn ≤ 1, C(t1 ,
...
, Fn−1 (tn )), where
F (x1 ,
...

Similar as before,
F (x1 ,
...
, Fn (xn ))
Again, Joint distribution F (x1 ,
...
, tn )
...


E
...
xn ) such that xi ∼ Fi (x) for any distribution Fi (•), ∀i = 1, 2,
...
For Multivariate normal distribution: see (1
...

2
...
D)
...
Marginal distributions Fi (x), i = 1, 2,
...

2
...


Algorithm
Call Cholesky Decomposition algorithm to get A, where AAT = Σ
...
Simulate z1 ,
...
Set z ∗ = (z1∗ ,
...
, zn )T
z∗
3
...
, n, set ui = Φ( σii ) ∼ U (0, 1), xi = Fi−1 (ui ) ∼ Fi (x)
...

xn1

x12
x22

...
x1p

...


...
, x
˜10
...

Covariance Matrix: Σ = Cov(x∗ ), where x∗i = Φ−1 (F (xi ))
Then call Gaussian Copula to generate x
˜1 ,
...


22

F
...
, xn ) such that marginal distribution xi ∼
Fi (x)
...


Algorithm
Call Cholesky Decomposition algorithm to get A, where AAT = Σ
...
Simulate z1 ,
...

2
...
, zn∗ )T = A(z1 ,
...
Simulate s ∼ χ2r
√
z∗ / σ
For i = 1, 2,
...
, n
...

Note: You can use step function to estimate Fi (ui ), and then calculate Fi−1 (ui )
...
Archimedean Copulas
For a given strictly decreasing function, h(t) : [0, 1] 7→ [0, +∞), we can get a copula
C(t, s) = h−1 (h(t) + h(s))
Examples
1
...
Clayton Copula: hC (t) =

−θt

−1
3
...
Define Brownian Motion
One dimensional Standard Brownian Motion
{W (t)|t ∈ [0, T ]}, has the following properties:
1
...
W(t) is continuous in t
3
...
W (t) − W (s) ∼ N (0, t − s)
Notation: W (t) ∼ BM (0, 1) W (t) = W (t) − W (0) ∼ N (0, t)
Brownian Motion BM(µ, σ 2 )
drift: µ
diffusion coefficient: σ 2
X(t) = µt + σW (t), X(0) = 0
dX(t) = µ + σdW (t)
Brownian Motion BM(µ(t), σ 2 (t))
R
R
X(t) = µ(t)dt + σ(t)dW (t)
dW (t) = µ(t)dt + σ(t)dW (t)

B
...
, W (tn ) at a fixed time path t0 = 0 < t1 <
...

In essence, we need to simulate

W (t1 ) − W (0)
 W (t2 ) − W (t1 )



...


...
 = D 

...


...


...

tn − tn−1

t1
0


 0 t2 − t1
 ∼ N 0, D 



...

0
0
24


...


...




0
 T
0
D 



...

1



0

0 
,Σ = D



...

1
...

1
...


...

√
√
...


...


...

tn − tn−1



...

0



...

√

...
Simulate z = (z1 ,
...
Set (W (t1 ),
...
, zn )T
3
...

Claim: When n is large, a good approximation to BM(0,1)
Algorithm 1-b (Random Walk)
1
...
, n,
√
iteratively update W (ti ) = W (ti−1 ) + ti − ti−1 zi , zi ∼ N (0, 1)
2
...

Simulate from BM (µ, σ 2 )
X(t) = µt + σW (t), ∀i = 0, 1,
...
< tn = T
X(ti ) − X(tj ) = µ(ti − tj ) + σ(W (ti ) − W (ti−1 ))
Algorithm
1
...

2
...
, n,
X(ti ) = X(ti−1 ) + µ(ti − ti−1 ) + σ

p

ti − ti−1 Zi , Zi ∼ N (0, 1)

3
...

Claim: When n is large, the path is from BM (µ, σ 2 )
...
Set X(t0 ) = 0
...
For i = 1, 2,
...
Connecting neighboring points to become a path
...


C
...

Basic task: For a given W(u) and W(t) at time u and t, 0 ≤ u < t ≤ T
...

Recall
 1×1   1×1
 1×1 

1×k
µX
X
σ11
σ12
∼
N
,
k×1
k×k
Y k×1
µk×1
σ21
σ22
Y
2
X|Y ∼ N (µX|Y , σx|Y
)

µX|Y = µX + σ12 Σ−1
22 (Y − µY )
2
σX|Y
= σ11 − σ12 Σ−1
22 σ21



X 1×1
Y 2×1




  
u
0
W (s)
↔  W (u)  ∼ N  0  ,  u
u
0
W (t)


u
s
s

µX = 0, µY = 0
2
σX

= s, σ12 = (u, s)


u u
Σ22 =
u t

Thus,
2
W (s)|W (u), W (t) ∼ N (µs|u,t , σs|u,t
)

(t − s)W (u) + (s − u)W (t)
t−u
(s − u)(t − s)
=
,∀ 0 ≤ u < s < t ≤ T
t−u

µs|u,t =
2
σs|u,t

For a given BM(0,1), say W (t0 ) = X0 , W (t1 ) = X1 ,
...
Multidimensional Brownian Motion



X1 (t)
X(t) = 
...

Wd (t)


has the following properties:
1
...
Continuous sample paths
3
...
Set W(0)=0
...
For i = 1,
...
t
...
Set X(t0 ) = 0
...
For i = 1,
...
Set X(t0 ) = 0
...
For i = 1, 2,
...
Geometric Brownian Motion
• A continuous-time stochastic process in which the logarithm of the random varying quantity
follows a Brownian Motion
...

• A stochastic process S(t) is GBM(µ, σ 2 ) if it follows the differential equation
dS(t)
= µdt + σdW (t)
S(t)

(∗)

Advantages
:
• Mathematical simple and trackable
...

• Percentages of change

)−S(tn−1 )
S(t2 )−S(t1 )
,
...


Ito’s integration solution of (*)
S(t) = S(0)e(µ−σ

2

/2)t+σW (t)

E[S(t)] = S(0)eµt
2

V ar[W (t)] = [S(0)]2 e2µt (eσ t − 1)
∀0 ≤ r < t ≤ T, S(t) = S(r)e(µ−σ

2

/2)(t−r)+σ[W (t)−W (r)]

Algorithm for GBM(µ, σ 2 ) (Random Walk Construction)
For i = 1,
...
CIR (Interest Rate Model)
By Cox, Ingersoll and Ross (1985)
...


dr(t) = a(b − r(t))dt + σ

p
r(t)dW (t)

Where W (t) ∼ BM (0, 1), a > 0, b > 0
...
If 2ab > σ 2 , then r(t) remains strictly non-negative for all t
(almost surely)
...

p
• The diffusion term σ r(t) decreases to 0 as r(t) approaches the origin and it also prevents
r(t) from taking negative values
...
, n, set
r(ti ) = r(ti−1 ) + a(b − r(ti−1 ))(ti − ti−1 ) + σ

p

p
r(ti−1 ) ti − ti−1 Zi , Zi ∼ N (0, 1)

G
...

N(t) is a counting process, and Y1 ,
...

In particular, suppose the random arrival times of the jumps are 0 < τ1 <
...

We often model (not always):
• τj − τj−1 ∼ exp(λ) ⇔ N (t) ∼ P oisson(λt)
• Yt ∼ lognormal, i
...
d
Ito’s representation of (*) (omit derivatives)
N (t)

S(t) = S(0)e

(µ−σ 2 /2)t+σW (t)

Y
j=1

29

Yj

Algorithm to simulate S(t) for 0 = t0 < t1 <
...
For i = 1,
...
, YN (ti−1 )+D=YN (ti ) ∼ lognormal

Alternatively, we can simulate the jump exactly
...

Within each (τk , τk+1 ), we can just use the regular way to simulate from GBM(µ, σ 2 ) for the
grids
...
The estimation uncertainty
is quantified by ”confidence” which is related to ”repeated experiments”
...

To account the uncertainty, a Bayesian approach treat parameter to be random
...
Bayesian Approach Steps
1
...

2
...

3
...
The posterior
distribution could be very complex, but we can simulate from the distribution
...


B
...

31

1
σ2

Inproper prior: approximated by a proper prior
e
...
Normal mean µ: π(µ) ∝ c
Approximation: π(µ) ∼ N (0, 1000000)
But uniform prior depends on parameterigation
...
Now if my interest is ψ = θ2 , uniform
prior ψ ∼ U (0, 1) is not equivalent to θ ∼ U (0, 1)

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Title: Statistical simulation lecture notes
Description: There are chapters: simulation from a known distribution, simulation techniques and methods(rejection sampling, importance sampling, etc.), Random vectors and copulas, generating sample path, Bayes, Bootstrap. It's from Rutgers University in USA.

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