Stochastic Volatility Model and Option Pricing Methodology

This study material discusses the stochastic volatility model and option pricing methodology. It explains the equations and numerical parameters involved in the model. It explores the characteristic function, call price values, Fourier transform, and its practical implementation. It also covers the probability density function and its transform. The material provides insights into the option pricing methodology and analyzes the results for different values of ρSS.

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Contents
Question 1.............................................................................................................................................1
Question 2.............................................................................................................................................5
Question 3.............................................................................................................................................7
Question 4.............................................................................................................................................8
Question 5.............................................................................................................................................9
Question 6...........................................................................................................................................11

Question 1
Let St
(0):=1 denote the underlying asset process and Stein model of the variance process. The
general form of stochastic volatility model is characterized by σ t is being the expected rate of
return process, where Xt is denoting the dividend yield βSS being the volatility of variance.
KSS is the speed of mean reversion BQ and W Qare two independent Q-Brownian motions
generating the filtration Ft, dθSS, which is the mean reversion level. Ft , t∈[0,T] with T>0
possibly has correlated with EQ[(ST −k )+¿∨Ft ¿]. The general form of a stochastic volatility
model is,
dSt:=σ t St dBt
Q, S0>0,
dσ t:= KSS(θSS−σ t)dt+ βSS( ρSS dBt
Q+√ 1−ρSS
2 dW t
Q), σ 0>0,
The equation is obtained by forming a risk-free portfolio, where this time it involves a
position in an extra derivative as compared to the constant volatility case, due to the second
source of randomness, or namely the variance stochastic dynamics. The price of a derivative
contract V should be satisfied.
dσ t:= KSS (θSS−σ t)dt+βSS(ρSS dBt
Q+√1−ρSS
2 dW t
Q), σ 0>0
Let us apply the values of numerical parameters as follows,
S0:= 100, σ 0:=θSS ≔0.1, KSS ≔2 , ρSS:=0.5 βSS:=0.05, T:=1 ρS:=0.5
dσ t: = 2 (0.1−σt ) dt+0.05 (0.5 dBt
Q+√1−(0.5)2 dW t
Q), σ 0>0
∂ σ
∂ t =2 ( 0.1−σt ) ∂ σ
∂ t +0.05 (0.5 ∂ B
∂ t +
√ 1−(0.5)2 ∂W
∂t ), σ 0>0
∂ σ
∂ t =2 ( 0.1−σt ) ∂ σ
∂ t +0.05 (0.5 ∂ B
∂ t + 1
2(1-(0.5)2) ∂W
∂t
Where, βSS( ρSS dBt
Q+√ 1−ρSS
2 dW t
Q), is the market price of volatility risk and r is the risk free
rate. For bearing the additional volatility risk, investors require extra return of amount ψSS(u;t,
σ t , xt )
The general definition will be presented. The option pricing methodology, which will be
discussed in the Fourier transform of a real function is ψSS(u;t,σ , x,x.
ψSS(u;t,σ t , xt ): = EQ[eiu X T
∨Ft]

With i being the imaginary unit, and u being a real number (u ∈ R). In case F( xt)=f XT ( x ) is a
probability density function of a random variable X, the transform above is the characteristic
function of X to be denoted by ϕX(u).
Random variables are fully described by their characteristic functions, that is, if ϕX(u) is
known then the distribution of X is completely defined. Also, by knowing the characteristic
function,
CSS(t, S, σ):= EQ[(ST −k )+¿∨Ft ¿], t∈[0,T]
C= EQ[eiu X T
∨Ft]
f XT ( x )=F−1[ϕX(u)]du= 1
√ 2 π ∫
0
T
eiu X T
σ X(u)du
CT(k)= EQ[eiu X T
∨Ft]
Thus, we can consider the call Price's model of the ODE for C, D and E such as for all
u∈R and all t∈ [0, T] the equation is,
ψSS(u;t,σ t , xt )= exp(c(u, T-t)+ D(u, T-t) σ + 1
2 E ( u ,T −t ) σ2 +iux)
Let us consider the value of C (u,0) where u∈R
C (u,T-t)=EQ[(ST −k )+¿¿| 1
2 σ ∫
0
T
eiu X T
σ X(u)du
X=log ST k=log k σ (X)are the risk natural values of the call price,
C (u,T-t)=EQ[(ST −k )+¿¿| 1
2 σ ∫
0
T
ex−ek q(x)dx
lim
k → ∞
CT (k )=s0
Let us consider the limit values as, T>0
ψSS = σt (S−(T +1) xt)
T 2+T −s2+ i(2T +1) s

=∫
0
T
eisk ψ ( s)ds
Let us consider the local variable,
^w(T-k)=∫
0
T
ei (T −t)k ψ (T −k )ds
^w(T-k)=∫
0
T
ei ( T −t ) k (eT −K )+¿dt
C(u,T-t)= K i ( T−k ) +1
(T −t)2 −i( T −t)
Let us consider the value of D (u,0) where u∈R
D (u,T-t)=EQ[(ST −k )+¿¿| 1
2 σ ∫
0
T
eiu X T
σ X(u)du
X=log ST k=log k σ (X) are the risk natural values of the call price,
D (u,T-t)=EQ[(ST −k )+¿¿| 1
2 σ ∫
0
T
ex−ek q(x)dx
lim
k → ∞
CT (k )=s0
Let us consider the limit values as, T>0
ψSS = σt (S−(T +1) xt)
T 2+T −s2+ i(2T +1) s
=∫
0
T
eisk ψ ( s)ds
Let us consider the local variable,
^w(T-k)=∫
0
T
ei (T −t)k ψ (T −k )ds

^w(T-k)=∫
0
T
ei ( T −t ) k (eT −K )+¿dt
D (u,T-t)= K i ( T−k ) +1
(T −t)2 −i( T −t)
Let us consider the value of E (u,0) where u∈R
1
2 E ( u ,T −t ) σ2 +iux= E (u,T-t) σ 2=EQ[(ST −k )+¿¿| 1
2∫
0
T
eiu X T
+iux σ 2X(u)du
X=log ST k=log k σ (X) is there risk natural values of the call price,
E (u,T-t)=EQ[(ST −k )+¿¿| 1
2∫
0
T
ex−ek +iux q(x)dx
lim
k → ∞
ET (k )=s0
Let as consider the limit values is, T>0
ψSS = σt (S−(T +1) xt)
T 2+ T −s2+ i(2T +1) s + 1
2 iux
=∫
0
T
eisk + 1
2 iux ψ (s)ds
Let us consider the local variable,
^w(T-k)=∫
0
T
ei (T −t)k + 1
2 iux ψ (T −k )ds
^w(T-k)=∫
0
T
ei ( T −t ) k ( eT −K ) + 1
2 iux
E (u ,T-t) = K i (T−k )+1
(T −t)2 −i(T −t)+ 1
2 iux
= K i (T−k )+1
(T −t )2 −i(T −t) + Ki (T −k )+ 1
( T −t)2−i(T −t) + Ki ( T−k )+1
(T −t)2−i(T −t )+ 1
2 iux

Xt :=log ( St
S0
) being the characteristic function of the standardized log-price at maturityXT - X 0
= log ( St
S0
), well as other important results regarding the practical implementation of the
Fourier transform apparatus can be found.
Question 2
a) f XT ( y ) y∈[-0.5,0.5] for ρSS ∈[−0.5,0,0 .5]
f XT ( y ) = 1
2 π ∫
❑
❑
e−iuy ψSS ( u ; 0 , σ0 , x0 )du ,y∈ R
f XT ( y )=F−1[ϕX(u)]du= 1
√ 2 π ∫
0
T
eiuy σ X(u)du
We can consider the equation as,
dσ t:= KSS (θSS−σ t)dt+βSS(ρSS dBt
Q+√1−ρSS
2 dW t
Q), σ 0>0,
f XT ( y ) y ∈ [-0.5,0.5] for ρSS ∈[−0.5,0,0 .5]
f XT ( y ) = 1
2 π ∫
❑
❑
e−(0.5)+√ 1−(−0.5)2 =1.818
f XT ( y ) = 1
2 π ∫
❑
❑
e−(0.5)+√1−(0)2 =3.589

f XT ( y ) = 1
2 π ∫
❑
❑
e−(0.5)+√ 1−(0.5)2 =1.818
f XT ( y ) = 1
2 π ∫
❑
❑
e0.5+√ 1−(−0.5)2 =3.455
f XT ( y ) = 1
2 π ∫
❑
❑
e0.5+√1−(0)2 =3.5898
f XT ( y ) = 1
2 π ∫
❑
❑
e0.5+√ 1−(0.5)2 =3.455
ρSS −0.5 0 0.5
-0.5 1.818 3.589 1.818
0.5 3.455 3.5898 3.455
0 2 4 6 8 10 12
0
2
4
6
8
10
12
𝑓_ ( )𝑋𝑇 𝑦
𝜌_𝑆𝑆
Y axis
b)
f XT ( y ) y ∈[-0.5,0.5] for ρSS ∈[0.01,0.05,0 .09]
f XT ( y ) = 1
2 π ∫
❑
❑
e−(0.5)+√ 1−(0.01)2 =1.952
Y

f XT ( y ) = 1
2 π ∫
❑
❑
e(0.5 )+√1−(0.01)2 =3.589
f XT ( y ) = 1
2 π ∫
❑
❑
e−(0.5)+√1−(0.05)2 =1.951
f XT ( y ) = 1
2 π ∫
❑
❑
e(0.5 )+√1−(0.05)2 =3.588
f XT ( y ) = 1
2 π ∫
❑
❑
e−(0.5)+√1−(0.09)2 =1.948
f XT ( y ) = 1
2 π ∫
❑
❑
e(0.5 )+√1−(0.09)2 =3.585
ρSS 0.01 0.05 0.09
-0.5 1.952 3.589 1.951
0.5 3.588 1.948 3.585
0 2 4 6 8 10 12
0
2
4
6
8
10
12
pss
Y axis
Question 3
Probability measure ~
Q by the Radon-Nikodym derivative.
We can find the values of the current price's value and measure the ~
Q probability by using
the Radon –Nikodym derivative.
Y

d ~
Q=ST =σ t St dBt
Q
d Q= S0 =σt S0 dBt
Q=σ t dBt
Q
We can apply the dynamic values of Girsanov's theorem Let us consider the {dBt
Q} as the
wiener process on the probability space as {σ , F , ~
Q } and measure the process of the adapted
natural filtration of the values as S0 =¿{Ft
dB}
d ~
Q
d Q := ST
S0
=EQ ¿ ¿
0 2 4 6 8 10 12
0
2
4
6
8
10
12
probability measure of call price
Series1 Series2 Series3 Series4 Series5 Series6
dQ'=s0
dQ=st
Question 4
CSS(t, S, σ)= St
~
Q(ST ≥ k ∨Ft )-KQ ¿ ¿)
CSS(t, S, σ):=EQ[(ST ≥ k )+¿∨ Ft ¿] t∈[0,T]
~
C=EQ[eiu X T
∨Ft]- KQ ¿ ¿)
f XT ( x )= F−1[ϕX(u)]du= 1
√2 π ∫
0
T
eiu X T
σ X(u)du

CT(k)=EQ[eiu X T
∨Ft]- KQ ¿ ¿)
Forward price ~
C of the option has to satisfy the forward version.
Bayes' rule to derive the relation
Question 5
CSS(t, S, σ):=EQ[(ST −k )+¿∨Ft ¿], t∈[0,T]
~
C=EQ[eiu X T
∨Ft]
f XT ( x )= F−1[ϕX(u)]du= 1
√2 π ∫
0
T
eiu X T
σ X(u)du
CT(k)=EQ[eiu X T
∨Ft]
Thus, we can consider the call Price's model of the ODE for C, D and E such as for all
u∈R and all t∈ [0, T] equation as,
ψSS(u;t, σ t , xt )= exp( ~
C(u, T-t)+ ~
D(u, T-t) σ + 1
2
~
D (u ,T −t ) σ2 +iux)
Let us consider the value of ~
C (u,0) where u∈R
~
C (u,T-t)= EQ[(ST −k )+¿¿| 1
2 σ ∫
0
T
eiu X T
σ X(u)du
X=log ST k=log k σ (X) are the risk natural values of the call price,

~
C (u,T-t)=EQ[(ST −k )+¿¿| 1
2 σ ∫
0
T
ex−ek q(x)dx
lim
k → ∞
~
CT (k )=s0
Let us consider the limit values as, T>0
ψSS = σt (S−(T +1) xt)
T 2+T −s2+ i(2T +1) s
=∫
0
T
eisk ψ ( s)ds
Let us consider the local variable,
^w(T-k)=∫
0
T
ei (T −t)k ψ (T −k )ds
^w(T-k)=∫
0
T
ei ( T −t ) k (eT −K )+¿dt
~
C(u,T-t)= K i ( T−k ) +1
(T −t)2 −i( T −t)
Let us consider the value of D (u,0) where u∈R.
~
D (u,T-t)=EQ[( ST −k )+¿¿| 1
2 σ ∫
0
T
eiu X T
σ X(u)du
X=log ST k=log k σ (X) are the risk natural values of the call price,
~
D(u,T-t)=EQ[( ST −k )+¿¿| 1
2 σ ∫
0
T
ex−ek q(x)dx
lim
k → ∞
~
DT (k )= s0
Let us consider the limit values as, T>0

ψSS = σt (S−(T +1) xt)
T 2+T −s2+ i(2T +1) s
=∫
0
T
eisk ψ ( s)ds
Let us consider the local variable,
^w(T-k)=∫
0
T
ei (T −t)k ψ (T −k )ds
^w(T-k)=∫
0
T
ei (T −t ) k (eT −K )+¿dt
~
D (u,T-t)= K i (T−k )+1
(T −t )2 −i(T −t)
Let us consider the value of ~
E (u,0)where u∈R.
1
2
~
E ( u ,T −t ) σ2 +iux= ~
E(u,T-t) σ 2=EQ[(ST −k )+¿¿| 1
2∫
0
T
eiu X T
+iux σ 2X(u)du
X=log ST k=log k σ (X) are the risk natural values of the call price,
~
E(u,T-t)=EQ[(ST −k )+¿¿| 1
2∫
0
T
ex−ek +iux q(x)dx
lim
k → ∞
~
ET (k )= s0
Let us consider the limit values as, T>0.
ψSS = σt (S−(T +1) xt)
T 2+ T −s2+ i(2T +1) s + 1
2 iux
=∫
0
T
eisk + 1
2 iux ψ (s)ds
Let us consider the local variable,
^w(T-k)=∫
0
T
ei (T −t)k + 1
2 iux ψ (T −k )ds

^w(T-k)=∫
0
T
ei ( T −t ) k ( eT −K ) + 1
2 iux
~
E (u ,T-t)= K i ( T−k ) +1
(T −t)2 −i( T −t)+ 1
2 iux
The values of Q measure is,
= K i (T−k )+1
(T −t )2 −i(T −t) + Ki (T −k )+ 1
( T −t)2−i(T −t) + Ki ( T−k )+1
(T −t)2−i(T −t )+ 1
2 iu
Question 6
To find the value of the call price, call prices for the strikes and expirations use the the
inversion theorem as follows,
= K i (T−k )+1
(T −t )2 −i(T −t) + Ki (T −k )+ 1
( T −t)2−i(T −t) + Ki ( T−k )+1
(T −t)2−i(T −t )+ 1
2 iux
Let us consider the the values of S0=100.
k ϵ {80, 90, 100, 110, 120}, T ∈{0.1, 0.5,1,2,5}
compute 25 prices in total value by using the inversion theorem.