Chapter 12: Stochastic differential equation Flashcards
Making sense of:
Xt = X0 + ∫{0,t} F_u du + ∫{0,t} G_u dB_t
- B_t is Brownian motion
- We allow cases in which the stochastic processes F_u and G_u depend on X_u e.g F_u = X_t^2
- X_t is an unknown stochastic process
eq called STOCHASTIC DIFFERENTIAL EQUATIONS SDEs
SOLVING SDES
STOCHASTIC DIFFERENTIAL EQUATIONS SDEs
solving Xt = X0 + ∫{0,t} F_u du + ∫{0,t} G_u dB_t
Solution involves finding X_t in terms of B_t
- If F_t and G_t only dep on t and B_t then SDE is just an explicit formula for a solution to it
- if F_t and or G_t dep on X_t ( e.g F_t=2X_t) then its no an explicit formula and no guarantee a sol for X_t exists
SDE
INTEGRAL FORM
STOCHASTIC DIFFERENTIAL FORM
INTEGRAL FORM
X_t - X_0 = ∫_{0,t} F_u du + ∫_{0,t} G_u dB_t
commonly write
STOCHASTIC DIFFERENTIAL FORM
dXt = Ft dt + Gt dBt
- drop integral signs
- converting from stoch diff to int form we add limits (any as long as consistent)
X_t - X_v = ∫{v,t} F_u du + ∫{v,t} G_u dB_t
DF 12.0.2
Stochastic process ito in stoch diff form
A stochastic process Xt is an Ito process if it satisfies
dXt = Ft dt + Gt dBt
for some G ∈ H^2 and a continuous adapted stochastic process F.
updated in new form*
NOTATION
ito process (X_t)and stoch process (H_t)
Given an Ito process Xt , as in Definition 12.0.2, and a
stochastic process H_t , we will often write
dZ_t = H_t dX_t
which (as a definition) we interpret to mean that
dZ_t = H_t(F_t dt + G_t dBt)
= H_tF_t dt + H_tG_t dB_t
In integral form this represents
Zt = Z0 + ∫{0,t} H_uF_u du + ∫{0,t} H_uG_u dB_u.
LEMMA 12.1.1 ITOS FORMULA
Suppose that, for t ∈ R and x ∈ R, f(t, x) is a deterministic function that is differentiable in t and twice differentiable in x.
Then Zt = f(t, Xt) is an Ito
process and
dZt =
[∂f/∂t (t, X_t) + F_t ∂f/∂x(t, X_t) + (1/2)G^2_t ∂^2f/ ∂x^2 (t, X_t).dt + G_t ∂f/∂x(t, Xt) dB_t
*notation:
usually:
dZt = {∂f/∂t + F_t∂f/∂x+ (1/2)G^2_t ∂^2f/∂x^2 } dt + G_t ∂f/∂x dB_t
sometimes:
dZt = {∂f/∂t +(1/2)G^2_t ∂^2f/∂x^2 } dt + ∂f/∂x dX_t
EXAMPLE 12.1.2.
Let us apply Ito’s formula to calculate dZ_t where Zt = B^2_t
Z_t= f(t,B_t) f(t,x)=x^2 ∂f/∂t =0 ∂f/∂x = 2x ∂^2 f/∂x^2 =2
B_t is an ito process satisfying dB_T = 0 dt + 1 dB_t
From Ito’s formula:
dZ_t = ( 0 +(0)(2B_t) + 0.5(1^2)(2))dt + (1)(2B_t)dB_t
= 1 dt + 2B_t dB_t
in integral form:
Z_t = Z_0 + ∫_{0,t} 1 du + 2B_u dB_u
rearranged
∫_{0,t} B_u dB_u = B^2_t/2 - t/2
IMPORTANT
*B^2t - t = ∫{0,t} 2B_u dB_u
we proved RHS was martingale ( we have that LHS is ito integral and thus its a martingale and thus RHS is a martingale) and now we have this w/o using conditional expectations
EXAMPLE 12.1.3
Suppose that X satisfies
dX_t = µ dt + σ dB_t, where µ ∈ R and σ bigger than 0 are deterministic constants. Let Zt = X_te^t
We want to find dZt
constant stoch processes
dZ_t =(X_te^t + (µ)(e^t) + (1/2)(σ^2)(0))dt + (σ)(e^t) dBt
Z_t = f(t, X_t) where f(t,x) = xe^t
F_t=µ
G_t= σ
by ITO’s formula
(Xt + µ) e^t dt + σe^t dBt
EXAMPLE 12.1.2.
Let us apply Ito’s formula to calculate dZ_t where Zt = B^2_t
NOTES
IMPORTANT
*B^2t - t = ∫{0,t} 2B_u dB_u
we proved RHS was martingale ( we have that LHS is ito integral and thus its a martingale and thus RHS is a martingale) and now we have this w/o using conditional expectations
EXAMPLE 12.1.4
Suppose that we want to calculate E[B^4_t]
Define Z_t= B^4_t and use itos formula:
Z_t = f(t,B_t) where f(t,x) = x^4
B_t is also an ito process with
dB_t = 0 dt + 1 dBt. So,
dZt =[0 + (0)(4B^3_t) + (1/2)(1^2)(12B^2_t)]
dt + (1)(4B^3_t) dBt
= 6B^2_t dt + 4B^3_t dBt
in integral form:
Z_t = Z_0 + ∫_{0,t} 6B^2u du + ∫{0,t} 4B^3_u dBu
taking expectations:
E[Z_t] = E[∫_{0,t}6B^2_u du] + E[∫_{0,t} 4B^3_u dBu]
(lemma to swap E and int 11.4.2 and ito integral is zero-mean martingale 11.2.1)
=
∫_{0,t} E[6B^2_u du] + 0
= ∫_{0,t} 6u .du
= 6t^2/2
= 3t^2
The result we have obtained matches that from exercise 10.4, but with much less work!
Geometric Brownian motion
dXt = αXt dt + σXt dBt
where α ∈ R and σ ≥ 0 are deterministic constants. The parameter α is known as the drift, and σ is known as the volatility
GBM
dXt = αXt dt + σXt dBt
*consider σ=0 and solve
dx_t=αx_t dt
in int form:
x_t - x_0 = ∫_{0,t} αx_u du
By FTOC:
dx/dt = αx ODE
To solve: sub z= log x
(we assume a sol exists and never negative)
dz/dt = (dz/dt)(du.dt) = (1/x) αx = α
int gives sol
z_t = αt+ C
e^z_t = C’e^{αt}
So we try substitution z=log(X)
dXt = αXt dt + σXt dBt
solving with z=log(X)
dZ_t = {(0) + (αX_t)(1/X_t) + 0.5 (σX_t)^2 (-1/X_62_t)dt + σX_t (1/X_t)dB_t
We saw σ=0 case solved
Z_t = f(t,x_t) where f(t,x)= log(x)
∂f/∂t =0
∂f/∂x = 1/x
∂^f/∂x^2 = -1/x^2
(α − 1/2 σ^2)dt + σ dBt
Integral form:
Z_t = Z_0 + ∫{0,t} (α − 1/2 σ^ 2)du + ∫{0,t} σ dBt
= Z_0 + (α − 1/2 σ^2)t + σBt
Z_t=log X_T so
Xt = X0 exp((α −1/2 σ^2)t + σBt)
As we said, this was all based on the assumption that a (strictly positive) solution exists
SOL OF GBM for times t and T
X_T = X_t exp((α −1/2 σ^2)(T − t) + σ(BT − Bt))
*see everything happened uo to time t and make predictions about stock prices until time T
CONSIDER SDE
dXt = σtXt dBt
where (σ_t) is an adapted stochastic process
solve
To solve we use same idea as before and set Z_t = log (X_t)
dZ_T= {0 + (0)(1/X_t) + 0.5(σtX_t)(X^2_t)} dt + (σ_tX_t)(1/X_t)dB_t
= -0.5 σ^t dt + σ_t dB_t
Z_t- Z_0 = ∫{0,t} -0.5 σ^u du
+∫{0,t}σ_u dB_u
(we cant do as we dont know anything about stoch process σ_t)
X_t = X_0 exp( ∫{0,t} -0.5 σ^u du
+∫{0,t}σ_u dB_u)
