Stochastic Differential Equation and a Martingale

[Alexei]

W(t) - 1d Brownian Motion
W(0) = 1
r: R
T = min[t>0;W(t) = 0], Let:

(R(t)%20=%20W_{t}^{T})
1) Find a(R(t)) and b(R(t)) such that:
(dR(t)%20=%20a(R(t))dt%20+%20b(R(t))dW(t))
2) Find function F such that M(t) is a martingale:
(M(t)%20=%20R(t)*e^{\int_{0}^{t}F(R(t))ds})

 

[quotes]

W(t) - 1d Brownian Motion
W(0) = 1
r: R
T = min[t>0;W(t) = 0], Let:

(R(t)%20=%20W_{t}^{T})
1) Find a(R(t)) and b(R(t)) such that:
(dR(t)%20=%20a(R(t))dt%20+%20b(R(t))dW(t))
2) Find function F such that M(t) is a martingale:
(M(t)%20=%20R(t)*e^{\int_{0}^{t}F(R(t))ds})

you mean R(t)=W(T)-W(t) ???

 

[Alexei]

R(t) is defined as above and it is valid for t<T.

 

[DStahl]

I think quotes is asking if ( w_t ^T =(w_t)^T ) or is notation for w(T)-w(t)

 

[Alexei]

R(t) is not a difference of BM values - it is W(t) which is valid for t<T

 

[DVV]

W(t) - 1d Brownian Motion
W(0) = 1
r: R
T = min[t>0;W(t) = 0], Let:

(R(t)%20=%20W_{t}^{T})
1) Find a(R(t)) and b(R(t)) such that:
(dR(t)%20=%20a(R(t))dt%20+%20b(R(t))dW(t))
2) Find function F such that M(t) is a martingale:
(M(t)%20=%20R(t)*e^{\int_{0}^{t}F(R(t))ds})

what does this notation means (R(t)%20=%20W_{t}^{T})? What is the idea behind big T in upper index?

 

[Kangxi]

I also have the same question,What does W_t^T mean?

 

[bubblegum]

Im not sure but W_t^T may mean, a wiener process stopped at time T, that is T is a stopping time, which by definition is less than t, i.e {T<=t}.

However It may be nothing to do with stopping times.

 

[cox-ross]

I also have the same question,What does W_t^T mean?

maybe he is tryning to find the SDE of the brownian bridge between t and T ??!!

(dW_t ^T = ( 0 - W_t ) / ( T - t ) dt + dW_t ) , but Here T is a random stopping time.

if W_t ^T =W_T - W_t it will be a Brownian motion starting from -1 and end at 0, but will have the same SDE, a = 0 and b =-1