Geometric Brownian Motion Drift

[WilliamY]

What's the drift of the ratio of two geometric Brownian motion? Thanks.

($(\frac{S_{T_{2}}}{S_{T_{1}}})$)

 

[parisjohn]

Hi,
Ito Formula :
If mu1 = Drift(S1), s1=Vol(S1)
mu2 = Drift(S2), s2=Vol(S2)
rho*dt=d(W1(t),W2(t))
Drift(S1/S2) = mu1 - mu2 + s2*s2 - rho*s1*s2
J

 

[AlexesDad]

What's the drift of the ratio of two geometric Brownian motion? Thanks.

($(\frac{S_{T_{2}}}{S_{T_{1}}})$)

Assuming

($S_t = S_0e^{\sigma W_t + \mu t} $)

and

($T_2>T_1$)

then

($\frac{S_{T_2}}{S_{T_1}} = e^{\sigma \large(W_{T_2} - W_{T_1}\right) + \mu \large({T_2} - {T_1}\right) }$)

and it follows from Ito that the drift is

($ \large(\mu - \frac{1}{2}\sigma^2\right)\large({T_2} - {T_1}\right) $)

 

[parisjohn]

Sorry,
i read S2/S1 instead of S(T2)/S(T1)

 

[WilliamY]

Assuming

($S_t = S_0e^{\sigma W_t + \mu t} $)

and

($T_2>T_1$)

then

($\frac{S_{T_2}}{S_{T_1}} = e^{\sigma \large(W_{T_2} - W_{T_1}\right) + \mu \large({T_2} - {T_1}\right) }$)

and it follows from Ito that the drift is

($ \large(\mu - \frac{1}{2}\sigma^2\right)\large({T_2} - {T_1}\right) $)

What's the ($dt$) then? It can not be the drift term in the SDE right? ($T_1$) and ($T_2$) are not random.