Let f(St) = log(St), then using a
Taylor Series in one variable, we have:
df(St) = d(log(St) = df/dSt * dSt + 0.5 * d^2f/dSt^2 * (dSt^2)
Here:
df/dSt = 1/St and d^2f/dSt^2 = -1/St^2
So, we have:
d(log(St) = 1/St * dSt + 0.5 * -1/St^2 * (dSt^2)
Finally, as:
dSt= a*St*dt + sig*St*dBt
and:
dSt^2 = (a*St*dt)^2 + (sig*St*dBt)^2 + 2*a*St*dt*sig*St*dBt
= 0 + sig^2*St^2*dt + 0 = sig^2*St^2*dt
we have:
d(log(St) = 1/St *(a*St*dt + sig*St*dBt) + 0.5 * -1/St^2 * sig^2*St^2*dt
So, cancelling all the St terms gives:
d(log(St) = a*dt + sig*dBt - 0.5 *sig^2*dt
ie:
d(log(St) = (a - 0.5 *sig^2)*dt + sig*dBt
I hope this helps.
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