We know that the relevant SDE for the geometric Brownian motion X(t) is as follows:
$$
dX(t) = \mu X(t)dt + \sigma X(t)dB(t),
$$
where B(t) is the (Standard?) Brownian motion, $\mu$ is a drift constant and $\sigma$ is the standard deviation.
I have also seen the following SDE for Brownian motion with drift (Wiener Process)
$$
dX(t) = \mu dt + \sigma dB(t),
$$
1) How X(t) was omitted in the second equation? under which condition?
2) We know that the variance of the Standard BM is $\sigma^2= 2Dt$ and so the standard deviation is $\sqrt{2Dt}$. Also, I have seen in some references that $dB=\sqrt {dt}$.
Does it mean that we can write : $\sigma dB(t)=\sqrt{2Dtdt}$ ?
The second equation is important, because it seems that the stochastic process of the (standard) Brownian motion
$$
X(t) = \sigma B(t),
$$
was derived from the second equation(for $X(0)=0$).
3) According to what i wrote in my second question, is it correct to write: $ X(t) = \sqrt{2Dt}\sqrt t$?
I would appreciate if you can correct any probable mistake in the above equations, or add more about the differences between Geometric BM, BM with drift and standard BM(Wiener process).
Best Answer
1) Both are two different equations. But solutions of such equation are know as Ornstein-Uhlenbeck processes.
2) No. The writing $f(t,\cdot )\mathrm d B_t=f(t,\cdot )\sqrt{dt}$ is a formal writing to denotes that the quadric variation of $$Y_t:\omega \mapsto Y_t(\omega ):=\int_0^t f(s,\omega )\,\mathrm d B_s(\omega )$$ is given by $$\left<Y\right>_t=\int_0^t f(s,\omega )^2\,\mathrm d s.$$
3) If $$\mathrm d X_t=\mu\,\mathrm d t+\sigma \,\mathrm d B_t,$$ then $$X_t=X_0+\mu t+\sigma B_t,$$ and not what you wrote. And no, this doesn't means that $X_t=X_0+\mu t+\sigma \sqrt{t},$ since $B_t\neq \sqrt t$. But even if $$\mathrm d X_t=\mathrm d B_t,\tag{E}$$ how can $X_t=\sqrt t$ ? There is no randomness in the last equality, so why using such a technical notation in $(E)$ to denotes simply $t\mapsto \sqrt t$ ? Mathematicians are not that Sado-Maso :-)