Brownian motion with zero volatility

Question

is it possible for Brownian motion to deal with zero volatility? and if it does, does it mean that the fund experiencing deterministic increment in value?

Answer 1

Judging from your comments, what I think you really want to ask is "what happens when I have a geometric Brownian motion with constant drift and zero volatility?"

Let $m$ be a constant and consider the stochastic differential equation $$dS_t = m S_t dt + 0 \cdot S_t dW_t = m S_t dt.$$ Because the drift can be written $\mu(t, S_t) \equiv m S_t$, the above is actually an ordinary differential equation. Apply Ito's lemma (which in this case degenerates to the ordinary chain rule) to $f(S_t) = \log S_t$ to get \begin{multline*} \log S_T = f(S_T) = f(S_0) + \int_0^T f^\prime(S_t) dS_t = \log S_0 + \int_0^T \frac{1}{S_t} m S_t dt \\= \log S_0 + \int_0^T m dt = \log S_0 + m T. \end{multline*} Exponentiating both sides, $$ S_T = S_0 \exp(m T). $$ It seems to me that you are using some finance terminology, so if it helps, you can think of the above as a bond that pays out continuously compounded interest at rate $m$.