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?


closed as off-topic by parsiad, Eevee Trainer, max_zorn, Cesareo, Javi Apr 18 at 10:21

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  • $\begingroup$ This question is missing a lot of context/is ill-specified. Brownian motion (a.k.a. Wiener process) has variance $\operatorname{Var}(W_t) = t$, which is nonzero for any $t > 0$. Are you asking if it's possible to make an Ito process with variance zero? In this case, the answer is yes: $dX_t = \mu dt + 0 dW_t$. $\endgroup$ – parsiad Apr 18 at 1:09
  • $\begingroup$ Thanks for the reply. Will improve that. So, we are left with dXt = mdt. Let's have m = 0.03. In this case, does its value same as the future value of a fund, for instance, St+1 = St(1.03)? $\endgroup$ – Raihana Nasir Apr 18 at 1:14

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 $m$.


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