I'm very new to Stochastic processes and SDEs, I have a question related to solutions to a SDE, The famous SDE for financial assets is:

$dS_{t} = \mu S_{t}dt + \sigma S_{t}dW_{t}$

the solution for it is: $S_{t} = S_{0} \exp^{(\mu - \frac{1}{2}\sigma^2)t + \sigma W_{t}}$

if I want to get the return for 1 period then $$ \int_{0}^{t}\frac{dS_{u}}{S_{u}} = \int_{0}^{t}\mu dt + \int_{0}^{t}\sigma dW_{u} = \log\left(\frac{S_{t}}{S_{0}}\right) = \mu t + \sigma W_{t} = \log\left(\frac{S_{1}}{S_{0}}\right) = \mu + \sigma W_{1} $$ Which is correct according to the assumptions:

Returns are i.i.d normal with mean = $\mu$ and var = $\sigma^2$

So the asset price follows a random walk

But if I used the solution above and I plug t = 1 to calculate the position of the asset at time t = 1 and calculate the return from there, I can't get exactly the return from 0 -> 1 due to the random component but my expected return should be the mean $\mu$ $$ S_{1} = S_{0} \exp^{(\mu - \frac{1}{2}\sigma^2) + \sigma W_{1}} = \log\left(\frac{S_{1}}{S_{0}}\right) = (\mu - \frac{1}{2}\sigma^2) + \sigma W_{1} $$ so how can I get rid of the $-(\frac{1}{2}\sigma)$ or I just can't use the solution this way and I have to go through the whole process of $dS_{t}$ using Ito lemma things Thank you guys so much!


1 Answer 1


You need to understand that $dW_t$ is much larger, in general, than $dt$. The random increments are so large (in the infinitesimal scale) that the accumulation of their squares is still proportional to $dt$ and will thus contribute to the "observable" behavior.

In more detail, you can calculate that $$ 1+σ\,dW_t+μ\,dt=e^{\ln(1+σ\,dW_t+μ\,dt)}=e^{σ\,dW_t+μ\,dt-\frac12(σ\,dW_t+μ\,dt)^2+\frac13(σ\,dW_t+μ\,dt)^3\mp...} $$ and the terms $σ\,dW_t$, $μ\,dt$ and $-\frac12σ^2\,(dW_t)^2$ all accumulate to appreciable sizes in $σ\,ΔW_t$, $μ\,Δt$ and $-\frac12σ^2\,Δt$ over some larger time span $Δt$. Only the terms following them are of a size $\sim (dt)^{3/2}$ that under accumulation still remain infinitesimal like $\sqrt{dt}$ or smaller.

  • $\begingroup$ Hi Lutz, tks for your reply but I don't really think i get it because as i said I'm new to SDE and still learning. I think the problem I'm having because I can't just simply plug t = 1 into S(t) not realizing S(t) is actually a function of 4 variables, 2 of them are constants but Wt and t arent $\endgroup$
    – Hoanheoo
    Oct 10, 2021 at 23:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .