In Bernt Oksendal's Stochastic Differential Equations, Chapter 4, one has the following stochastic differential equation (whose solution is geometric Brownian motion): $$dN_t=rN_tdt+\alpha N_tdB_t\;\;\;\text{ ie } \;\;\; N_t-N_0=r\int_0^t N_sds+\alpha\int_0^tN_sdB_s,$$ where $\alpha,r\in\mathbb{R}$ and $B_t$ is standard Brownian motion (ie $B_0=0$). After assuming that $N_t$ solves the above equation, the author abruptly deduces that $$\frac{dN_t}{N_t}=rdt+\alpha dB_t \;\;\;\text{ie}\;\;\; \int_0^t\frac{1}{N_s}dN_s= rt+\alpha B_t. \;\;\;(*)$$

I don't understand how he obtained this directly.

What I understand for sure is that if we seek to compute $$\int_0^t\frac{1}{N_s}dN_s $$ we apply Itô's formula for $Y_t=\ln(N_t)$ (assuming $N_t$ satisfies all the needed conditions). After some computation this yields
$$\frac{1}{N_t}dN_t=d\ln N_t+\frac{1}{2}\alpha^2dt \;\;\text{ i.e } \;\;\int_0^t\frac{1}{N_s}dN_s=\ln(N_t)-\ln(N_0)+\frac{1}{2}\alpha^2t.\;\;\;(**)$$ But at first glance, it does not seem that $(**)$ implies $(*)$. How did he obtain $(*)$? Did he use a method other than the Ito formula or am I missing something?

Thank you for the helpful answers! I've upvoted both and will accept whichever has more votes (in case of tie I'll just leave them be).

It turns out that what I was missing is the definition of the Itô integral with respect to an Itô process, which I could not find in the book. So actually by definition one has that for any Itô process of the form $$dX_t=\alpha dt+\sigma dB_t,$$ and $Y_t$ an appropriate integrand that $$\boxed{\int_0^tY_tdX_t:=\int_0^t\alpha Y_t dt + \int_0^t\sigma Y_tdB_t}$$ This justifies the formal notation $$\frac{1}{N_t}dN_t=\frac{1}{N_t}(rN_tdt+\sigma N_tdB_t)= \alpha dt+\sigma dB_t,$$ and automatically gives $(*)$ when $Y_t=1/N_t$, of course, while assuming $Y_t$ meets all the necessary requirements.

  • 1
    $\begingroup$ I don't know anything about this stuff, but I'm surprised that the integral of $dB_t$ shouldn't turn out to be $B_t$ (this seems to be what the author is doing). $\endgroup$ – Olivier Bégassat Feb 24 at 14:19
  • $\begingroup$ Indeed Olivier is right $\endgroup$ – Chaos Feb 24 at 14:42
  • 1
    $\begingroup$ For this specific SDE, you can solve for $N$ and find out that $N_t = N_0\exp \left(\alpha B_{t}+\left(r -{\tfrac {\alpha^{2}}{2}}\right)t\right)$ which will then imply the equality. For a more rigorous justification of why you can do such manipulations in a general setting, you should look up the associativity of Itô's integral (check out this question for instance) $\endgroup$ – StratosFair Feb 24 at 14:50
  • $\begingroup$ @StratosFair thanks! The link is indeed helpful. Might I ask though how you can solve for $N_t$ ? $\endgroup$ – UserA Feb 24 at 15:03
  • 1
    $\begingroup$ I would accept @NullUser's answer, it's better than mine. You can find mine on every website or in every textbook, but the other answer points to associativity of stochastic integrals, which is only found in grad-level measure theory / advanced probability lecture notes: so the other answer is "nicer" in that it reveals a more advanced (but more elegant) way to deal with the problem. $\endgroup$ – Jan Stuller Feb 25 at 9:58

Typically associativity of the integral is proved early on. If $X$ is a semimartingale and the integral $K \cdot X = \int K dX$ makes sense then the integral $(HK) \cdot X = \int HK dX$ makes sense if and only if the integral $H \cdot (K\cdot X) = \int H d(\int K dX)$ makes sense, in which case they are equal. In "differential form", this is $H\, d(K dX) = (HK) dX$. I am assuming you are already comfortable with linearity of stochastic integrals.

Thus taking $dN_t = N_t r dt + N_t\alpha dB_t$ and integrating $1/N_t$ with respect to the semimartingales defined by either side gives $$ (1/N_t) dN_t = (1/N_t) d(N_t r dt + N_t\alpha dB_t) = (1/N_t) d(N_t r dt)+ (1/N_t)d( N_t\alpha dB_t) = r dt + \alpha dB_t. $$

There is no need to use Ito here, it is simply associativity of the integral. The key idea is that we are not "dividing by $N_t$", instead we are integrating $1/N_t$ with respect to two (equal) semimartingales. In differential form, it just looks like dividing.

  • $\begingroup$ Cool. I learned something new! $\endgroup$ – Jan Stuller Feb 24 at 18:26

I find notation such as $\frac{dN_t}{N_t}=rdt+\alpha dB_t$ incredibly unfortunate, it leads to the type of confusion you have encountered (and that I used to encounter myself).

That is why I advise against using short-hand notation (at least until one becomes very confident in stochastic calculus techniques).

In long-hand notation:

$$N_t=N_0+\int_{h=0}^{h=t}rN_hdh+\int_{h=0}^{h=t}\alpha N_hdB_h$$

From the above, it should be obvious that $N_h$ cannot be taken out of the integral and brought to the LHS (which is not obvious in the short-hand notation). Ito's Lemma needs to be used, applied to $ln(N_t)$ as you point out. Let $F(N_t,t):=ln(N_t)$, then taking the derivatives:

$$\frac{\partial F}{\partial N_t}=\frac{1}{N_t}, \frac{\partial^2 F}{\partial N_t^2}=\frac{-1}{N_t^2}, \frac{\partial F}{\partial t}=0$$


$$F_t=F(N_0)+\int_{h=0}^{h=t}\left(\frac{\partial F}{\partial h}+\frac{\partial F}{\partial N_h}rN_h+0.5\frac{\partial^2 F}{\partial N_h^2}\alpha^2 N_h^2\right)dh+\int_{h=0}^{h=t}\frac{\partial F}{\partial N_t}\alpha N_hdB_h=\\=ln(N_0)+\int_{h=0}^{h=t}\left(r-0.5\alpha^2 \right)dh+\int_{h=0}^{h=t}\alpha dB_h=\\=ln(N_0)+(r-0.5\alpha^2)t+\alpha B_t$$

Because $F_t=ln(N_t)$, the final answer is (by exponentiating both sides):

$$N_t=N_0e^{(r-0.5\alpha^2)t+\alpha B_t}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.