A question about SDE and geometric Brownian motion.

Question

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?

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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 $$\boxed{\int_0^tY_sdX_s:=\int_0^t\alpha Y_s ds + \int_0^t\sigma Y_sdB_s}$$ 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.

Answer 1

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.

Answer 2

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:

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

Then:

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

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

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