"Dividing" sides of stochastic differential equation

Question

I'm looking at example 5.1.1 of Oksendal's book http://th.if.uj.edu.pl/~gudowska/dydaktyka/Oksendal.pdf My question is: why we can divide the differential equation by $N_t$? I know that this is just a notation, but when we use the "integral" notation, it's not very clear to me. We have $$ N_t - N_0 = \int_0^tdN_s = \int_0^trN_sds+\int_0^t\alpha N_sdB_s $$ However, after dividing differential form of equation by $N_t$, as it is stated in book, we get new integral equation: $$ \int_0^t\frac{dN_s}{N_s} = \int_0^trds+\int_0^t\alpha dB_s $$ Could you please let me know why these two above equations are equivalent? How can we remove $N_s$ from integrals on the left side when working on integral notation? Thank you in advance.

Answer 1

Applying your notations, it is possible to solve the SDE

$$dN_t=rN_tdt+\alpha N_tdB_t$$

with an initial condition $N_0>0$ as follows: Preliminarily, this SDE satisfies the preconditions of Theorem 5.2.1 in Oskendal's book, i.e., it maintains a unique solution $N_t$. Let's guess that $N_t$ is a positive process. In such a case, it is fine to apply Ito's Lemma with $g(t,x)=\ln x,\forall x\in(0,\infty)$ and $X_t=N_t$ (see Theorem 4.2.1 with the remark in page 48 of Oskendal's book). Thus, deduce that

$$\text{(I)}\ d\ln N_t=\frac{dN_t}{N_t}-\frac{1}{2}\cdot\frac{\left(dN_t\right)^2}{N_t^2}\,.$$

It is known that $dt\cdot dt=dt\cdot dB_t=dB_t\cdot dt=0$ and $dB_t\cdot dB_t=dt$ and hence, since $N_t$ satisfies the SDE it turns out that

$$\left(dN_t\right)^2=\left(rN_tdt+\alpha dB_t\right)^2=\alpha^2N_t^2dt\,.$$

Thus, an insertion of this identity and the SDE into (I) implies that

$$d\ln N_t=rdt+\alpha B_t-\frac{\alpha^2N_t^2}{2N_t^2}dt=\left(r-\frac{\alpha^2}{2}\right)dt+\alpha dB_t\,.$$

Using the initial conditions $N_0$ and $B_0=0$, this equation can be Rephrased in an integral form as follows

$$\ln N_t-\ln N_0=\left(r-\frac{\alpha^2}{2}\right)t+\alpha B_t\,.$$

Note that $N_t$ can be extracted, i.e., for every $t\geq0$

$$N_t=N_0\exp\left[\left(r-\frac{\alpha^2}{2}\right)t+\alpha B_t\right]\,.$$

Since $N_0$ is positive, then $N_t$ is a positive process which makes the result follows. This solution may be easily extended to the case where $N_0$ is nonnegative random variable.