# Integrating log-normal

The usual log normal model in differential form is:

$dS = \mu S dt + \sigma S dX$

where $dX$ is the stochastic part, so

$\frac{dS}{S} = \mu dt + \sigma dX$ (1)

and we normally solve this by subbing in $Y=\log(S)$. What's to stop us just integrating (1) to get

$S = \exp\left(\mu t + \sigma\mathcal{N}(0,1)\right)$ ?

Why do we have to through all the business of subbing in for $Y$ and using Ito's lemma?

• An intuitive explanation can be found in Financial Calculus by Rennie and Baxter. Sep 29, 2012 at 18:13
• The obvous answer is that you can't simply evaluate $\int dS / S$ to $ln(S)$ as in the deterministic case. The reason why this is not allowed is the Ito integral, which is a different thing than, say, the Riemann integral where it works. Naively, the stochastic part of the Ito integral brings in a factor $\text{exp}(\sigma^2/2)$ on the lhs. here. Ito's lemma is a shorthand for not having to calculate this explicitly. Aug 9, 2019 at 6:18

$$dS_t=\mu S_t dt + \sigma S_t dW_t$$

where $W_t$ is a standard brownian motion (SBM).

You want to solve for $S_t$, so how would you proceed?

If you integrate both sides of the equation between 0 and $T$, you get:

$$S_T - S_0= \mu \int_0^T S_t dt + \sigma \int_0^T S_t dW_t$$

Okay and then what? The fact that you have $S_t$ in both integral is problematic.

The thing is, to solve for $S_t$, you in fact need to use a bit of trickery, and the substitution and the application of Ito's lemma allows you to get rid of the $S_t$ as you get:

$$dY = d(\ln S_t)=\left(\mu - \frac{\sigma^2}{2} \right) dt +\sigma dWt$$

The integration afterwards is straightforward.

So, you use the trick to get rid of the $S_t$ in the integrals an to be able to solve easily.