How does one think of the "trick" in solving the GBM equation

Question

The trick in solving the GBM equation $dX=\mu Xdt + \sigma X dW$ is to consider the logarithm and apply Ito lemma to it.

What is the idea behind this approach, is it a change of variables simple or something else? How would you describe the solution technique?

As I see it, we know that for any process $X$ we have $d(\ln X)=\frac{dX}{X}-\frac{1}{2X^2}dXdX$ and if $X$ follows a GBM dynamic then we can conclude that $X=e^{t(\mu-\frac{\sigma^2}{2})+\sigma W_{t}}$. But how do we check or know that this satisfy the equation, we only proven that if $X$ satisfy the equation then it looks like this, not the converse.

Answer 1

If $(X_t)_{t \geq 0}$ is an Itô process, i.e. a process of the form $$dX_t = b(X_t) \, dt + \sigma(X_t) \, dW_t, \tag{1}$$ then formally, by Itô's formula, $$d(\ln X_t) = \frac{dX_t}{X_t} - \frac{1}{2} \frac{(dX_t)^2}{X_t^2}.$$ Equivalently, by $(1)$, $$d(\ln X_t) = \frac{\sigma(X_t)}{X_t} \, dW_t + \left( \frac{b(X_t)}{X_t} - \frac{1}{2} \frac{\sigma(X_t)^2}{X_t^2} \right) \, dt. \tag{2}$$ Now if $(X_t)_{t \geq 0}$ is a solution to the SDE $$dX_t = \mu X_t \, dt + \sigma X_t \, dW_t, \tag{3}$$ then $b(x)=\mu x$ and $\sigma(x)=\sigma x$, and so $(2)$ reads

$$d(\ln X_t) = \sigma dW_t + \left( \mu- \frac{\sigma^2}{2} \right) \, dt.$$

Integrating both sides we get

$$\ln(X_t)-\ln(X_0) = \sigma W_t + \left(\mu- \frac{\sigma^2}{2} \right) t,$$

i.e.

$$X_t = X_0 \exp \left( \sigma W_t + \left[\mu-\frac{\sigma^2}{2} \right]t \right).$$

It remains to check that this process is indeed a solution to $(3)$. To this end, you can apply once more Itô's formula.

Remark: An alternative way to solve the SDE $(3)$ is the variation of constants approach (...which you might know from ODEs), see this answer.

Answer 2

Check your formula, a $\sigma^2 X^2$ is missing.

The idea is to cancel $X$ with the denominator, which comes from differentiating. One function having this property is the logarithm. Applying the Itô-formula leads to it.

Answer 3

The resemblance of the SDE to the basic exponential ODE might inspire one to try a solution of the form $X_t =X_0\exp(\mu t+\sigma W_t)$. The Ito differential of this $X$ is, however, off by a finite variation term. Tweaking our guess we arrive at $X_t=X_0\exp(\mu_t+\sigma W_t + K_t)$, where $K$ is adapted and of finite variation, but otherwise to be determined. Now Ito's formula yields to $$ {d X_t\over X_t } = \mu \,dt +\sigma \,dW_t +dK_t+{1\over 2}\sigma^2\,dt, $$ provoking the choice $K_t = -\sigma^2t/2$.