Proving that Geometric Brownian Motion is positive

Question

An Ito process is defined as:

$$X_t = X_0 + \int_{h=0}^{h=t}\left(a(X_h,h)\right)dh + \int_{h=0}^{h=t}\left(b(X_h,h)\right)dW_h$$

Where $a(X_t, t)$ and $b(X_t, t)$ are square integrable processes.

If one chooses $a(X_t,t)=\mu X_t$ and $b(X_t,t)=\sigma X_t$, with $\mu$ and $\sigma$ being constant, we obtain the well known Geometric Brownian Motion (where $\mu$ can be regarded as the "drift" and $\sigma$ as the "volatility"):

$$X_t = X_0 + \int_{h=0}^{h=t}\left(\mu X_h\right)dh + \int_{h=0}^{h=t}\left(\sigma X_h\right)dW_h$$

The above is of course still an Ito process. Ito's lemma states that for any (well-behaved), twice-differentiable function $F()$ of $X_t$ and $t$, the function $F(X_t, t)$ follows the process:

$$F(X_t,t)= F(X_0,t_0) + \int_{h=0}^{h=t}\left(\frac{\partial F}{\partial t}+\frac{\partial F}{\partial X}a(X_h,h)_{=\mu X_h}+\frac{1}{2}\frac{\partial^2 F}{\partial X^2}b(X_h,h)^2_{=\sigma^2 X_h^2}\right)dh + \int_{h=0}^{h=t}\left(\frac{\partial F}{\partial X}b(X_h,h)_{=\sigma X_h}\right)dW_h$$

Using the Ito's lemma above, we can easily find the solution to the GBM equation for $X_t$ by taking $F(X_t,t)=ln(X_t)$ to obtain:

$$X_t=X_0e^{\left(\mu t -0.5 \sigma^2 t + \sigma W_t\right)}$$

Question: I just got asked how come we can take the log of $X_t$, without knowing whether the equation $X_t = X_0 + \int_{h=0}^{h=t}\left(\mu X_h\right)dh + \int_{h=0}^{h=t}\left(\sigma X_h\right)dW_h$ potentially produces negative values for $X_h$ along some specific paths.

Looking at the solution $X_t=X_0e^{\left(\mu t -0.5 \sigma^2 t + \sigma W_t\right)}$, it is obvious that for any initial positive $X_0$, $X_t$ cannot be negative. However, using this argument to justify taking the log is a bit sketchy because we are using the solution to validate the method we use to derive that solution.

Is there a way to show that the GBM SDE for $X_t$ produces strictly positive values if the initial value $X_0>0$ for all $X_h: 0\leq h \leq t < \infty$?

I suppose we could start by writing out the integrals by definition:

$$X_t = X_0 + \int_{h=0}^{h=t}\left(\mu X_h\right)dh + \int_{h=0}^{h=t}\left(\sigma X_h\right)dW_h= \\= \mu \lim_{n\to\infty}\sum_{h=t_0}^{h=t_{n-1}}X_h((h+1)-h)+\sigma \lim_{n\to\infty}\sum_{h=t_0}^{h=t_{n-1}}X_{h}(W_{h+1}-W_{h})=\\=\mu \lim_{n\to\infty} \left(x_0(t_{1}-t_{0})+...+x_{t_{n-1}}(t_n-t_{n-1}) \right) +\sigma \lim_{n\to\infty}\left(x_0(W_{t_1})+...+x_{t_{n-1}}(W_{t_n}-W_{t_{n-1}})\right)$$

Not sure where to go from here and whether the above is a good idea?

Answer 1

Let $X$ solve the GBM SDE with $X_0>0$. Let $\tau$ be the first time $t$ at which $X_t=0$. Clearly $\tau>0$. By Ito's formula, $$ \log(X_t/X_0) = (\mu-\sigma^2/2)t+\sigma B_t,\qquad 0\le t<\tau, $$ almost surely. On the event $\{\tau<\infty\}$, the left side of this equality goes to $-\infty$ as $t\uparrow \tau$ while the right side converges to the finite quantity $(\mu-\sigma^2/2)\tau+\sigma B_\tau$. This contradiction means that $P[\tau<\infty]=0$. That is, a solution $X$ satisfies $P[X_t>0,\,\forall t]=1$.

(Alternatively, you can exhibit one strictly positive solution, and the GBM SDE has a unique solution.)

Answer 2

We don't need to justify taking the log beforehand because we get an explicit solution, i.e. "using the solution to validate the method we use to derive that solution" is perfectly fine. A good way to think about it is that applying Ito's lemma to $F(t,X_t) = \ln(X_t)$ doesn't give us the solution, it just gives us something we think is likely to be a solution. When we use that to obtain $X_t = X_0 e^{\sigma W_t + (\mu - \frac 12 \sigma^2)t}$, we haven't shown that $X_t$ is actually the solution yet. Once we apply Ito to $X_t$ we can verify that it does indeed satisfy $dX_t = \mu X_t dt + \sigma X_t dW_t$ and now can say that $X_t$ is a solution.

Note that if $X_0 < 0$ then we couldn't actually take $\ln(X_t)$, but the "candidate solution" we get from doing the same derivation is $X_t = X_0 e^{\sigma W_t + (\mu - \frac 12 \sigma^2)t}$ and we can use Ito's lemma to show that this is still a solution, even though the method we used to obtain it wasn't fully justified.