Why Geometric Brownian Motion Stays Positive

# Proving $\mathbb{P}(S_t<0|S_0=s_0)=0$ for Geometric BM


# Proving $\mathbb{P}(S_t<0|S_0=s_0)=0$ for Geometric BM












I am trying to prove that for the geometric Brownian motion of a stock $\textrm{d}S_t=\mu S_t\textrm{d}t+\sigma S_t\textrm{d}B_t$ with strictly positive constants $\mu$ and $\sigma$ and and $S_0=s_0>0$, we have $\mathbb{P}(S_t<0|S_0=s_0)=0$. The conditional probability looks oddly like what we should get from the Feynman-Kac formula, with $\mathbb{P}(S_t<0)=\mathbb{E}(\mathbf{1}_{S_t<0})$. However, I am unsure how to construct the PDE to solve and therefore derive the required probability. Am I even on the right track, and if so, how should the PDE be constructed? If not, what steps should be taken? Thank you!

## Answer by Jan Stuller (score 5, accepted)

https://quant.stackexchange.com/a/59516

I think the easiest way to derive the solution to the GBM is via Ito's Lemma.

The GBM: $dS_t = \mu S_t dt + \sigma S_t dW_t$ is a short hand for:

$$ S_t = S_0 + \int_{h=0}^{h=t}\left(\mu S_h\right)dh + \int_{h=0}^{h=t}\left(\sigma S_h\right)dW_h $$

Ito process is defined as:

$$ X_t = S_0 + \int_{h=0}^{h=t}\left(a(X_h,h)\right)dh + \int_{h=0}^{h=t}\left(b(X_h,h)\right)dW_h $$

(where $a(X_t,t)$ and $b(X_t,t)$ must be square integrable). In the GBM case, $X_t = S_t$, $a(X_t,t)=\mu S_t $ and $b(X_t,t) = \sigma S_t$, so GBM is an Ito Process.

Ito's lemma states that for any well behaved function $F()$ of $X_t$ and $t$, where $X_t$ must be an Ito Process, the process for $F(X_t,t)$ will be as follows:

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

To derive the solution to the GBM, set $F(S_t,t)=ln(S_t)$ (how come we can take the log, without "a priori" knowing whether the GBM SDE for $S_t$ can potentially produce negative $S_t$ values? see bottom *). Then, computing the derivatives, we get: $\frac{\partial F}{\partial t}=0$ (because $F=ln(S_t)$ is only a function of $S_t$ and not $t$ explicitly), $\frac{\partial F}{\partial S}=\frac{1}{S_t}$, $\frac{\partial F^2}{\partial S^2}=-\frac{1}{S_t^2}$.

Substituting the above derivatives into the equation for $F$, we get:

$$F(X_t,t)= ln(S_0) + \int_{h=0}^{h=t}\left(0+\frac{1}{S_h}a(X_h,h)_{=\mu S_h}-\frac{1}{2}\frac{1}{S_h^2}b(X_h,h)^2_{=\sigma^2 S_h^2}\right)dh + \int_{h=0}^{h=t}\left(\frac{1}{S_h}b(X_h,h)_{=\sigma S_h}\right)dW_h=\\=ln(S_0) + \int_{h=0}^{h=t}\left(\mu-\frac{1}{2}\sigma^2\right)dh + \int_{h=0}^{h=t}\left(\sigma \right)dW_h=\\=ln(S_0)+(\mu - 0.5 \sigma^2)t + \sigma W_t$$

With $F(X_t,t)=ln(S_t)$, we now just need to exponentiate both sides to get:

$$S_t=S_0e^{(\mu-0.5 \sigma^2)t+\sigma W_t}$$

Now we can move on to the probability problem:

$$\mathbb{P}(S_t<0|S_0=s_0)=\mathbb{P}(s_0e^{(\mu-0.5 \sigma^2)t+\sigma W_t}<0)=\\=\mathbb{P}(e^{(\mu-0.5 \sigma^2)t+\sigma W_t}<0)=\\=\mathbb{P}(e^{(\mu-0.5 \sigma^2)t}e^{\sigma W_t}<0)=\\=\mathbb{P}(e^{\sigma W_t}<0)$$

Now $\sigma W_t \epsilon \mathbb{R}$ and $e^x>0 \forall x\epsilon \mathbb{R}$, so we can deduce that:

$$\mathbb{P}(e^{\sigma W_t}<0)=0$$.

Edit: * a very nice proof was given here. Borrowing that proof:

With $S_0>0$, set $\tau$ to be the first time that the SDE for $S_t$ makes $S_t$ hit zero. Suppose $\tau < \infty$. Then, for some $0<t<\tau$, take the log to get: $$ln(S_t)=ln(S_0)+\mu t -0.5 \sigma^2t + \sigma W_t$$ As $t\uparrow \tau$, the LHS goes to $-\infty$, whilst the RHS converges to a finite quantity. The contradiction proves that $\mathbb{P}(\tau < \infty)=0$.

## Answer by jherek (score 1)

https://quant.stackexchange.com/a/59582

Another sketch of proof:

If you move to the equivalent PDE (using Feynman-Kac), you can assume that S is positive, find the solution by log-transfomation. Then as the solution is unique given initial conditions, and it is the solution of the original PDE, S must be positive.