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!

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    $\begingroup$ You can solve $S_t=S_0 exp((\mu-0.5\sigma^2)t+\sigma W_t)$ for $S_0>0$. You can see that $S_t\geq0$ irrespective of $W_t$. $\endgroup$
    – fes
    Commented Nov 22, 2020 at 8:22

2 Answers 2


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 $\mathbb{P}(\tau < \infty)=0$.

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    $\begingroup$ I think there's a typo when you calculate $F(X_t,t)$. The final line (and all lines that follow it) should read as $\mu\color{red}{\mathbf{-}}0.5\sigma^2$ and not $\mu\color{red}{\mathbf{+}}0.5\sigma^2$. $\endgroup$
    – Kevin
    Commented Nov 22, 2020 at 10:59
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    $\begingroup$ @Kevin: you're totally right :) Careless algebra has plagued me through the years. My brain processor tends to work on multiple tasks in parallel, and somehow the task of algebra doesn't get enough processing power dedicated to it. I think I might have to shift a bit more power to it :P $\endgroup$ Commented Nov 22, 2020 at 11:09
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    $\begingroup$ It does not look very clean that we take the logarithm of S to prove that S is positive. S must be positive first to be able to take the logarithm. $\endgroup$
    – jherek
    Commented Nov 24, 2020 at 22:56
  • $\begingroup$ @jherek: thank you for the feedback. I'd normally agree, but the question says that we can assume that the initial state $s_0$ is strictly positive, and we are then meant to show that $\mathbb{P}(S_T<0|s_0)=0$, i.e. we condition the probability on the initial state. So It's ok to use that initial state and just work with the equation for $S_T$ that way, including taking a log. $\endgroup$ Commented Nov 25, 2020 at 7:29
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    $\begingroup$ Alternatively you can refer to some standard uniqueness conditions, see here en.wikipedia.org/wiki/…. $\endgroup$
    – fes
    Commented Nov 29, 2020 at 14:34

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.

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    $\begingroup$ hi, I don’t exactly understand what you mean, could you describe it a bit more? $\endgroup$
    – user107224
    Commented Nov 25, 2020 at 15:27

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