Deriving a Barrier-Hitting Probability PDE for Geometric Brownian Motion

# The PDE of the probability hitting the barrier before T


# The PDE of the probability hitting the barrier before T












Suppose: $$d S=\mu S dt+\sigma Sd W$$ $Q(t,S)$ is the probability that $S$ hit the barrier $B(S_t<B)$ before $T,$ then $Q$ satisfies following `PDE` $$Q_t+\dfrac{1}{2}\sigma^2S^2_{SS}Q+\mu S Q_S=0.$$ Could I prove that this way

`Proof:` $$Q(t,S)=\mathbb{P}(\tau_B\leq T)$$ here $\tau_B$ is the `first passage time` at level $B$.

Then use the `reflection principle` for a Wiener process:

We have $$\mathbb{P}(\tau_B\leq T)=2\mathbb{P}(S_T>B)=2\int^{\infty}_Bp(t,S,T,y)d y$$ Here $p(t,S,T,y)$ is the `transition function` of $S_T$

From `Kolmogorov backward equation` we know $$p_t+\dfrac{1}{2}\sigma^2S^2p_{SS}+\mu S p_S=0.$$ then take the derivatives into the integral, we done.

I am not sure is the whole process correct? And is there any standard way to calculate the such PDE of probability, since the default probability also meet this pde

## Answer by Gordon (score 2)

https://quant.stackexchange.com/a/33992

May be I have overlooked something, but I believe that \begin{align*} Q(t, S) = \mathbb{P}\left(\tau_{B} \le T \mid \mathcal{F}_t\right). \end{align*} Then $\{Q(t, S), \, 0<t < T\}$ is a martingale, and the PDE follows immediately, by noting that \begin{align*} dQ &= Q_t dt + Q_S dS + \frac{1}{2}Q_{SS} d\langle S, S\rangle_t\\ &=\Big(\underbrace{Q_t + \mu S Q_S + \frac{1}{2}\sigma^2Q_{SS} S^2}_{=0}\Big)dt + \sigma S Q_S dW_t. \end{align*}

## Answer by M. Jeunesse (score 1)

https://quant.stackexchange.com/a/32786

Reflection principal ? Reflection principle.

It holds for the Brownian process, not the GBM. [Reflection principle is quite specific to symmetric random walks].

By chance, if $\mu-\frac{\sigma^2}{2}=0$ and $\sigma>0$, then you have : $$\mathbb{P}(\tau^S_B<T)=\mathbb{P}(\tau^W_{\frac{1}{\sigma}\ln(B)}<T)$$ and you can apply reflection principle.