# 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

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*}

• @noob2: Thanks for the enhancement. – Gordon May 1 '17 at 20:10
• Could you please bring more precision on the justification of $Q(t,S)$ being a martingale? Otherwise your answer is great. Thanks! – JejeBelfort May 4 '17 at 12:37
• @JejeBelfort: By tower law, it is easy to see that $\{Q(t, S), \, 0<t<T\}$ is a martingale. – Gordon May 4 '17 at 12:53

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.

• yeah, this is one of my confusion, reflection principal doesn't hold for any process? Another confusion is that, if the barrier is moving, say $B(t),$ does it only depend on the final value $B(T)$ i.e $\mathbb{P}(\tau_{B(t)}<T)=2\mathbb{P}(W(T)>B(T))?$ – A.Oreo Mar 3 '17 at 1:20
• maybe, reflection principal holds for all Levy process with $X_0=0,$ I think – A.Oreo Mar 3 '17 at 3:36
• no it does not hold for all levy processes. A poisson process is a levy process. Reflection principle is quite specific to symmetric random walk. – MJ73550 Mar 6 '17 at 10:05