2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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*}

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

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.

$\endgroup$
  • $\begingroup$ 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 $\mathbb{P}(\tau_{B(t)}<T)=2\mathbb{P}(W(T)>B(T))?$ $\endgroup$ – A.Oreo Mar 3 '17 at 1:20
  • $\begingroup$ maybe, reflection principal holds for all Levy process with $X_0=0,$ I think $\endgroup$ – A.Oreo Mar 3 '17 at 3:36
  • $\begingroup$ 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. $\endgroup$ – MJ73550 Mar 6 '17 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.