suppose $$dA = \mu Adt + \sigma AdX.$$ is a geometric Brownian motion. One says that the Probability $P(A,t)$ of $A$ reashing the critical level $K(t)$ before maturity: $$\dfrac{\partial P}{\partial t} + \dfrac{1}{2}\sigma^2A^2\dfrac{\partial^2 P}{\partial A^2}+\mu \dfrac{\partial P}{\partial A} = 0$$ I know this is the Kolomogorov backward equation for transition density, but why this is true for probability function here? It seems we should replace CDF by PDF i.e replace $P$ by $\dfrac{\partial P}{\partial x}$?

Actually the background is the probability of default, Paul Wilmott on Quantitative Finance Volumne II Page641 39.2.2 enter image description here

  • $\begingroup$ @Quantuple sorry, here is $P(A\geq K(t))$ $\endgroup$ – A.Oreo Feb 13 '17 at 15:12
  • $\begingroup$ Ah, now this makes more sense :) Then you could equivalently see that equation as the pricing PDE for a digital option. $\endgroup$ – Quantuple Feb 13 '17 at 15:19
  • $\begingroup$ the price PDE of digital option is same as the black-schole equation, how does a probability become this form? $\endgroup$ – A.Oreo Feb 14 '17 at 1:23
  • $\begingroup$ Because your $P(a, t)$ is actually $Q(A_T > K \mid A_t = a) = \Bbb{E}^\Bbb{Q}[ 1\{ A_T > K \} \mid A_t = a]$. Which can be seen as the undiscounted price of a digital option. The discounted price should verify the usual pricing PDE since it should emerge as a $\Bbb{Q}$-martingale (Feynman-Kac formula). $\endgroup$ – Quantuple Feb 14 '17 at 6:37
  • $\begingroup$ If I did not interpret your $P(A,t)$ correctly would you please insert a copy of the page 641 you refer to since I do not own that book. $\endgroup$ – Quantuple Feb 14 '17 at 8:41

Your Answer

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

Browse other questions tagged or ask your own question.