1
$\begingroup$

I have a pay-off function for a derivative which is defined by the Heaviside difference between $G$ and $B$ shifted by $-F$. To find the value of $V_{t=0}$, I need to find $\tau$ when $\frac{dV}{dt} = 0$.

I start by setting up as a classic diffusion argument for the expected value of $\frac{dV}{dt}$.

$\mathbb{E}\left[\frac{dV}{dt}\right] = \max{(-F,\, G_t -B_t-F)} \approx \mathbb{H}\left(G_t - B_t\right) - F = \mathbb{E}\left[G_t\right]^+ - \mathbb{E}\left[B_t\right]^- -F$

where $\mathbb{H}$ is the Heaviside step function: $f(\frac{dV}{dt}, \,\tau) =0 \quad \forall \; G_t < B_t$

Assume $G_t$ and $B_t$ are both drifting proccess with drift $= -d$; $G_t$ is a $\mathbb{P}$ Brownian Motion (i.e, Wiener process) with variance $= \sigma^2$. Therefore, we can set up set up the diffusion process as follows:

$\mathbb{E}\left[\frac{dV}{dt}\right] =\varPhi\left(\frac{\log{\left(\frac{G}{B}\right)}+\frac{\sigma ^2 \tau }{2}}{\sigma \sqrt{\tau}}\right)G e^{-d \tau } - \varPhi\left(\frac{\log{\left(\frac{G}{B}\right)}-\frac{\sigma ^2 \tau }{2}}{\sigma \sqrt{\tau}}\right)B e^{-d \tau } - F $

$\mathbb{E}\left[\frac{dV}{dt}\right] = 0 \to \frac{e^{-d \tau}}{\sqrt{2 \pi}} \left[G\left(\int_{-\infty }^{\frac{\log \left(\frac{G}{B}\right)+\frac{\sigma ^2 \tau }{2}}{\sigma \sqrt{\tau }}} e^{-\frac{Z^2}{2}} \, dZ\right)-{B \left(\int_{-\infty }^{\frac{\log \left(\frac{G}{B}\right)-\frac{\sigma ^2 \tau }{2}}{\sigma \sqrt{\tau }}} e^{-\frac{Z^2}{2}} \, dZ\right)}\right] = F$

where $\varPhi(X)$ is the CDF of the standard normal (Gaussian) distribution for i.i.d. $X_i$.

I can solve for the roots of $\frac{dV}{dt}$ graphically and/or recursively, but I require an analytical and/or closed-form approach.

How can I find $\tau$ when $\frac{dV}{dt} = 0$?

$\endgroup$

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.