# Black-Scholes and Markovian contingent claim

Background information:

Proposition 4.1 - For a European Markovian contingent claim, the Black-Scholes price satisfies $$\Theta(\tau,S) = -\frac{\sigma^2 S^2}{2}\Gamma(\tau,S) - rS\Delta(\tau,S) + rV(\tau,S)$$

Problem - Use Proposition 4.1, to show the above equality. That is let $\tilde{S}_t = e^{-rt}S_t$ and $\tilde{V}(t,\tilde{S}_t) = e^{-rt}V(t,S_t)$ be respectively the discounted underlying price and discounted option price. Then, we can show that $$\partial_t\tilde{V}(t,\tilde{S}) = -\frac{\sigma^2\tilde{S}^2}{2}\partial_{\tilde{S}\tilde{S}}V(t,\tilde{S})$$

I am confused where to begin, I think I can manage showing this if I have a formula or just a set up to work from, any suggestions is greatly appreciated.

• What is the relationship between $t$ and $\tau$? Is $\tau = T-t$? Feb 25, 2016 at 14:00
• yes $\tau = T - t$ @Gordon Feb 25, 2016 at 16:24
• @Gordon do you know how to prove this, I am completely lost Feb 25, 2016 at 19:28
• I would think your $\tau$ and $t$ are the same, by comparing with the Black-Scholes equation such as that given by en.wikipedia.org/wiki/Black%E2%80%93Scholes_model Feb 25, 2016 at 20:03
• If you assume they are the same can you prove it? My professor makes many errors so let us just assume that is the case Feb 25, 2016 at 20:34

## 2 Answers

We assume the following Black-Scholes equation: \begin{align} \frac{\partial V}{\partial t} = -\frac{\sigma^2 S_t^2}{2}\frac{\partial^2 V}{\partial S_t^2} -r S_t \frac{\partial V}{\partial S_t} +r V.\tag{1} \end{align} From the assumption, \begin{align} V(t,\, S_t) = e^{rt}\tilde{V}(t,\, \tilde{S}_t).\tag{2} \end{align} Then \begin{align*} \frac{\partial V}{\partial t} &= re^{rt}\tilde{V}+e^{rt}\frac{\partial \tilde{V}}{\partial t} + e^{rt}\frac{\partial \tilde{V}}{\partial \tilde{S}_t}\frac{\partial \tilde{S}_t}{\partial t}\\ &= re^{rt}\tilde{V}+e^{rt}\frac{\partial \tilde{V}}{\partial t} -rS_t \frac{\partial \tilde{V}}{\partial \tilde{S}_t},\tag{3}\\ \frac{\partial V}{\partial S_t} &= e^{rt}\frac{\partial \tilde{V}}{\partial S_t}\\ &=e^{rt}\frac{\partial \tilde{V}}{\partial \tilde{S}_t}\frac{\partial \tilde{S}_t}{\partial S_t}\\ &=\frac{\partial \tilde{V}}{\partial \tilde{S}_t},\tag{4}\\ \frac{\partial^2 V}{\partial S_t^2} &=\frac{\partial^2 \tilde{V}}{\partial \tilde{S}_t^2}\frac{\partial \tilde{S}_t}{\partial S_t}\\ &=e^{-rt}\frac{\partial^2 \tilde{V}}{\partial \tilde{S}_t^2}.\tag{5} \end{align*} Now, plugin (2)-(5) into (1), we obtain that \begin{align*} \frac{\partial \tilde{V}}{\partial t} = -\frac{\sigma^2 \tilde{S}_t^2}{2} \frac{\partial^2 \tilde{V}}{\partial \tilde{S}_t^2}. \end{align*}

• Gordon, you are a champion Feb 25, 2016 at 21:23

A good place to start is to calculate the $\Delta$ for your option $\tilde{V}$. Once you calc your delta, plug it back into the black scholes PDE. Now check which terms cancel each other out. After you remove the terms that cancel each other out you have the desired result.