Assume a risk free bond $B_t$and the stock St follow the dynamics of the Black & Scholes model. (with interest rate r, stock drift $\mu$ and volatility $\sigma$). Find $\beta$ such that the process $e^{-\beta t}S_{t}^3$ is a martingale under the risk neutral measure Q. How should I use martingale on the function to find the value for $\beta$.
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Under $\mathbb{Q}$, let $\mathrm{d}S_t=rS_t\mathrm{d}t+\sigma S_t\mathrm{d}B_t$. Define the function $f(t,x)=e^{-\beta t}x^3$ with partial derivatives $f_t(t,x)=-\beta f(t,x)$, $f_x(t,x)=3e^{-\beta t}x^2$ and $f_{xx}(t,x)=6e^{-\beta t}x$.
You are interested in the process $X_t=f(t,S_t)=e^{-\beta t}S_t^3$. By Ito's Lemma, \begin{align*} \mathrm{d}f(t,S_t) &= \left( f_t(t,S_t) +rS_tf_x(t,S_t)+\frac{1}{2}\sigma^2S_t^2 f_{xx}(t,S_t) \right)\mathrm{d}t+ \bigg( \sigma S_t f_x(t,S_t) \bigg)\mathrm{d}B_t \\ &= \left(-\beta f(t,S_t) + 3rf(t,S_t)+\frac{1}{2}\sigma^26f(t,S_t) \right)\mathrm{d}t + \bigg(\sigma 3f(t,S_t)\bigg)\mathrm{d}B_t \\ &= \left( -\beta+3r+3\sigma^2 \right)f(t,S_t)\mathrm{d}t+3\sigma f(t,S_t)\mathrm{d}B_t. \end{align*} This firstly proves that $f(t,S_t)$ is again a geometric Brownian motion. In general, any powers of a GBM is a GBM again. Finally, for $f(t,S_t)$ to be a martingale, you need to have zero drift (i.e. no $\mathrm{d}t$ term). (Recall that the simple Ito Integral $\mathrm{d}B_t$ is already a martingale and adding a deterministic time trend would violate the martingale condition).
So, you need to set $$\beta=3(r+\sigma^2).$$
You can easily include a dividend yield by replacing $r$ with $r-q$.
