Using Black-Scholes to price a geometric average price call

Question

Sorry if this is the wrong exchange for this question. It seems to be the most relevant, anyway.

I'm trying to learn and understand the Black-Scholes framework, with a focus on the stochastic differential equation approach (the exam I will be taking focuses on it). So I set out a challenge for myself. I'd like to price a special geometric average price call, where the average is taken on $S_0$ and $S_1$.

My intuition is that what I'm "really" trying to price is a European call, where the underlying is the geometric average of the stock price. I defined a process $G(t)$ by

\begin{equation*} G(t) = \left(S_0 S_t\right)^{\frac{1}{2}}. \end{equation*}

The intention is to apply Ito's lemma, so I took derivatives: \begin{align*} G_t &= 0 & G_S &= \frac{1}{2}S_0S_t^{-\frac{1}{2}} & G_{SS} = -\frac{1}{4}S_0^{-\frac{3}{2}}. \end{align*}

After applying Ito's lemma, I end up with the stochastic differential equation \begin{equation*} \frac{\mathrm{d}G(t)}{G(t)} = \frac{1}{2}\left[\left(\alpha - \delta - \frac{1}{4}\sigma^2\right) \mathrm{d}t + \sigma \mathrm{d} Z_t \right], \end{equation*} where $\alpha$ is the stock's expected rate of return.

So I see that $G(t)$ is a geometric Brownian motion. But this is where I become deeply confused, since it is a derivative of the stock $S_t$. So when I do risk-neutral pricing, do I have to assume that $S$ earns the risk-free rate (which amounts to setting $\alpha = r$, in the stochastic differential equation above), or do I assume that G earns the risk-free rate? Or something else?

My intuition is telling me that once I figure out which rates to use and where, I can just use the Black-Scholes formula for a call to get this claim price done. Am I on the right track?

Answer 1

This is not the way to do it. The Black-Scholes argument requires the underlying to be tradable. $S_{t}^{1/2}$ is not tradable.

Instead, recognize that the underlying is still $S_t$ but the pay-off has changed to $$ (\alpha S_{t}^{1/2} - \beta)_+ $$ for appropriate constants $\alpha,\beta.$ So, the derivation of the BS equation still holds and the boundary condition is different.

To solve the easiest route is risk-neutral expectation. $$ e^{-rT}E((\alpha S_{T}^{1/2} - \beta)_+ ). $$ To get the distribution of $S_{T}^{1/2}$ get that of $S_T$ which is lognormal and so has lognormal square root.

Answer 2

if we forget about $S_0$, you are just trying to price a power option, i.e. an option on $S^\alpha$.

By Ito $$ d \log S^\alpha = \alpha d\log S = \alpha (r- q - \frac{1}{2}\sigma^2 ) dt + \alpha\sigma dW_t $$ This can be rewritten $$ d \log S^\alpha = (r-q'-\frac{1}{2}\sigma'^2 ) dt + \sigma' dW_t $$ If you set

$\sigma' = \alpha \sigma$

$q' = r -\frac{1}{2}(\alpha\sigma)^2 - \alpha (r- q - \frac{1}{2}\sigma^2 ) = (1-\alpha)r - \frac{1}{2}\alpha(\alpha - 1)\sigma^2 + \alpha q$

Then $e^{-rt}S^\alpha_te^{q't}$ is also a martingale and the BS formula applies with the new parameters (the interest rate stays the same but the vol and div yield change).

$$ E^Q_t[e^{-rT}(S^\alpha_T - K)_+] = C_{BS}(S_t^\alpha,T-t,K,r,q',\sigma') $$