Pricing a Geometric Average Call Through the Stock Payoff

# Using Black-Scholes to price a geometric average price call


# Using Black-Scholes to price a geometric average price call












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 by Mark Joshi (score 6)

https://quant.stackexchange.com/a/17452

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 by AFK (score 3)

https://quant.stackexchange.com/a/17453

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).

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