# Price of Geometric basket call option

I wonder if someone can explain how this should be solved:

Compute the arbitrage free price at t=0 of the Geometric basket call option (My remark: the payoff function is $\max\left(\left( \prod_{i=1}^n S_i(T) \right)^{1/n} - K , 0)\right)$ , on the stock $S_t$.

Hint: $\left( \prod_{i=1}^n S_i(T) \right)^{1/n}$ has the same distribution as $e^X$ where $X \sim N(a, b)$. Start by calculating $a$ and $b$. After that, you can find the price in the same fashion as in the derivation of the Black Scholes formula.

I think I'm supposed to use the Feynman-Kac formula for the price process: $$F(t,s)= e^{-r(T-t)}\mathbb{E}[\Phi(S(T)) \ | \ S_t =s]$$ (or something like this), but I'm confused by the expectation and w.r.t what I'm integrating. I'd be really grateful if someone could help me with this one!

• The key here is that the product of two lognormal random variabels is still lognormally distributed.
– will
Sep 24 '17 at 20:12
• Try to write out $\left( \prod_{i=1}^n S_i(T) \right)^{1/n}$ explicitly, then show the exponent is a normal random variable. Sep 25 '17 at 13:01
• Thanks for your help! If I put $Y= (\prod_{i=1}^nS_i(T))^{1/n}$ then the density of Y is $f_Y(y)=f_X(ln(y)), \ X \sim \mathcal{N}(a,b)$ and then the price becomes $\Pi_t = e^{-r(T-t)}\mathbb{E}[(Y-K)^+]=\int_{\mathbb{R}}(y-K)^+f_y(y)dy$, is this procedure correct so far? Sep 26 '17 at 13:55
• Why not use the density of $X$? Sep 26 '17 at 14:12
• You mean plug it into the integral? $\int (y-K)^+f_Y (y)dy= \int (y-K)^+ \frac{f_X(ln(y))}{y} dy$ ? Sep 26 '17 at 14:46

We assume that, under the risk-neutral probability measure, \begin{align*} dS_i(t) = S_i(t) \big(r dt + \sigma_i dW_i(t)\big), \end{align*} for $i=1, \ldots, n$, where $r$ is the interest rate, $\sigma_i$ is the volatility and $\{W_i(t), t\ge 0\}$ is a standard Brownian motion such that $d\langle W_i, W_j\rangle(t) = \rho_{i, j}dt$ for $i\ne j$. Then, \begin{align*} \Big(\Pi_{i=1}^nS_i(T)\Big)^{\frac{1}{n}} = \Big(\Pi_{i=1}^nS_i(0)\Big)^{\frac{1}{n}}\exp\bigg(\Big(r -\frac{1}{2n}\sum_{i=1}^n\sigma_i^2\Big)T + \frac{1}{n}\sum_{i=1}^n \sigma_iW_i(T)\bigg). \end{align*} Let \begin{align*} \sigma = \frac{1}{n}\sqrt{\sum_{i=1}^n \sigma_i^2 + 2\sum_{i\ne j} \rho_{i, j} \sigma_i \sigma_j}, \end{align*} and \begin{align*} \xi = \frac{\frac{1}{n}\sum_{i=1}^n \sigma_iW_i(T)}{\sigma \sqrt{T}}. \end{align*} Then, $\xi$ is standard normal, that is, $\xi\sim N(0, 1)$. Let \begin{align*} F = \Big(\Pi_{i=1}^nS_i(0)\Big)^{\frac{1}{n}}\exp\bigg(\Big(r -\frac{1}{2n}\sum_{i=1}^n\sigma_i^2 + \frac{1}{2}\sigma^2\Big)T\bigg). \end{align*} Then, \begin{align*} \Big(\Pi_{i=1}^nS_i(T)\Big)^{\frac{1}{n}} = F e^{-\frac{1}{2}\sigma^2 T + \sigma \sqrt{T} \xi}. \end{align*} Moreover, as the derivation of the Black-Scholes formula, \begin{align*} E\left(\max\bigg(\Big(\Pi_{i=1}^nS_i(T)\Big)^{\frac{1}{n}} -K, 0\bigg) \right) &=E\left(\max\bigg(F e^{-\frac{1}{2}\sigma^2 T + \sigma \sqrt{T} \xi} -K, 0\bigg) \right)\\ &=F\Phi(d_1) - K \Phi(d_2), \end{align*} where $\Phi$ is the cumulative distribution function of a standard normal random variable, \begin{align*} d_1 = \frac{\ln \frac{F}{K}+\frac{1}{2}\sigma^2 T}{\sigma \sqrt{T}}, \end{align*} and \begin{align*} d_2 = d_1 - \sigma \sqrt{T}. \end{align*} The option value is then given by \begin{align*} e^{-rT} \big[F\Phi(d_1) - K \Phi(d_2) \big]. \end{align*}