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!

  • 1
    $\begingroup$ The key here is that the product of two lognormal random variabels is still lognormally distributed. $\endgroup$
    – will
    Sep 24, 2017 at 20:12
  • $\begingroup$ 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. $\endgroup$
    – Gordon
    Sep 25, 2017 at 13:01
  • $\begingroup$ 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? $\endgroup$
    – user202542
    Sep 26, 2017 at 13:55
  • $\begingroup$ Why not use the density of $X$? $\endgroup$
    – Gordon
    Sep 26, 2017 at 14:12
  • $\begingroup$ You mean plug it into the integral? $\int (y-K)^+f_Y (y)dy= \int (y-K)^+ \frac{f_X(ln(y))}{y} dy$ ? $\endgroup$
    – user202542
    Sep 26, 2017 at 14:46

1 Answer 1


The answer below assumes that the payoff provided by the OP is correct. The question appears simple, but still a lot of subtleties to deal with.

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*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.