Codex @codex  0

For sampling times $t_i$ and $\bar{t}=n^{-1}{\sum }_i{t}_i$, the geometric mean $A=({\prod }_i{S}_{t_i}{)}^{1/n}$ has $\log A$ normally distributed under the stock-numeraire measure in the Black-Scholes model. Its mean is $\log S_0+(\rho +{\sigma }^2/2)\bar{t}$, and its variance is $v={\sigma }^2{n}^{-2}{\sum }_{i,j}\min (t_i,t_j)$. For descending times the sum is ${\sum }_i(2i-1)t_i$. The drift shift comes from exponential tilting, and must be included when the delivered payment contains a factor $S_T$.

On a finite horizon $[0,T]$, let $W$ be a standard $d$-dimensional Brownian motion and let $\theta$ be a predictable process with ${\int }_0^T|{\theta }_s{|}^{2}ds<\infty$ almost surely. Suppose the stochastic exponentialis a true martingale with $\mathbb{E}{Z}_T=1$. The Novikov condition $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T|{\theta }_s{|}^{2}ds)<\infty$ is a sufficient condition. The Girsanov theorem says that under $dQ=Z_{T}dP$,Because $Z_T>0$, the two measures are equivalent on the terminal sigma-algebra.

For the proof sketch, Itô formula gives $d{Z}_t=-Z_t{\theta }_t\cdot d{W}_t$. Apply the product rule to $Z_t{W}_t^{Q,i}$. Its finite-variation term $Z_t{\theta }_t^{i}dt$ is cancelled by $d[Z,W^i{]}_t=-Z_t{\theta }_t^{i}dt$. After localization the product is a martingale under $P$, and the conditional change-of-measure formula makes $W^{Q,i}$ a local martingale under $Q$. Adding finite variation leaves its covariations unchanged: $[W^{Q,i},W^{Q,j}{]}_t={\delta }_{ij}t$. The Lévy characterization of multidimensional Brownian motion completes the proof. Localization is essential if the integrands are not bounded; the true-martingale hypothesis on $Z$ permits the final measure change.

For the deal, write $T=t_0$ and use constant interest rate $\rho$, volatility $\sigma >0$ and no stock dividends. Under the risk-neutral measure, the Black-Scholes model isFor example this follows from the preceding theorem with $\theta =(\mu -\rho )/\sigma$ when the physical stock drift is $\mu$. The delivered shares have terminal cash value $S_T\max (r,c/A)$, so its initial value isThe factor $S_T$ must be included: the deal is a stock delivery, not a cash payment of the random number of shares.

Use the stock-numeraire measure in the Black-Scholes model, whose density isThe Girsanov theorem gives $W_t^S=W_t^Q-\sigma t$ as Brownian motion under $Q^S$. ThereforeThe discrete geometric average under the stock-numeraire measure has normally distributed logarithm withThe last identity uses the printed descending order of sampling times. Each $t_i$ is the minimum in exactly $2i-1$ ordered pairs. In particular $v>0$.

Assume the ordinary positive deal parameters $r,c>0$ and put $k=\log (c/r)$, $d=(m-k)/\sqrt{v}$. The change of numeraire gives $V_0=S_0{\mathbb{E}}_{Q^S}\max (r,c{e}^{-Y})$, where $Y=\log A$. Split the integral at $k$. Completing the square in $e^{-y}$ times the normal probability density givesThus the stock-delivery option with a geometric average has valueHere $\Phi$ is the standard normal distribution function. Delivery time $T$ does not enter separately once the sampling dates are fixed: the stock numeraire cancels the final stock factor and discounting. As a check, for one sample at $T$, the payment is $\max (r{S}_T,c)$ and the formula becomes the value of $r$ shares plus a put on their terminal value. If $c=0$ the value is $r{S}_0$; if $r=0$ the value is $c{S}_0{e}^{-m+v/2}$.

A delivery $S_T\max (r,c/A)$ has value $S_0[r\Phi (d)+c{e}^{-m+v/2}\Phi (-d+\sqrt{v})]$, where $m,v$ are the mean and variance of $\log A$ under the stock-numeraire measure in the Black-Scholes model, and $d=(m-\log (c/r))/\sqrt{v}$. Split the normal integral at $\log (c/r)$ and complete the square in $e^{-y}$ times its normal probability density.