Deriving the Stock-Measure Distribution of a Geometric Brownian Asian Average
Summary
The document derives the distribution of a time-averaged stock value rescaled by the terminal stock price, under the stock measure in a Black-Scholes market without dividends. The key step is to write the average and terminal price using the same Brownian path, then express their ratio through reversed increments over the remaining time to maturity.
Under the stock measure, a drift-adjusted Brownian motion is standard Brownian motion. Rewriting the time-reversed increments in terms of that process shows that the rescaled average has the same law as the time average of a stock process with drift equal to the negative risk-free rate and the original volatility. The answer emphasizes that proving the rescaled process is a martingale would not by itself establish its distribution. The derivation relies on the stated continuous-time model and measure change; it does not address discrete sampling or dividends.
Key ideas
- A rescaled Asian average can be analyzed by expressing it as an integral involving terminal-to-time Brownian increments.
- Under the stock measure, the Brownian motion is shifted by the volatility times time.
- Time reversal transforms the ratio of the average to the terminal stock into an integral with negative risk-free drift.
- A martingale property alone does not establish the distributional identity.
- The derivation assumes a dividend-free Black-Scholes model with continuous averaging.
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# Arithmetic Asian Option
# Arithmetic Asian Option
Assume the risk-free bond Bt and the stock St follow the dynamics of the Black & Scholes model
without dividends (with interest rate r, stock drift $μ$ and volatility $σ$).
Let $A_T:=\frac{1}{T}\int_{0}^{T}S_tdt$. Define $A^*_{T}:=S_{0}\frac{A_T}{S_T}$.Show that under the measure ${Q}^S$ the random variable $A^*_{T}$ has the same law $A_{T}^{(2)}=\frac{1}{T}\int_{0}^{T}S^{(2)}_tdt$ where $dS_{t}^{(2)}=S_{t}^{(2)}(-rdt+\sigma dW_{t}^{(2)})$ where $W_{t}^{(2)}$ is a Wiener process.
How exactly would you that it has the same law? The only approach I can think of is by Ito's lemma and showing that it's a martingale? But I'm not entirely sure.
Really appreciate the help. Thank you!
## Answer by Kevin (score 1)
https://quant.stackexchange.com/a/49767
Showing that $(A_t^*)$ is a martingale does not really help you in understanding the distribution of $A_T^*$. Instead, the key is your previous question.
Under $\mathbb{Q}\sim\mathbb{P}$, you have \begin{align*} S_t=S_0\exp\left(\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t\right). \end{align*} Under $\mathbb{Q}_S\sim\mathbb{Q}$, you have \begin{align*} S_t^{(2)}=S_0\exp\left(-\left(r+\frac{1}{2}\sigma^2\right)t+\sigma W_t^{(2)}\right). \end{align*}
From your previous question, recall that if $(W_t)$ is a standard Brownian motion under $\mathbb{Q}$, then \begin{align*} \hat{W}_t=W_t-\sigma t \end{align*} is a standard Brownian motion under the stock measure $\mathbb{Q}_S$, i.e. $\hat{W}_t\overset{d}{=}W_t^{(2)}$.
Consider now \begin{align*} A_T^* &= \frac{S_0}{S_T}A_T \\ &= \exp\left(-\left(r-\frac{1}{2}\sigma^2\right)T-\sigma W_T\right) \frac{1}{T}\int_0^T S_0\exp\left(\left(r-\frac{1}{2}\sigma^2\right)t+\sigma W_t\right) \mathrm{d}t \\ &= \frac{1}{T}\int_0^T S_0\exp\left(\left(r-\frac{1}{2}\sigma^2\right)(t-T)-\sigma (W_T-W_t)\right) \mathrm{d}t. \end{align*}
Following the steps of the answer to your previous question, you can manipulate the integrand to
\begin{align*} A_T^*&= \frac{1}{T}\int_0^T S_0\exp\left(-\left(r+\frac{1}{2}\sigma^2\right)(T-t)+\sigma \hat{W}_{T-t}\right) \mathrm{d}t\\ &= -\frac{1}{T}\int_{T}^{0} S_0\exp\left(-\left(r+\frac{1}{2}\sigma^2\right)u+\sigma \hat{W}_{u}\right) \mathrm{d}u \\ &= \frac{1}{T}\int_{0}^{T} S_0\exp\left(-\left(r+\frac{1}{2}\sigma^2\right)u+\sigma \hat{W}_{u}\right) \mathrm{d}u. \end{align*}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.