# 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!

• What is the superfix (2) supposed to mean here? Aug 13, 2020 at 17:50

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*}
\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*}