2
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ What is the superfix (2) supposed to mean here? $\endgroup$ Aug 13, 2020 at 17:50

1 Answer 1

1
$\begingroup$

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*}

$\endgroup$

Your Answer

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

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