Assume the risk-free bond $B_t$ and the stock $S_t$ follow the dynamics of the Black & Scholes model

without dividends (with interest rate r, stock drift $\mu$ and volatility $\sigma$). Show that $S_{u;T} := \frac{S_{u}}{S_T}$ under the measure $Q^S$ (with the stock as a numeraire) can be written as $exp\{(-r-\frac{\sigma^2}{2})(T-u)+\sigma\hat{W}_{T-u}\}$ where $\hat{W}_t$ for $t\in[0,T]$ has the same law of a Wiener process under the $Q^S$ measure.

I'm stuck on how to solve this question. Would really appreciate the help.


1 Answer 1


Let $\mathbb{Q}$ be the risk-neutral probability measure which uses the risk-free bank account $(B_t)$ as numeraire. In general, $\mathrm{d}B_t=r_tB_t\mathrm{d}t$. In the Black-Scholes setting, $r_t\equiv r$, we have $B_t=e^{rt}$.

The stock measure $\mathbb{Q}_S$ uses the compounded stock price $S_te^{qt}$ as numeraire and is defined via the Radon Nikodym derivative \begin{align*} \frac{\mathrm{d} \mathbb{Q}_S}{\mathrm{d}\mathbb{Q}}(t) &= \frac{B_0}{B_t}\frac{S_te^{qt}}{S_0} \\ &= \frac{1}{e^{rt}}\exp\left(\left(r-q-\frac{1}{2}\sigma^2\right)t+\sigma W_t\right)e^{qt} \\ &=\exp\left(-\frac{1}{2}\sigma^2t+\sigma W_t\right) \\ &= \mathcal{E}\left(\sigma W_t\right), \end{align*}

using that $\mathrm{d}S_t=(r-q)S_t\mathrm{d}t+\sigma S_t\mathrm{d}W_t$. Recall that $(W_t)$ is a standard Brownian motion under $\mathbb{Q}$. Using Girsanov's theorem, we know that $\mathbb{Q}_S\sim\mathbb{Q}$ and that the process \begin{align*} \hat{W}_t &=W_t-\sigma t \end{align*} is a standard Brownian motion under $\mathbb{Q}_S$.

So, we conclude with \begin{align*} S_{u,T} &= \frac{S_u}{S_T} \\ &= \exp\left(\left(r-q-\frac{1}{2}\sigma^2\right)u+\sigma W_u -\left(\left(r-q-\frac{1}{2}\sigma^2\right)T+\sigma W_T\right)\right) \\ &= \exp\left(\left(r-q-\frac{1}{2}\sigma^2\right)(u-T)-\sigma (W_T-W_u)\right) \\ &= \exp\left(\left(r-q-\frac{1}{2}\sigma^2\right)(u-T)-\sigma \left(\hat{W}_T +\sigma T -\left(\hat{W}_u+\sigma u\right)\right)\right) \\ &\overset{d}{=} \exp\left(-\left(r-q-\frac{1}{2}\sigma^2\right)(T-u)-\sigma \big(\hat{W}_{T-u}+\sigma(T-u)\big)\right) \\ &= \exp\left(-\left(r-q+\frac{1}{2}\sigma^2\right)(T-u)-\sigma \hat{W}_{T-u}\right) \\ &\overset{d}{=} \exp\left(-\left(r-q+\frac{1}{2}\sigma^2\right)(T-u)+\sigma \hat{W}_{T-u}\right). \end{align*}


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