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Deriving a Stock-Numeraire Brownian Motion for a Price Ratio

Article Quant Q&A · Author: Anon

Summary

The answer derives the distribution of the ratio of a stock price at an earlier time to its value at a later time under a stock-based numeraire. It starts with the risk-neutral measure associated with the bank account, defines the stock measure using the compounded stock price, and writes the Radon–Nikodym density. In the Black–Scholes setting, Girsanov’s theorem shifts the Brownian motion so that the transformed process is Brownian under the stock measure.

The solution then expresses the stock-price ratio using the stock dynamics and rewrites the future Brownian increment in terms of the shifted process. It uses the distributional symmetry of Brownian increments to express the result with a Brownian term over the time gap. The answer includes a dividend yield parameter even though the question states no dividends; setting that yield to zero recovers the no-dividend case. The result relies on the stated Black–Scholes assumptions and constant rate and volatility.

Key ideas

  • The stock measure is defined by changing from the bank-account numeraire to the compounded stock numeraire.
  • Girsanov’s theorem shifts the Brownian motion under the stock measure.
  • The earlier-to-later stock-price ratio can be rewritten using the shifted Brownian increment over the time gap.
  • The derivation includes a dividend-yield parameter, which is set to zero for the question’s no-dividend setting.

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# Asian Options-Change of Numeraire


# Asian Options-Change of Numeraire












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.

## Answer by Kevin (score 5, accepted)

https://quant.stackexchange.com/a/49750

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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