I have $\frac{dS_t}{S_t} = rdt + \sigma dW_t$ as usual under the money-market numéraire and I need to price options with payoffs

$$(S_T f(S_T))^+$$

How do I express the stock dynamics using the stock as numéraire, and how do I get the stock distribution under the equivalent measure.


Let $P$ be the risk-neutral measure. We define the measure $P_S$ such that \begin{align*} \frac{dP_S}{dP}\big|_t &=\frac{S_t}{e^{rt}S_0}\\ &=e^{-\frac{1}{2}\sigma^2 t+\sigma W_t}. \end{align*} Then $\{\widehat{W}_t \mid t \ge 0\}$, where \begin{align*} \widehat{W}_t = W_t -\sigma t, \end{align*} is a standard Brownian motion under the measure $P_S$. Moreover, under $P_S$, \begin{align*} \frac{dS}{S} &= rdt + \sigma dW_t\\ &=\big(r+\sigma^2\big)dt + \sigma d \widehat{W}_t. \end{align*} That is, the stock price process $S$ is still log-normal. The option price is then given by \begin{align*} e^{-rT}E\Big(\big(S_Tf(S_T) \big)^+\Big) &= e^{-rT}E_S\bigg(\Big(\frac{dP_S}{dP}\big|_T\Big)^{-1}\big(S_Tf(S_T) \big)^+\bigg)\\ &=S_0E_S\left(\big(f(S_T) \big)^+\right), \end{align*} where $E$ and $E_S$ are respectively the expectation operators under the measures $P$ and $P_S$.

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    $\begingroup$ How to deduce that $\{\widehat{W}_t \mid t \ge 0\}$ is a standard Brownian motion? Girsanov's Theorem? $\endgroup$ – Idonknow Apr 6 '20 at 2:04
  • $\begingroup$ Yes. That is correct. $\endgroup$ – Gordon Apr 6 '20 at 2:19
  • $\begingroup$ May I know which version of Girsanov's Theorem are you applying? The Girsanov that I know involves risk-neutral measure. I just started learning forward measure recently so I am not familiar with Girsanov's Theorem for forward measure. $\endgroup$ – Idonknow Apr 6 '20 at 2:21
  • $\begingroup$ See this book by Oksendal. $\endgroup$ – Gordon Apr 6 '20 at 11:37
  • $\begingroup$ Thanks for your suggestion. However, I can't find any forward measure content relatead. $\endgroup$ – Idonknow Apr 6 '20 at 11:39

Let $\text{d}S_t = \mu S_t \text{d}t +\sigma S_t\text{d}W_t$. under the real-world measure

With $S_t$ being numeraire, then $e^{rt}/S_t$ must be a martingale under the equivalent martingale measure.

Under the real world measure, $\frac{e^{rt}}{S_t}= \exp(rt -\mu t-\sigma W_t+\frac{1}{2}\sigma^2t)$, where $W_t$ is a Brownian motion under this measure.

Now you need to make a Cameron-Martin-Girsanov transform to make $\frac{e^{rt}}{S_t}$ a martignale. This essential comes down to $r-\mu+\frac{1}{2}\sigma^2 = -\frac{1}{2}\sigma^2$, or $\mu = r+\sigma^2$.

so under the risk-neutral measure with $S_t$ being numeraire, $S_t=S_0\exp(r+\sigma^2t-\frac{1}{2}\sigma B_t)$, where $B_t$ is a Brownian motion under the risk-neutral measure.. To find time $t<T$ value $V_t$ of an asset with pay off $S_TF(S_T)$, then

$\frac{V_t}{S_t} = \mathbb E[\frac{S_TF(S_T)}{S_T}|\mathcal{F}_t]=E[F(S_T)|\mathcal{F}_t]$

Note, for example, if $F(\cdot) = (K-\cdot)^+$, you can still use Black-Scholes formula though you need to figure out the appropriate parameter and might need to multiply by a factor. Essentially, this is because $S_t$ is still log-normal distributed.

  • $\begingroup$ What do you mean "under the risk-neutral measure with $S_t$ being numeraire"? $\endgroup$ – Gordon Nov 4 '15 at 16:47
  • $\begingroup$ You need to have the dynamics for $S_t$ with $S_t$ being the numeraire. $\endgroup$ – Gordon Nov 4 '15 at 16:57
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    $\begingroup$ It is unclear, why $\frac{\exp(rt)}{S_{t}}$ must be a martingale? From my POV, according to Radone-Nikodime theorem $\frac{N(t)D(t)}{N(0)}$ defines a new measure, where $N(t)=S_{t}$ and $D(t)=\exp(-rt)$, and this must be a martingale under the real measure. Therefore, $\mu=r+\sigma^{2}$ and this measure is defined with $\exp((r+\sigma^{2})t+\frac{1}{2} \sigma W_{t})$. $\endgroup$ – user18387 Nov 22 '15 at 10:40

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