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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.

SRKX
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jojo
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2 Answers2

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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 $P_S$.

Gordon
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    How to deduce that ${\widehat{W}_t \mid t \ge 0}$ is a standard Brownian motion? Girsanov's Theorem? – Idonknow Apr 06 '20 at 02:04
  • Yes. That is correct. – Gordon Apr 06 '20 at 02:19
  • 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. – Idonknow Apr 06 '20 at 02:21
  • See this book by Oksendal. – Gordon Apr 06 '20 at 11:37
  • Thanks for your suggestion. However, I can't find any forward measure content relatead. – Idonknow Apr 06 '20 at 11:39
  • There is no forward measure involved in this question. However, no matter what kind of measures, the principles are the same. – Gordon Apr 06 '20 at 12:08
  • Last question, how to show that $P_S$ defined in your answer above is indeed a measure? Can we use Radon-Nikodym Theorem? – Idonknow Apr 06 '20 at 12:16
  • Yes, Randon-Nikodym theorem. – Gordon Apr 06 '20 at 12:21
  • Is it possible that you provide some details on how to use Randon-Nikodym Theorem to prove that $P_S$ is a measure? Or does it follow directly from the statement of the theorem? – Idonknow Apr 06 '20 at 12:23
  • You need to show it is a measure with a measure one for the whole sample space. I think you need to ask this as another question. – Gordon Apr 06 '20 at 12:30
  • I posted the question in this post. https://quant.stackexchange.com/questions/53078/how-to-use-radon-nikodym-theorem-to-show-that-forward-measire-is-indeed-measure – Idonknow Apr 06 '20 at 12:38
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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.

Lost1
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  • What do you mean "under the risk-neutral measure with $S_t$ being numeraire"? – Gordon Nov 04 '15 at 16:47
  • You need to have the dynamics for $S_t$ with $S_t$ being the numeraire. – Gordon Nov 04 '15 at 16:57
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    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 $\exp((r+\sigma^{2})t+\frac{1}{2} \sigma W_{t})$. –  Nov 22 '15 at 10:40