I've been having a hard time to applicate Girsanov theorem with Radon-Nikodym derivative in the demonstration of German-El Karoui-Rochet formule.

I know that $\Pi_0:=S_0\mathbb{Q}^S(S_T\geq K)-K\mathbb{Q}^T(S_T\geq K)P(0,T)$. I have to calculate these two probabilities. I start from the second assuming that the procedure is the same for the first.

Let $Z_t:=\frac{S_t}{P(t,T)}$ a process under physical measure $\mathbb{P}$ and dynamics $dZ_t:=Z_t[u_t^zdt+\sigma_t^zdW_t]$. Since the process is $(\Omega,F,\mathbb{P})$-definited while the probability to calculate is $\mathbb{Q}^T$-definited, i have to apply a change of measure with Girsanov theorem. I know that Girsanov allows me to construct (through Radon-Nikodym derivative expressed in terms of exponential martingale: $L(\omega):=\frac{d\mathbb{Q}^T}{d\mathbb{P}}:=M_t\Rightarrow E^{\mathbb{Q}^T}[X]=E^{\mathbb{P}}[M_tX]$) a martingale measure $\mathbb{Q}^T$ equivalent to the physical measure in such a way that the process $\tilde{W_t}$ under the new measure is a Brownian Motion standard. Let $P(T,T):=1$, i start saying that:

$\mathbb{Q}^T(S_T\geq K)=\mathbb{Q}^T(\frac{S_T}{1}\geq K)=\mathbb{Q}^T(\frac{S_T}{P(T,T)}\geq K)=\mathbb{Q}^T(Z_T\geq K)$

Now the problems.

Given that to do the change of measure it's true that $E^{\mathbb{Q}^T}[Z_T]=E^{\mathbb{P}}[M_tZ_T]$, i thought that the $\mathbb{Q}^T$-martingale resulting should have, for definition, a dynamics contains only diffusive part. Formerly: $d\mathbb{Z}_t^{\mathbb{Q}^T}=\sigma_t^zd\tilde{W}_t$ . However the text (very incomplete, actually) says the dynamics of $Z_t$ under $\mathbb{Q}^T$ measure is $d\mathbb{Z}_t^{\mathbb{Q}^T}=Z_t\sigma_t^zd\tilde{W}_t$.

I thought to applicate the Fundamental Theorem of change of numeraire saying that, since $P(t,T)$ fulfils the conditions of theorem (always assumes values strictly positive and is a $\mathbb{Q}$-martingale for the First Fundamental Theorem of APT), for the change of numeraire $Z_t:=\frac{S_t}{P(t,T)}$ is a $\mathbb{Q}^T$-martingale. But this would contradict the hypothesis in which the process is $\mathbb{P}$-definited.

I tried to apply the Ito's formule to $Z_T^{\mathbb{Q}^T}=\left ( Z_Te^{\int_{0}^{T}\sigma_s^zdW_s-\frac{1}{2}\int_{0}^{T}(\sigma_s^z)^2ds} \right )^{\mathbb{P}}$ (in according to the change of measure) but i dont'understand why the text fixed the stochastic integral $dW_s$ under $\mathbb{Q}^T$: this contradict not only the change of measure with Radon-Nikodym measure but the definition of exponential martingale, that is a process $\mathbb{P}$-definited.

Anyway, I can't derive the result that is $\mathbb{Q}^T(S_T\geq K)=\mathbb{Q}^T(-\tilde{Y}\leq \frac{ln(\frac{S_0}{KP(0,T)})-\frac{1}{2}\sum ^2}{\sqrt{\sum ^2}})$, where:

  1. $\sum ^2=\int_{0}^{T}(\sigma_s^z)^2ds$;
  2. $\tilde{Y}$ is the standardization of $Y:=e^{\int_{0}^{T}\sigma_s^zdW_s-\frac{1}{2}\int_{0}^{T}(\sigma_s^z)^2ds}$ (with $dW_s$ under $\mathbb{Q}^T$).

Any help would be really welcome. Thanks!


The definition of $Z$ tell us that the process $\lbrace{Z_t\rbrace}_{t\geq0}$ is a martingale under the $T-$forward measure $\mathbb{Q}^T$. Given the dynamic of $Z$, we can find out the Radon-Nikodym derivative. \begin{align*} dZ_t &= Z_t\left(u_t^zdt + \sigma_t^zdW_t \right) \\ &= Z_t\sigma_t^z\left(dW_t + \frac{u_t^z}{\sigma_t^z}dt \right) \\ &= Z_t\sigma_t^zd\bar{W}_t \quad \quad (1) \end{align*} where $\bar{W}$ is a $\mathbb{Q}^T$ Brownian motion. Therefore, we can construct the R-N measure which is \begin{equation*} M_t = \exp\left(-\frac12\int_0^t \left(\frac{u_s^z}{\sigma_s^z}\right)^2ds - \int_0^t \frac{u_s^z}{\sigma_s^z}dW_s\right) \end{equation*} Now we have all the ingredients to compute $\mathbb{Q}^T(S_T \geq K)$. \begin{align*} \mathbb{Q}^T(S_T \geq K) &= \mathbb{Q}^T(Z_T \geq K) \\ &=\mathbb{Q}^T\left(Z_0\exp\left(-\frac12\int_0^T(\sigma_s^z)^2ds + \int_0^T \sigma_s^zdW_s\right) \geq K\right)\\ &=\mathbb{Q}^T\left(\int_0^T \sigma_s^zd\bar{W}_s \geq \log\left(\frac{K}{Z_0}\right) + \frac12\Sigma^2\right) \\ &=\mathbb{Q}^T\left(\tilde{Y} \geq \frac{\log\left(\frac{K}{Z_0}\right) + \frac12\Sigma^2}{\Sigma}\right) \\ &=\mathbb{Q}^T\left(-\tilde{Y} \leq \frac{\log\left(\frac{S_0}{KP(0,T)}\right) - \frac12\Sigma^2}{\Sigma}\right) \end{align*} The fourth equality follows from the fact that $\lbrace{\int_0^t \sigma_s^zd\bar{W}_s\rbrace}_{t\geq0}$ is a Wiener process under the measure $\mathbb{Q}^T$. Hence, it follows a gaussian distribution with mean $0$ and variance $\Sigma$.

Some observations on your post :

  1. $Z$ is a lognormal process under $\mathbb{P}$ as well as under $\mathbb{Q}^T$. Hence, one cannot have $dZ_t = \sigma_t^zd\bar{W}_t$ (which is btw a normal process). That being said, you are right that under $\mathbb{Q}^T$ the process $Z$ is a real martingale and should only have the diffusive part (no finite variation process) and it is case in (1).
  2. Often a numeraire, say $Y$, is chosen for a given process, say $X$, such that $X$ becomes a martingale under a new measure called the martingale measure associated to the numeraire $Y$. As noted, $Y$ has to fulfill some conditions to be a numeraire.
  • $\begingroup$ @ Sesame: Thanks so much for your answer. I take this opportunity to clearing another aspect. If passing from $\mathbb{P}$ to $\mathbb{Q}$ means applying $\frac{d\mathbb{Q}}{d\mathbb{P}}=M_t=e^{\int_{0}^{t}z_sdW_s^{\mathbb{Q}}-\frac{1}{2}\int_{0}^{t}z_s^2ds}$ with $W_t^{\mathbb{Q}}=W_t-\int_{0}^{t}z_sds$, is correct saying that for passing from $\mathbb{Q}$ to $\mathbb{P}$ means applying $\frac{d\mathbb{P}}{d\mathbb{Q}}=M_T^{-1}=e^{-\int_{0}^{t}z_sdW_s^{\mathbb{P}}+\frac{1}{2}\int_{0}^{t}z_s^2ds}$ with $W_t^{\mathbb{Q}}=W_t+\int_{0}^{t}z_sds$? I get confused with the signs. $\endgroup$ – Marco Pittella Apr 26 at 14:09
  • 1
    $\begingroup$ If $\mathbb{P} \sim \mathbb{Q}$, then there exists a strictly positive process $(\mathbb{P}, F)$-martingale $M_t$ for $t \leq T$ such that $\frac{d\mathbb{Q}}{d\mathbb{P}} = M_t$. Plus, $M_0 = 1$ and $E_{ \mathbb{P}}[M_t] = 1, \forall t \leq T$. Thus, we need to have $E_{\mathbb{P}}[\frac{d\mathbb{Q}}{d\mathbb{P}}]=1$ or $E_{\mathbb{Q}}[\frac{d\mathbb{P}}{d\mathbb{Q}}]=1$. In your case, we don't have this unless we express the Brownians in the right measure. Informally, if $d\mathbb{Q} = M_td\mathbb{P}$ then we use $W_t^{{P}}$ and if $d\mathbb{P} = M_t^{-1}d\mathbb{Q}$ then we use $W_t^{{Q}}$ $\endgroup$ – Sesame Apr 26 at 14:49
  • $\begingroup$ For precision, we have 1.$\frac{d\mathbb{Q}}{d\mathbb{P}} = M_t = e^{- \frac12\int_0^tz_s^2ds +\int_0^tz_sdW_s^{\mathbb{P}}}$ 2. $\frac{d\mathbb{P}}{d\mathbb{Q}} = M_t^{-1} = e^{- \frac12\int_0^tz_s^2ds -\int_0^tz_sdW_s^{\mathbb{Q}}}$ $\endgroup$ – Sesame Apr 26 at 14:57
  • $\begingroup$ @ Sesame: I don't understand two things: 1) If $dW_t^{\mathbb{Q}^T}=dW_t+\frac{\mu_t^z}{\sigma_t^z}dt\Rightarrow W_t^{\mathbb{Q}^T}=W_t+\int_{0}^{t}\frac{\mu_s^z}{\sigma_s^z}ds$, so $\frac{d\mathbb{P}}{d\mathbb{Q}^T}=e^{-\int_{0}^{t}z_sdW_s^{\mathbb{Q}}-\frac{1}{2}\int_{0}^{t}z_s^2ds}=e^{-\int_{0}^{t}\frac{\mu_s^z}{\sigma_s^z}(dW_s+\frac{\mu_s^z}{\sigma_s^z}ds)-\frac{1}{2}\int_{0}^{t}(\frac{\mu_s^z}{\sigma_s^z})^2ds}=e^{-\int_{0}^{t}\frac{\mu_s^z}{\sigma_s^z}dW_t-\frac{3}{2}\int_{0}^{t}(\frac{\mu_s^z}{\sigma_s^z})^2ds}$. Where I wrong? 2) What happened to $\mu_s^z$ in the numerator? $\endgroup$ – Marco Pittella Apr 26 at 18:27
  • $\begingroup$ 1. This is correct. But I don't see why you need to express the BM in the measure $\mathbb{P}$. In your precise case, we are only interested in the measure $\mathbb{Q}^T$. Read my previous comment. 2. What do you mean by that ? $\endgroup$ – Sesame Apr 26 at 18:37

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