5
$\begingroup$

At the beginning of all the stuff about Girsanov theorem, we introduced the Radon-Nikodym derivative as $Z_\infty := \frac{d \mathbb{Q}}{d \mathbb{P}}\vert_{\mathcal{F}_\infty}$.

Next, we considered the following:

\begin{align} Z_t = \mathbb{E}[Z_\infty\vert\mathcal{F}_t]=\frac{d \mathbb{Q}}{d \mathbb{P}}\vert_{\mathcal{F}_t}. \end{align} So this last line does not seems so obvious for me. I do not understand the last equality. The conclusion is that the derivative is a martingale. Do I have to work with the definition of conditional expectation?

$\endgroup$
5
$\begingroup$

Let $(\Omega, \mathcal{F}_{\infty}, \mathbb{P})$ be a probability space and let $(B_t)$ be a standard Brownian motion with its natural filtration $(\mathcal{F}_t)_{t\geq 0}$. Let $\mathbb{Q}$ be a probability measure which is absolutely continuous w.r.t $\mathbb{P}$ on $(\Omega, \mathcal{F}_{\infty})$ then Radon-Nikodym theorem yields the existence of measurable function $Z_{\infty} =\left. \frac{d\mathbb{Q}}{d\mathbb{P}}\right|_{\mathcal{F}_{\infty}}$ (the Radon-Nikodym derivative).

Now define $Z_t = \mathbb{E}[Z_{\infty}|\mathcal{F}_t]$ we want to show that $Z_t$ is a martingale with respect to $\mathcal{F}_t$. Let $s \leq t$ then $$\mathbb{E}[Z_t | \mathcal{F}_{s}] = \mathbb{E}[\mathbb{E}[Z_{\infty}|\mathcal{F}_t] | \mathcal{F}_{s}]$$ the tower property of conditional expectation yields that $$\mathbb{E}[\mathbb{E}[Z_{\infty}|\mathcal{F}_t] | \mathcal{F}_{s}] = \mathbb{E}[Z_{\infty}|\mathcal{F}_s]=Z_s.$$ Hence, $(Z_t)$ is a martingale.

Now we show that $Z_t = \left.\frac{d\mathbb{Q}}{d\mathbb{P}}\right|_{\mathcal{F}_t}$, indeed, for any bounded $\mathcal{F}_t$-measurable function $f$ we have \begin{align*} \int_{\Omega}f \ d\mathbb{Q}=& \int_{\Omega} f Z_{\infty} \ d\mathbb{P} = \mathbb{E}_{\mathbb{P}}[fZ_{\infty}]\\=&\mathbb{E}[\mathbb{E}[fZ_{\infty} | \mathcal{F}_t]]= \mathbb{E}[f \mathbb{E}[Z_{\infty}|\mathcal{F}_t]] \\=& \mathbb{E}[fZ_t] = \int_{\Omega} fZ_t d\mathbb{P}. \end{align*} Hence, $Z_{\infty}d\mathbb{P} = Z_t d\mathbb{P}$ on $\mathcal{F}_t$ for each $t$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.