Let $\lambda$ be a probability measure on $\Omega$ (finite), with filtration $\{\mathcal{F}_t\}$. Define $\nu(X) = \lambda\left(X\frac{d\nu}{d\lambda}\right)$, where $\frac{d\nu}{d\lambda}$ is a random variable i.e., $\nu(\omega) = \lambda(\omega)\frac{d\nu}{d\lambda}(\omega)$, all $\omega\in\Omega$. Show that $$E\nu[X|\mathcal{F_t}] = \frac{E_{\lambda}\left[X\frac{d\nu}{d\lambda}|\mathcal{F_t}\right]}{E_{\lambda}\left[\frac{d\nu}{d\lambda}|\mathcal{F}_t\right]}$$

Recall from the second fundamental theorem of asset pricing $$\frac{d\nu}{d\lambda} = \frac{S_T^{0}}{\lambda(S_T^{0})}$$ if $S_T^{0}$ is a constant then $$\frac{d\nu}{d\lambda} = 1 \ \ \Rightarrow \ \ \lambda = \nu$$ The change of measure formula is $$E_{\nu}[X] = E_{\lambda}\left[X\frac{d\nu}{d\mu}\right]$$

For some attainable claim $X$ let $\phi$ be a self financing strategy replicating $X$ then by the first fundamental theorem of asset pricing $$V_t(\phi) = E_{\nu}\left[X\frac{S_t^{0}}{S_T^{0}} |\mathcal{F_t}\right]$$

I am pretty sure the result will follow from one of these fundamental theorems of asset pricing but I am not sure where to go from here. Sorry for the messy start, also if you need me to write the three fundamental theorems I would be happy to do so. Any comments or suggestions is greatly appreciated.

Alternative Solution - For all $\omega\in \Omega$, let $\mathcal{F}_t(\omega) = \mathcal{F}_t$ be the partition element containing $\omega$. Then

\begin{align*} E_{\nu}[X|\mathcal{F}_t](\omega) &= \frac{\sum_{\omega\in\mathcal{F}_t(\omega)} X(\omega)\nu(\omega)}{\sum_{\omega\in\mathcal{F}_t(\omega)} \nu(\omega)}\\ &= \frac{\sum_{\omega\in\mathcal{F}_t(\omega)} X(\omega)\lambda(\omega)\frac{d\nu}{d\lambda}(\omega)}{\sum_{\omega\in\mathcal{F}_t(\omega)}\lambda(\omega)\frac{d\nu}{d\lambda}(\omega)}\\ &= \frac{\left( \frac{\sum_{\omega\in\mathcal{F}_t(\omega)} X(\omega)\lambda(\omega)\frac{d\nu}{d\lambda}(\omega)}{\sum_{\omega\in\mathcal{F}_t(\omega)} \lambda(\omega)} \right )}{\left(\frac{\sum_{\omega\in\mathcal{F}_t(\omega)} \lambda(\omega)\frac{d\nu}{d\lambda}(\omega)}{\sum_{\omega\in\mathcal{F}_t(\omega)} \lambda(\omega)} \right )}\\ &= \frac{E_{\lambda}\left[X\frac{d\nu}{d\lambda}|\mathcal{F}_t\right](\omega)}{E_{\lambda}\left[\frac{d\nu}{d\lambda}|\mathcal{F}_t\right](\omega)} \end{align*}

  • $\begingroup$ If $\lambda$ and $\nu$ are measures, then $\lambda(\omega)$ and $\nu(\omega)$ do not make good sense. What do you mean $\lambda(S_T^0)$? Is $S_t^0$ a numeraire process? $\endgroup$
    – Gordon
    Oct 26 '16 at 16:21
  • $\begingroup$ @Gordon I may have written the question down wrongly I will double check. $S_t^{0}$ is a numeraire process. Apologies if I made some notation mistake, I will correct it. $\endgroup$
    – Wolfy
    Oct 27 '16 at 0:18
  • $\begingroup$ @Gordon in regards to the $\lambda(S_T^{0})$ question, it is just a normalization factor to ensure that $\nu$ has total mass $1$. We define $\nu$ by means of the Radon-Nikodym derivative. $\endgroup$
    – Wolfy
    Oct 27 '16 at 1:02
  • $\begingroup$ See also here. $\endgroup$
    – Gordon
    Oct 27 '16 at 13:49
  • $\begingroup$ @Gordon posted the solution but the latex generator here is messed up $\endgroup$
    – Wolfy
    Nov 2 '16 at 15:57

Let define $\mathbb{Q}$ and $\mathbb{P}$ two equivalent probabilities on a filtered space $(\Omega,(\mathcal{F}_t)_{t\geq 0})$

Let define $Z_T=\frac{d\mathbb{Q}}{d\mathbb{P}}$ restricted to $\mathcal{F}_T$ measurable events.

It means that for $X_T$ being $\mathcal{F}_T$ measurable we have: $$\mathbb{E}^{\mathbb{Q}}[X_T] = \mathbb{E}^{\mathbb{P}}\left[Z_TX_T\right]$$

Let $t\leq T$.

We want to define the change of probability measure on $\mathcal{F}_t$. i.e we want to find $Z_t$ being $\mathcal{F}_t$ measurable such that for $X_t$ being $\mathbb{F}_t$ measurable, we have:

$$\mathbb{E}^{\mathbb{Q}}[X_t]= \mathbb{E}^{\mathbb{P}}\left[Z_tX_t\right]$$

By definition of $Z_T$, and since $X_t$ is also $\mathcal{F}_T$ measurable, we have: $$\mathbb{E}^{\mathbb{Q}}[X_t]= \mathbb{E}^{\mathbb{P}}\left[Z_TX_t\right]$$


for any $X_t$ being $\mathcal{F}_t$ measurable we have $Z_t$ being $\mathcal{F}_t$ measurable such that:

$$\mathbb{E}^{\mathbb{P}}[Z_T X_t]=\mathbb{E}^{\mathbb{P}}[Z_t X_t]$$

so $Z_t = \mathbb{E}^{\mathbb{P}}[Z_T|\mathcal{F}_t]$ by definition of conditional expectation.

Let $Y_T$ being $\mathcal{F}_T$ measurable, then we want to compute $\mathbb{E}^{\mathbb{Q}}[Y_T|\mathcal{F}_t]$.

We denote $Y_t = \mathbb{E}^{\mathbb{Q}}[Y_T|\mathcal{F}_t]$

We look for $Y_t$ such that for any $X_t$ being $\mathcal{F}_t$ measurable, we have :

$$\mathbb{E}^{\mathbb{Q}}[Y_TX_t]=\mathbb{E}^{\mathbb{Q}}[Y_t X_t]$$

By definition of $Z_T$ we have $\mathbb{E}^{\mathbb{Q}}[Y_TX_t]=\mathbb{E}^{\mathbb{P}}[Z_TY_TX_t]$

By definition of $Z_t$ we have $\mathbb{E}^{\mathbb{Q}}[Y_tX_t]=\mathbb{E}^{\mathbb{P}}[Z_tY_tX_t]$

so we have:


and again by definition of conditional expectation, we have:


we can now conclude using the definition of $Y_t$ and $Z_t$.

$$\mathbb{E}^{\mathbb{Q}}[Y_T|\mathcal{F}_t] = \frac{\mathbb{E}^{\mathbb{P}}[Z_TY_T|\mathcal{F}_t]}{\mathbb{E}^{\mathbb{P}}[Z_T|\mathcal{F}_t]}$$

  • $\begingroup$ Very good, indeed. $\endgroup$
    – Gordon
    Oct 26 '16 at 11:23
  • $\begingroup$ I wish this was clear for me haha $\endgroup$
    – Wolfy
    Oct 28 '16 at 20:15

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