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
    Commented Oct 26, 2016 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
    Commented Oct 27, 2016 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
    Commented Oct 27, 2016 at 1:02
  • $\begingroup$ See also here. $\endgroup$
    – Gordon
    Commented Oct 27, 2016 at 13:49
  • $\begingroup$ @Gordon posted the solution but the latex generator here is messed up $\endgroup$
    – Wolfy
    Commented Nov 2, 2016 at 15:57

1 Answer 1


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
    Commented Oct 26, 2016 at 11:23
  • $\begingroup$ I wish this was clear for me haha $\endgroup$
    – Wolfy
    Commented Oct 28, 2016 at 20:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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