# martingale question 3

Let $\{U(t)\}$ be a positive martingale, such that $\mathbb{E}(U(T))=1$. Then we define a new probability measure $Q$ such that $dQ=U(T)dP$. Prove that if $X\in\mathcal{F}_t$, $E_{Q}(X|\mathcal{F}_s)=E_{P}(\frac{U(T)}{U(S)}X|\mathcal{F}_t)$, for $s\leq t$.

I think we should use the tower property on $E_{Q}(X|\mathcal{F}_s)=E_{P}(\frac{U(T)}{U(S)}X|\mathcal{F}_t)$ and then take advantage of martingale properties. Can you please help me out?

• what is $T$ in your defintion? – Canardini Dec 19 '16 at 21:56
• @Canardini that s a specific time T – Susan_Math123 Dec 19 '16 at 22:16
• @Ita any idea how to solve this? – Susan_Math123 Dec 19 '16 at 22:17
• Yeah , I have but I do not thing your wording is correct . You must define $T$ and $S$ – Canardini Dec 20 '16 at 14:34

Let $(\Omega,\mathcal{F},\mathcal{F}_T,\mathbb{P})$ be a filtered Probability space and $(U_t)_{t\in[0,T]}$ a positive martingale w.r.t. $\mathbb{P}$, such that $\mathbb{E}_{\mathbb{P}}(U_T)=1$. Define a measure $\mathbb{Q}$ via the Radon Nikodym density $$\frac{d\mathbb{Q}}{d\mathbb{P}}=U_T$$
We start from an equivalent definition of conditional expectation, i.e. for all $A\in\mathcal{F}_s$ $$\mathbb{E}_{\mathbb{Q}}[X \mathbf{1}_A]=\mathbb{E}_{\mathbb{Q}}[\mathbb{E}_{\mathbb{Q}}[X\vert\mathcal{F}_s]\mathbf{1}_A]$$
So let $A\in\mathcal{F_s}$, then by definition of $\mathbb{Q}$ and measurability of $X$ and the indicator function of $A$ w.r.t. $\mathcal{F}_t$ we get \begin{eqnarray} \mathbb{E}_{\mathbb{Q}}[X \mathbf{1}_A]&=&\mathbb{E}_{\mathbb{P}}[U_T X \mathbf{1}_A] \\ &=&\mathbb{E}_{\mathbb{P}}[\mathbb{E}_{\mathbb{P}}[U_T X \mathbf{1}_A\vert\mathcal{F}_t]] \\ &=&\mathbb{E}_{\mathbb{P}}[\mathbb{E}_{\mathbb{P}}[U_T\vert\mathcal{F}_t]X \mathbf{1}_A] \end{eqnarray} by the martingale property of $U$ and the towerlaw of conditional expectation this equals $$=\mathbb{E}_{\mathbb{P}}[U_t X\mathbf{1}_A]=\mathbb{E}_{\mathbb{P}}[\mathbb{E}_{\mathbb{P}}[U_t X\vert\mathcal{F}_s]\mathbf{1}_A]$$ again by the definition of the measure $\mathbb{Q}$ we get \begin{eqnarray} &=&\mathbb{E}_{\mathbb{Q}}[(U_T)^{-1}\mathbb{E}_{\mathbb{P}}[U_t X\vert\mathcal{F}_s]\mathbf{1}_A]\\ &\overset{towerlaw}{=}&\mathbb{E}_{\mathbb{Q}}[\mathbb{E}_{\mathbb{Q}}[(U_T)^{-1}\mathbb{E}_{\mathbb{P}}[U_t X\vert\mathcal{F}_s]\mathbf{1}_A\vert\mathcal{F}_s]]\\ &\overset{measurability}{=}&\mathbb{E}_{\mathbb{Q}}[\mathbb{E}_{\mathbb{P}}[\mathbb{E}_{\mathbb{Q}}[(U_T)^{-1}\vert\mathcal{F}_s]U_t X\vert\mathcal{F}_s]\mathbf{1}_A]\\ \end{eqnarray} Note that the inverse of $U$ is a martingale w.r.t. $\mathbb{Q}$. Hence we have shown $$\mathbb{E}_{\mathbb{Q}}[X \mathbf{1}_A]=\mathbb{E}_{\mathbb{Q}}[\mathbb{E}_{\mathbb{P}}[(U_s)^{-1}U_t X\vert\mathcal{F}_s]\mathbf{1}_A]$$ which by definition of the conditional expectation (above) implies $$\mathbb{E}_{\mathbb{Q}}[X \vert \mathcal {F}_s]=[\mathbb{E}_{\mathbb{P}}[(U_s)^{-1}U_t X\vert\mathcal{F}_s].$$