Radon–Nikodym derivative with another variable is a martingale

Let $N > 0$ be an integer. Suppose $Y$ is a random variable on $\Omega$ and define, for $n = 0, 1, \ldots, N$, $$Y_n = \tilde{E}_n(Y)$$ (i.e. the conditional expectation with respect to the risk-neutral probability measure $\tilde{\mathbb{P}}$). Further, let $Z_0, Z_1, \ldots, Z_n$ be the Radon–Nikodym derivative process of $\tilde{\mathbb{P}}$ with respect to $\mathbb{P}$, so $Z_n = E_n(Z)$, with $Z$ the Radon–Nikodym derivative of $\tilde{\mathbb{P}}$ with respect to $\mathbb{P}$. Show that the process $$Z_0 Y_0, Z_1 Y_1, \ldots, Z_N Y_N$$ is a martingale with respect to $\mathbb{P}.$

Since I am stuck on this for hours, any help would be greatly appreciated. I have no idea how to deal with the conditional expectation in the expression we have to show, which is $E_n(Y_{n+1} Z_{n+1}) = Y_n Z_n.$

• It would help to define the "risk-neutral probability measure" since that is not apparent from context. – Math1000 May 25 '18 at 12:04

1 Answer

You want to show that $$E[1_AY_{n+1}Z_{n+1}] = E[1_AY_{n}Z_{n}]$$ for every $A \in \mathcal{F}_n$.

\begin{align} E[1_AY_{n+1}Z_{n+1}] &= E[1_AY_{n+1}E[Z\mid\mathcal{F}_{n+1}]]\\ &= E[E[1_AY_{n+1}Z\mid\mathcal{F}_{n+1}]]\\ &= E[1_AY_{n+1}Z]\\ &= \tilde{E}[1_AY_{n+1}]\\ &= \tilde{E}[1_AY_{n}]\\ &= E[1_AY_{n}Z]\\ \end{align}

On the other hand,

\begin{align} E[1_AY_{n}Z_{n}] &= E[1_AY_{n}E[Z\mid\mathcal{F}_{n}]]\\ &= E[E[1_AY_{n}Z\mid\mathcal{F}_{n}]]\\ &= E[1_AY_{n}Z]\\ \end{align}

Let me know if you don't see how I went from step to the other.