Let P and Q be two probability measures on the same space $(\Omega,\mathcal{F},\mathcal{P})$ and let $\mathcal{F_n}$ be filtration. Assume that $Q \ll P$. Let $X_n$ denote the Radon-Nikodym derivative of Q with respect to P on $\mathcal{F_n}$ i.e $X_n$ is $\mathcal{F_n}$-measurable and for any $A \in \mathcal{F_n}$, $Q(A) = \int_AX_ndP$. Show that $(X_n,\mathcal{F_n})$ is a martingale with respect to P, that is $E_P(X_n|\mathcal{F_{n-1}}) = X_{n-1},$ where for any random variable X, $E_P(X) = \int XdP$.

To Prove $(X_n,\mathcal{F_n})$ martingale

i) $X_{n}$ is $\mathcal{F_{n}}$ - measurable.

ii) $E[|X_n|] <\infty$

ii) $E[X_{n+1}|\mathcal{F_{n}}] = X_n$

what I don't understand what is the mean of $(X_n,\mathcal{F_n})$ is a martingale with respect to P, that is $E_P(X_n|\mathcal{F_{n-1}}) = X_{n-1},$

Can you help me please?

  • 1
    $\begingroup$ Both properties ii) and iii) depend on a probability measure, so when one is saying "a martingale with respect to P" we mean that all expectations and conditional expectations are with respect to the probability measure P. $\endgroup$ – Stefan Hansen Apr 6 '14 at 12:37
  • $\begingroup$ Thanks for the clarification can you help me start the prove please as I am not sure why we need Q and P here $\endgroup$ – Falcon-StatGuy Apr 6 '14 at 13:53

Measurability and integrability is a consequence of the Radon-Nikodym theorem. Since $\mathcal{F}_{n-1}\subseteq\mathcal{F}_n$ we have that $$ \int_AX_{n-1}\,\mathrm dP=Q(A)=\int X_n\,\mathrm dP $$ for all $A\in\mathcal{F}_{n-1}$ showing that ${\rm E}_P[X_n\mid\mathcal{F}_{n-1}]=X_{n-1}$.

  • $\begingroup$ Thank for the help is this all the answer or I still have to prove something $\endgroup$ – Falcon-StatGuy Apr 8 '14 at 2:58
  • $\begingroup$ What do you think? Have you shown that properties i), ii) and iii) are satisfied? $\endgroup$ – Stefan Hansen Apr 8 '14 at 14:35
  • $\begingroup$ No I don't think this enough. Can you please help? As I am new to martingale and measure $\endgroup$ – Falcon-StatGuy Apr 8 '14 at 18:50
  • $\begingroup$ What do you think is missing? Which of the properties i)-iii) is not clear to you? $\endgroup$ – Stefan Hansen Apr 9 '14 at 7:03
  • $\begingroup$ Can some one Explain this question please $\endgroup$ – Falcon-StatGuy Apr 11 '14 at 21:47

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