Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It's well known that the mgf (moment generating function) characterizes the distribution of a r.v., i.e. if the two mgf of two r.v. coincide then they have the same distribution.

Is the same true for the conditional distribution? This means, if, $E[\exp{(tX)}|\mathcal{F}_t] = E[\exp{(tX)}|\mathcal{F}_s]$ then we know $P[X\in A|\mathcal{F}_t] = P[X\in A|\mathcal{F}_s]$?

hulik

share|improve this question

1 Answer 1

up vote 1 down vote accepted

One should rather assume that $\mathrm E(\mathrm e^{\mathrm ixX}\mid\mathcal F_t)=\mathrm E(\mathrm e^{\mathrm ixX}\mid\mathcal F_s)$ for every $x$ in $\mathbb R$. Then the usual proof works, namely, one considers the class $\mathcal C$ of measurable (bounded) functions $u$ such that $\mathrm E(u(X)\mid\mathcal F_t)=\mathrm E(u(X)\mid\mathcal F_s)$ and one shows that $\mathcal C$ is in fact the whole class of measurable (bounded) functions.

A problem to get rid of is that one usually only assumes that $\mathrm E(\mathrm e^{\mathrm ixX}\mid\mathcal F_t)=\mathrm E(\mathrm e^{\mathrm ixX}\mid\mathcal F_s)$ almost surely, for every $x$ in $\mathbb R$, hence the underlying uncountable collection of negligible sets might become a nuisance. But, considering $L^1$--continuous versions of the mappings $x\mapsto\mathrm E(\mathrm e^{\mathrm ixX}\mid\mathcal F_t)$ and $x\mapsto\mathrm E(\mathrm e^{\mathrm ixX}\mid\mathcal F_s)$, one can show this does not happen.

In the end, for every measurable $B$, one has $\mathrm P(X\in B\mid\mathcal F_t)=\mathrm P(X\in B\mid\mathcal F_s)$ almost surely.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.