Given a random variable $X=(X_1,...,X_p)$ with $P(X \in M) = 1$ for compact $M$, do the values of $E[X_1]$, $E[X_2], ..., E[X_1^2], E[X_1 X_2],..., E[X_1^3],...,E[X_1 X_2 X_3]... $ determine the distribution $F(x)=P(X_1 \leq x_1 \wedge \dots\wedge X_p \leq x_p)$ uniquely?
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asked Apr 17, 2011 at 10:42
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$\begingroup$ Did you think of the particular case p=1? $\endgroup$– leonbloyApr 17, 2011 at 14:11
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$\begingroup$ The $p = 1$ case follows easily from Carleman's condition. $\endgroup$– cardinalApr 17, 2011 at 14:36
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$\begingroup$ For a general setting, you may want to search for "Hamburger moment problem" for $M$ noncompact. Like en.wikipedia.org/wiki/Hamburger_moment_problem $\endgroup$– Henry.LSep 18, 2015 at 15:45
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1 Answer
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The answer is "yes".
Every probability distribution $\mu$ on $M$ is characterized by integration against continuous functions, that is, if $\int_M f(x)\, \mu(dx)= \int_M f(x)\, \nu(dx)$ for all continuous $f$, then $\mu=\nu$.
Now the Stone-Weierstrass theorem says that any continuous $f$ on $M$ can be approximated uniformly by polynomials. Therefore if $\int_M p(x)\, \mu(dx)= \int_M p(x)\, \nu(dx)$ for all polynomials $p$, then $\mu=\nu$.