I would like to know if there exists a measure $\rho$ on the positive real line such that its moments $\int_0^{\infty} x^j d\rho(x)$ are equal to a constant (for example equal to one) for all $j=0,\dots,n,\dots$ (or for $j\leq n$ for any $n$). In other words, if there exists a density function (equal to zero on the negative real part) such that its characteristic function is $e^{it}$.
Continuous Random Variable with constant moments?
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probability
measure-theory
fourier-analysis
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1$e^{it}$ is a continuous map, and positive define, which is $1$ at $0$, so it's a characteristic function. Now, we have to see whether the law whose characteristic function is this map is concentrated on the positive real line. Well, $\rho=\delta_1$ works, but it seems too trivial. – 2012-07-19
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0and how to see that? Or any reference for that? – 2012-07-19
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0The first assertion comes from a theorem due to Bochner. But it's not necessary here. – 2012-07-19
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2Notice that constant moments imply zero variance. What does it tell you about the purported density function? – 2012-07-19
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0Got something from an answer below? – 2012-07-25
2 Answers
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(With no characteristic function.)
Note that $$\int(x-1)^2\,\mathrm d\rho(x)=\int x^2\,\mathrm d\rho(x)-2\int x\,\mathrm d\rho(x)+\int 1\,\mathrm d\rho(x). $$ Hence, if $$ \int x^2\,\mathrm d\rho(x)=\int x\,\mathrm d\rho(x)=\int 1\,\mathrm d\rho(x), $$ then the set $\mathbb R\setminus\{1\}$ has measure zero with respect to $\rho$. (No hypothesis on some other moments is necessary.)
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The (unique) random variable whose characteristic function is $t\mapsto e^{it}$ is associated to the measure $\delta_1$, that is $$\delta_1(A)=\begin{cases} 1&\mbox{ if }1\in A,\\ 0&\mbox{ otherwise}. \end{cases}$$ This is not a continuous random variable.