Probability distribution with shifted central binomial moments

Question

(Asked previously in MSE)

Suppose a probability distribution $p(x)$ has moments $m_n=\int p(x)x^ndx$ given by $m_0=1$, $m_1=1$, $m_2=2$ and, for $n>0$, $$m_{n+1}={2n \choose n}.$$

The moment generating function exists everywhere and is $$f(t)=1+\sum_{n=0}^\infty \frac{(2n)!}{n!^2(n+1)!}t^{n+1}=1+te^{2t}\left(I_0(2t)-I_1(2t)\right),$$ in terms of Bessel functions. An inverse Fourier transform $\int e^{-itx}f(it)dt$ then gives the function $$ \rho(x)=\frac{\delta(x)}{2\pi}+\frac{1}{\pi x^{3/2}\sqrt{4-x}}.$$

This function has all the correct moments, except for the norm, $m_0$, because it is in fact not normalizable.

Is this weird? If the moment generating function converges everywhere, a probability distribution is uniquely determined (right?), but this $\rho$ is not a probability distribution. Can I conclude that no random variable can possibly have those moments?

Answer 1

The function $\mathbb R\ni t\mapsto g(t):=f(it)$ is not the characteristic function of any probability distribution.

Indeed, if it were, then we would have $|g(t)|\le1$ for all real $t$. However, in fact, $|g(10)|=1.316\ldots>1$.

So, $f$ is not the moment generating function of any probability distribution, and hence there is no probability distribution with your "moments" $m_0,m_1,\dots$.

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You wrote

Can I conclude that no random variable can possibly have those moments?

It is now shown that, indeed, "no random variable can possibly have those moments".

You also wrote

If the moment generating function converges everywhere, a probability distribution is uniquely determined (right?), but this $\rho$ is not a probability distribution.

The uniqueness result is actually this: If two probability distributions (say on the real line) have the same moment generating function which is finite in a neighborhood of $0$, then the two distributions are the same.

However, in our case it has been shown that $f$ is not the moment generating function of any probability distribution. So, no contradiction whatsoever here.

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Finally, concerning this:

$$ \rho(x)=\frac{\delta(x)}{2\pi}+\frac{1}{\pi x^{3/2}\sqrt{4-x}}.$$

This function has all the correct moments, except for the norm, $m_0$, because it is in fact not normalizable.

Let $$p(x):=x\rho(x)=\frac{1}{\pi x^{1/2}\sqrt{4-x}}$$ for $x\in(0,4)$, with $p(x):=0$ for real $x\notin(0,4)$. Then $p$ is a true probability density, with moments $$\mu_n:=\int_{\mathbb R}x^n p(x)\,dx=m_{n+1}.$$ That is, your almost-moments $m_1,m_2,\dots$ are just shifted true moments $\mu_0,\mu_1,\dots$, respectively -- exactly with no place for your $m_0$. This completes the resolution of the puzzle.