(Asked previously in [MSE](https://math.stackexchange.com/questions/4625673/find-a-probability-density-from-the-moments))


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?



Probability distribution with shifted central binomial moments
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" $m\_0,m\_1,\dots$.




---


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.


