# Are these moments related to any usual distribution?

Knowing the moments of a random variable, we are wondering if we can express it in terms of some usual distribution (beta, uniform...), maybe as a product of distributions.

$$X$$ is a $$[0,1]$$-valued random variable with the following moments:

$$\mathbb{E}[X^k]=\frac{2}{3(k+2)}+\frac{8}{(k+1)(k+2)^2(k+3)}.$$

Is it correct to think of $$X$$ as a mixture of two distributions, one being a size biaised distribution? Is the second term related to some common distribution?

Note that for all real $$k\ge0$$ and $$a>0$$ we have $$\frac1{k+a}=\int_0^1 x^k x^{a-1}\,dx.$$ Differentiating in $$k$$, we further get $$\frac1{(k+a)^2}=-\int_0^1 x^k \ln x\, x^{a-1}\,dx.$$ So, doing the partial fraction decomposition, we have $$\frac2{3(k+2)}+\frac8{(k+1)(k+2)^2(k+3)} =\frac2{3(k+2)}-\frac4{k+3}+\frac4{k+1}-\frac8{(k+2)^2} =\int_0^1 x^k f(x)\,dx=EX^k$$ if the pdf of $$X$$ is given by $$f(x):=4+\frac{2x}3-4 x^2+8x\ln x \tag{1}$$ for $$x\in(0,1)$$.

Note also that any compactly supported distribution on $$\mathbb R$$ is uniquely determined by its moments, because then the characteristic function of the distribution is uniquely determined by the moments; here one can use the moment generating function instead of the characteristic one.

So, here the distribution of $$X$$ with its pdf $$f$$ given by (1) is uniquely determined by the moments.

Note further that the function $$f$$ defined by (1) is indeed a pdf. First here, $$\int_0^1 f(x)\,dx=1$$. Also, $$f''(x)=8/x-8>0$$ for $$x\in(0,1)$$ and $$f'(3/5)=-0.219\ldots<0$$. So, $$f$$ is convex on $$(0,1)$$ and hence $$f(x)\ge f(3/5)+f'(3/5)(x-3/5)\ge f(3/5)+f'(3/5)(1-3/5) =0.42006\ldots>0$$ for all $$x\in(0,1]$$. Here is the graph of $$f$$:

This method should work for any compactly supported distribution on $$\mathbb R$$ whose $$k$$th moments are given by a rational (in $$k$$) expression.

I just found what I was looking for: if I consider two independent random variables $$Y$$ and $$Z$$ such that $$Y$$ has law Beta(1,2) and $$Z$$ has law Beta(2,2), then their product $$W:=YZ$$ satifies

$$\mathbb{E}[W^k]=\frac{2}{(k+1)(k+2)}\frac{6}{(k+2)(k+3)}$$.

So the law of $$X$$ is a mixture of a size biaised distribution (which is also the Beta(2,1) distribution) and the product distribution of a Beta(1,2) and a Beta(2,2).

If you see anything that could complete this answer, please let me know.