Looking for the logic of a sequence from convolution of probability distributions

Question

I am trying to detect a pattern in the followin sequence from convolution of a probability distribution (removing the scaling constant $\frac{6 \sqrt{3}}{\pi }$: $$\left(\frac{1}{\left(x^2+3\right)^2},\frac{2 \left(x^2+60\right)}{\left(x^2+12\right)^3},\frac{3 \left(x^4+162 x^2+9477\right)}{\left(x^2+27\right)^4},\frac{4 \left(x^6+324 x^4+44928 x^2+2847744\right)}{\left(x^2+48\right)^5},\frac{5 \left(x^8+564 x^6+141750 x^4+19912500 x^2+1388390625\right)}{\left(x^2+75\right)^6},\frac{6 \left(x^{10}+900 x^8+366120 x^6+87829920 x^4+13038019200 x^2+998326798848\right)}{\left(x^2+108\right)^7}, \dots \right)$$

Answer 1

Well, at least the denominator seems to be $(x^2 + 3n^2)^{n+1}$...

Answer 2

Are you sure about the $564x^6$ term? All polynomial coeffs are divisible by $k$ with the exception of that one.

Answer 3

The denominators are fairly straightforward. I assume the next two are: (x^2 + 192)^9 and (x^2 + 243)^10? The recursion is: Given denominator (x^2 + k)^n, then the next denominator is (x^2 + k*(n/(n-1))^2)^(n+1). (Yes, this is same as the previous post, but I like to think in terms of recursions because it sometimes helps in decoding things.) As for the numerators I think have something based on the a common special function that's promising but it would help to know more specifics about the convolution. (@financequant)

Answer 4

Given that $\frac{6\sqrt{3}}{\pi}\cdot\frac{1}{(x^2+3)^2}$ is the pdf of a random variable $X$, we have: $$\varphi_X(t)=\mathbb{E}[e^{itX}]=(1+\sqrt{3}\,|t|)\,e^{-\sqrt{3}\,|t|}$$ hence the characteristic function of $Y=X_1+\ldots+X_n$, where $X_1,\ldots,X_n$ are independent random variables with the same distribution of $X$, is: $$\varphi_Y(t) = (1+\sqrt{3}|t|)^n\,e^{-n\sqrt{3}\,|t|}$$ and the pdf of $Y$ is given by: $$f_Y(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}(1+\sqrt{3}\,|t|)^n\,e^{-n\sqrt{3}\,|t|}e^{-itx}\,dt=\frac{1}{\pi}\int_{0}^{+\infty}(1+\sqrt{3}\,t)^n\,e^{-n\sqrt{3}\,t}\cos(tx)\,dt$$ that can be computed through the residue theorem or by integration by parts.

Answer 5

I played a little bit with Jack's answer. One can expand with the Newton formula $$ (1 + \sqrt{3} t)^n = \sum_{k=0}^n \binom{n}{k} (\sqrt{3}t)^k $$ then the integrals can be performed using $$ \int_0^\infty dt\, t^k e^{- t} \cos(s t) = \frac{T_{1+k}\bigl(1/\sqrt{1 + s^2}\bigr) k!}{(1+s^2)^{(k+1)/2}} $$ where $T_a(x)$ is the T-Chebyshev polynomial. Then, the given sequence can be rewritten setting $s = x/\sqrt{3}$ as $$ S_n = \frac{1}{18(s^2 + n^2)^{n+1}} \sum_{k=0}^n \frac{n!}{(n-k)!} (s^2 + n^2)^{n -\frac{k - 1}{2}} T_{1+k}\bigl(1/\sqrt{1 + s^2/n^2}\bigr) $$

Answer 6

Second non-zero coeff of long poly seems to be $3(n-1)(n^2+2n-12)$.

Could try subtracting $(x^2+3(n^2+2n-12))^{n-1}$ from each long poly to eliminate first two terms and see if the rest looks manageable.