Generalized univariate normal distribution with $k+1$ parameters

Final update on 11/28/2019: I have worked on this a bit more, and wrote an article summarizing all the main findings. You can read it here.

The goal here is to obtain a highly generic family of stable distributions (governed by $$k+1$$ parameters), that is, if $$X, Y$$ have the distribution in question, and are independent, then $$aX+bY$$ has a distribution belonging to the same family. A special case is the normal distribution.

Perhaps the easiest way would be a distribution with the following characteristic function:

$$\psi_X(t) = E[\exp(itX)] = \exp P_k(t)$$

where $$P_k$$ is a polynomial of degree $$k$$ with $$P(0)=0$$, with complex coefficients chosen so that $$\psi_X$$ is actually the characteristic function of a real distribution. I am particularly interested in $$P_4(t)=-(\alpha^2 t^2 + \beta^2 t^4)$$ where are $$\alpha, \beta$$ are two real numbers (the parameters). If this corresponds to an actual distribution, its mean would be zero (it's a symmetrical distribution) and its variance equal to $$4\alpha^2$$. The case $$\beta = 0$$ corresponds to a normal distribution. Even if $$\beta \neq 0$$, the distribution will look pretty much like normal (a good approximation.)

Question

Does this $$P_4$$ corresponds to an actual distribution? More generally, what are the constraints on the coefficients of the polynomial so that it yields an actual distribution?

Context

I am interested to find a distribution so that if $$X_1, X_2, X_3$$ and so one are i.i.d. and have that distrubution, then the successive differences (defined below) also have that same distribution:

• $$Y_1 = X_1 - X_2$$
• $$Y_2 = X_1 - 2X_2 + X_3$$
• $$Y_3 = X_1 - 3X_2 + 3 X_3 - X_4$$
• $$Y_4 = X_1 - 4X_2 + 6 X_3 - 4X_4 + X_5$$

and so on.

We have $$\mbox{Var}(Y_k) = \mbox{Var}(X_1) \cdot (2k)!/(k!)^2$$. Let's define $$Z = \lim_{k\rightarrow \infty} \frac{Y_k}{\sqrt{\mbox{Var}(Y_k)}}.$$ The convergence is in distribution, and all the $$Y_i$$'s and $$Z$$ have zero mean. Their belong to the same family distribution as $$X_1, X_2$$ and so on. For more details, see here. The context is to develop statistical tests for error detection.

Update

It seems that the only case where $$P_4(t)$$ yields a valid characteristic function is if $$\beta =0$$, which corresponds to a normal distribution. Thus, we must consider alternatives. One that works for sure: if $$Z_1$$ is Cauchy centered at 0, $$Z_2$$ is Normal centered at 0, and $$Z_1, Z_2$$ are independent, then $$X = Z_1+ Z_2$$ has a Normal-Cauchy distribution (I made up the word) but it has no expectation. Yet this constitutes a larger class than the Gaussian distribution, and this family is stable under addition.

Another potential candidate is $$X$$ with the following characteristic function:

$$\psi_X(t) = \exp(-(a^2 |t|^\alpha + b^2 |t|^\beta).$$

Here $$1< \alpha, \beta \leq 2$$. That distribution has a zero mean, but no variance unless $$\alpha=0, \beta = 2$$ (the Gaussian case.) The two parameters in this family are $$a$$ and $$b$$, while $$\alpha, \beta$$ are fixed. Note that the Normal-Cauchy distribution corresponds to $$\alpha=1, \beta = 2$$. And sampling from this distribution might not be easy (see here for a way to do it.)

For a stable one-parameter, non-Gaussian family of distributions with finite variance, I would investigate the one defined by the following characteristic function:

$$\psi_X(t) = \exp\Big(-a^2 (4-\sin^2 t) t^2\Big).$$

I am not sure yet if this is a valid characteristic function, but it looks like it might be, yielding a density that seems to be positive everywhere at first glance if $$a\geq 1$$, after a very quick check using WolframAlpha:

$$f_X(x) = \frac{1}{2\pi}\int_{-\infty}^\infty \cos (xt)\cdot \exp\Big(-a^2 (4-\sin^2 t) t^2\Big) dt.$$

If $$Z_1, Z_2$$ are i.i.d from that family, respectively with parameter $$a_1$$ and $$a_2$$, then $$Z_1 + Z_2$$ belongs to the same family, and has parameter $$\sqrt{a_1^2 + b_1^2}$$.

Yet, as far as I know, the only stable distribution with a finite variance is the Gaussian one. So either there is something wrong in my reasoning, or I discovered something new. The issue here is proving that $$f_X(x) \geq 0$$. One way to do it is by vanishing the derivative of $$f_X$$ to find its minimum value. In other words, find $$x_0$$ such that

$$\int_{-\infty}^\infty t \cdot \sin (x_0 t)\cdot \exp\Big(-a^2 (4-\sin^2 t) t^2\Big) dt =0.$$

If $$f(x_0)=0$$ then we are dealing with a real density. If $$f(x_0)< 0$$ then this is not a density. The case $$f(x_0) > 0$$ is not possible. Another test is to check if all the even moments $$E(X^{2k})$$ are positive. These moments can be easily derived from the characteristic function:

$$E(X^{2k}) = (-1)^k\cdot\frac{d^{2k}\psi_X}{dt^{2k}}(0).$$

Assuming we are dealing with an actual distribution, its mean would be zero, and its variance would be $$8 a^2$$. Its odd moments are all equal to 0. It can be generalized as follows (with two parameters $$a, b$$ rather than just $$a$$) to include the Gaussian distribution (corresponding to $$b=0$$):

$$\psi_X(t) = \exp\Big(-a^2 (4-b\sin^2 t) t^2\Big).$$

In this case, if $$Z_1, Z_2$$ are i.i.d. with that distribution, respectively with parameter $$(a_1,b_1)$$ and $$(a_2, b_2)$$, then $$Z_1 + Z_2$$ belongs to the same family, with parameter $$(\sqrt{a_1^2+b_1^2}, a_1^2 b_1 + a_2^2 b_2)$$.