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.)


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?


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.


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 $(\sqrt{a_1^2+b_1^2}, a_1^2 b_1 + a_2^2 b_2)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.