0
$\begingroup$

I have posted this in the math and stats sites, but I am not sure where the proper forum for this question is. If it is not here, please go on and delete it.

Suppose we have a stochastic linear process:

$$x_{k+1} = Ax_{k} + Bw_{k} \qquad \text{with} \qquad x_{0} = a\in\mathbb{R}^{n}$$

Here, $A,B$ are known matrices and $a$ is a known vector. Moreover, the eigenvalues of $A\in\mathbb{R}^{n\times n}$ are in the open unit disk. Finally, the elements of the sequence $\{w_{k}\}_{k\in\mathbb{N}}$ are iid random variables uniformly distributed on $[-1,1]^{m}$

  1. What would be the distribution of $x_{k}$ as $k\to\infty$? Notice that the linear recursion above can indeed be arranged as:

$$ x_{N} = \sum^{N-1}_{k=0} A^{k}Bw_{N-k-1} \qquad \text{for each}\quad N\in\mathbb{N}.$$

Thus, for $x_{N}$ with $N\to\infty$ we are looking at some form of the central limit theorem, where the distribution of $x_{\infty}$ seems to be bell shaped but can not possibly be normally distributed. This, of course since the the distribution has a compact support---due to the fact that each $w_{k}$ is compactly supported and the eigenvalues of $A$ are in the open unit disk.

As a plus I would like to know if the family of distributions generated by the sum above, for each $N$, has some sort of name (the first is a multidimensional trapezoidal distribution, the second is piecewise quadratic, then piecewise cubic,...).

  1. As an example, lets take a scalar system with $A,B = 1/2$, and $x_0 = 0$. For $k=0$, $x_0$ has a dirac distribution. For $k=1$, we have a uniform distribution supported on $[-\frac{1}{2},\frac{1}{2}]$. For $k=2$ we would get some trapezoidal distribution supported on $[-\frac{3}{4},\frac{3}{4}]$. For each $k$, the distribution of $x_{k}$ is some piecewise polynomial function compactly supported which indeed looks increasingly bell-shaped. Even for this case, I can't really figure out what the limit distribution would be.

Thanks again in advance.

$\endgroup$
7
  • $\begingroup$ Could this question help? $\endgroup$
    – Dabed
    Commented Mar 18, 2021 at 16:10
  • $\begingroup$ @DanielD. Thanks for the link. I had a brief look but I fail to see how it helps. As far as I can see, the question linked asks about jointly normal random variables (and their moments). Here, the variables $w_{k}$ are independent and uniformly distributed. If I were to add normally distributed variables, I would get back a normally distributed variable. Could you perhaps expand? maybe I am not seeing something that you are. Thanks. $\endgroup$
    – NoobNoob
    Commented Mar 18, 2021 at 19:05
  • $\begingroup$ Sorry I just though maybe it could help to see a similar problem to work your way trough this one but I really didn't thought much about it so maybe there is nothing helpful there, I guess it crossed my mind it could be helpful to obtain the expectation as $E(x_t)=E(E(x_t|x_{t-1},..,x_0))=E(E(x_t|x_{t-1}))$ then do something similar to the variance and apply CLT but in the link is important that the errors being normal instead of uniform as the projection $\beta$ is obtained trough ols $\endgroup$
    – Dabed
    Commented Mar 18, 2021 at 20:12
  • $\begingroup$ @DanielD. No.worroes, comment is more than appreciated (only comment so far). Funny thing is that you can't apply the CLT as the limit distribution has compact support. As you mention, the key in the previous question is the unbounded support of the resulting PDF. I'll keep thinking and waiting! Thanks. $\endgroup$
    – NoobNoob
    Commented Mar 19, 2021 at 8:17
  • $\begingroup$ If you think of this as a discrete-time continuous-state Markov process. What you are looking for is its stationary distribution. It is not too difficult to write down the recursion the stationary distribution satisfies. In the case $A,B=1/2$ for example, this is $f(z)=\int_{-1}^{2z+1}f(x)dx$ ($0\leq z\leq0$) where $f(.)$ is the pdf of the stationary distribution. Or $F'(z)=1/2F(2z+1)$ in terms of the cdf. This is difficult to solve explicitly as it is a pantograph equation, but maybe this is a start. $\endgroup$ Commented Mar 22, 2021 at 9:48

0

You must log in to answer this question.