# Invariant distributions for iterated random variables (stochastic dynamical systems)

This is related to discrete dynamical systems, with the initial condition $$X_1$$ being a random variable with a non singular distribution. The system is driven by the iteration $$X_{n+1} = g(X_n)$$ for some rather smooth mapping $$g$$. The purpose here is to find a mapping $$g$$ so that the invariant distribution (also called attractor or invariant measure or fixed point distribution solution of an integral equation) is pre-specified. This is an inverse problem in the sense that typically $$g$$ is known and we are looking for the invariant distribution. Here the opposite is true: we assume the invariant distribution is known, and we want to find $$g$$ that will guarantee that no matter what the non-singular distribution of $$X_1$$ is, we end up with the the target invariant distribution. There are typically many possible solutions for $$g$$. The reason I am asking this question is because I believe I got the wrong solution in a simple case, and would like to know how to get it right.

The most basic example is when the invariant distribution is uniform on $$[0, 1]$$, that is, $$F_X(x) = x$$. In that case, $$g(x) = \{ bx \}$$ (the fractional part of $$bx$$), regardless of $$b$$ ( an integer $$>1$$) will work. Here I am trying to solve the case where the invariant distribution (the CDF) is $$F_X(x) = x^2$$, also with the same support domain being $$[0, 1]$$. Of course, this is a fist step towards dealing with much more complicated $$F_X(x)$$.

So, here is my investigation about $$F_X(x)=x^2$$, and my question is about what is wrong with my analysis. I know something is wrong. First, $$g$$ must be a many-to-one mapping. The easiest case is when $$g$$ is a two-to-one mapping, as in the logistic map where $$g(x)=4x(1-x)$$, with $$x\in [0, 1]$$. Besides that, if the density $$f_X$$ is symmetric around $$x_0$$, then $$g(\cdot)$$ can not be symmetric around $$x_0$$. The reason I say this is because I am focusing on mappings $$g$$ of $$[0, 1]$$ that are symmetric around some point $$x_0$$, and in this case, $$x_0=\frac{1}{2}$$. That is, $$g(x)=g(1-x)$$.

My wrong solution that needs a fix

Let $$g$$ be a two-to-one mapping, with $$g(x)=g(1-x)$$. If the (known) invariant distribution is $$F_X(x)$$, then $$g$$ must satisfy the functional equation $$f_X(x)= f_X(h_1(x))-f_X(h_2(x))$$ with $$h_1(x)=g^{-1}(x)$$ being one of the two inverses of $$g(x)$$ when $$x<1/2$$, and $$h_2(x)=1-h_1(x)$$ being the other inverse. Based on this, we must have, using the notation $$y=g^{-1}(x)$$,

$$x^2 = y^2 - (1-y^2)$$

which results in $$y = (x^2+1)/2$$. Inverting this again, we get $$x=\sqrt{2y-1}$$, valid if $$y\geq \frac{1}{2}$$. It follows that $$g(x)=\sqrt{2x-1}$$ if $$\frac{1}{2}\leq x \leq 1$$, and $$g(x) = \sqrt{1-2x}$$ if $$0\leq x \leq \frac{1}{2}$$. Yet the true mean of the invariant distribution ($$F_X(x) = x^2$$ on $$[0, 1]$$) does not seem to match that resulting from $$g(\cdot)$$ at equilibrium, when $$n=\infty$$. The system in question is ergodic.

• Shouldn't it be $x^2 = y^2 - (1-y)^2$ in your centered formula ? Aug 23 at 8:19
• I think the correct formula should be $f_X(x)= f_X(h_1(x))h_1 '(x)-f_X(h_2(x)h_2 '(x))$ Aug 23 at 10:13
• If you add some noise to $g$ it can be done in some cases. For example, it is known that any Markov chain can be described as iterations of a noisy map. You can design Markov chains with prescribed stationary measure using the Metropolis method Aug 23 at 10:21

## 1 Answer

To have here the invariant distribution with cdf $$F$$ given by $$F(x)=x^2$$ for $$x\in[0,1]$$, all that is needed is a change of variables.

More generally, let $$F$$ be the cdf of any non-atomic distribution supported on an interval $$I$$ in $$\mathbb R$$. Let $$F^{-1}$$ denote the inverse of the restriction of $$F$$ to $$I$$. Let $$U$$ be a random variable uniformly distributed on $$[0,1]$$. Then the cdf of the random variable $$X:=F^{-1}(U)$$ is $$F$$, and $$U=F(X).$$

Suppose that a function $$g_*$$ is such that the uniform distribution on $$[0,1]$$ is invariant for the dynamical system $$U_{n+1}=g_*(U_n)$$ -- that is, $$U\overset D=g_*(U)$$, where $$\overset D=$$ means the equality in distribution. Then $$F(X)\overset D=g_*(F(X))$$, that is, $$X\overset D=g(X),$$ where $$g:=F^{-1}\circ g_*\circ F.$$ So, the distribution with the prescribed cdf $$F$$ is invariant for the dynamical system $$X_{n+1}=g(X_n)$$, as desired.

In particular, if $$F(x)=x^2$$ for $$x\in[0,1]$$ and $$g_*(u)=\{bu\}$$ for $$u\in[0,1]$$, then $$g(x)=\sqrt{\{bx^2\}}$$ for $$x\in[0,1]$$.

• Thank you! How did you pick up $g_*(u)=\lfloor bu\rfloor$? I assume $b$ is an integer $> 1$. Aug 23 at 17:02
• I think $g_*(u) = \{bu\}$. I got it wrong (and you probably used my wrong solution) but I fixed it. So $g(x)=\sqrt{\{bx^2\}}$. Aug 23 at 17:28
• @VincentGranville : Oops! Indeed, I had just thoughtlessly copied the incorrect expression for $g_*$. This is now fixed. Aug 23 at 17:34