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Assume that I have a binary distribution with probability $p$ determined by the previous outcome. For instance, for random variable $X$, define it's initial distribution to be $X_0 \sim Bernoulli(p_0) $ and subsequent values of $p$ to be $ (\forall\ n \in \Bbb N)$:$$ p_{n+1} = \left\{ \begin{array}{c} f(p_n)\qquad (X_n=0) \\ g(p_n)\qquad (X_n=1) \\ \end{array} \right. $$ and $X_n \sim Bernoulli(p_n) $.

Is there a way to determine if the distribution is stable based on $f$ and $g$? If it's stable, is it possible to describe the stable distribution of $X$ (or its characteristics) via $f$ and $g$, possibly using a Markov Chain?

In a broader sense, are there methods of describing this type of distribution where the next sampled value is generated from a varying probability distribution, which depends on the current sampled value?

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  • $\begingroup$ It looks like a Markov process. $\endgroup$ – herb steinberg Dec 24 '19 at 0:52
  • $\begingroup$ I imagine for general $f,g$ it is pretty hopeless. Yes it is Markov, but if $f,g$ are completely arbitrary one can imagine e.g. $f(f(g(g(p_0)))) \neq f(g(f(g(p_0))))$ and so on...i.e. the $16$ possible ways to make a chain of $4$ functions would result in $16$ different $p_4$ values. For such "bad-behaving" $f,g$ after $n$ steps you would need $2^n$ states just to account for the $2^n$ possible different values of $p_n$. $\endgroup$ – antkam Dec 24 '19 at 5:08
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I'm not sure I understand what "stable distribution" means. But I'll attempt to share some of my understanding on this problem.

Through your equations, you seem to have described a sequence $p_0, p_1, p_2... $ of real numbers. For any given $f, g$, one might ask whether this sequence converges to a limit. Of course, this depends on what $f$ and $g$ are. For instance, we can define $f(p) = g(p) = 1 - p$. If we start off with $p_0 \neq \frac{1}{2}$, then this sequence alternates between $p_0$ and $1-p_0$ and it does not converge. The convergence is trivial if $p_0=\frac{1}{2}$.

Even if the sequence does not converge, it is possible that the values it takes are from a particular set of values (state space). This is because the process of generating $p_i$'s is essentially a Markov Chain. The state space, however, depends on the range of $f$ and $g$. In the example above, the state space of the Markov Chain is $\{p_0, 1-p_0 \}$. This state space might well be finite, countably infinite or even uncountably infinite! Of course, $p_n$ will be able to take on only one of a countable set of values (see comment by @antkam), but this is conditional on the fact that $p_0$ is given. If $p_0$ is unknown, then in the most general case, the state space is $[0,1]$ which is uncountable.

Let's say that the range is countable and finite and you index this using the function $I$ (i.e., $I(0), I(1), ...$ are the possible states). Then, you can define the state transition matrix $M_i$ for each stage $i$ such that $M_i (x, y)$ is the probability that $p_{i}=I(x)$ and $p_{i+1} = I(y)$, for all $x,y \in domain(I)$. Then you'll need to find a way of checking whether the sequence of vectors $Q_i = (\prod_{k=1}^{i}M_k)v$ converges, where $v$ is the vector satisfying $v[k] = Pr(I(k) = p_0)$. In Markov Chain terminology, we are trying to decide if the distribution is $stationary$.

As an exercise, you can try doing it for the example we had above. It should be very easy.

I hope this helps. Feel free to ask any questions about this. :)

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  • $\begingroup$ +1 for nice explanations. A minor technical detail: Suppose $p_0$ is given. Then for any specific $n$, $p_n$ can have at most $2^n$ possible values. This set is not only countable, but finite. However, if you take all $n$, i.e. consider the set $\{p_n: n \in \mathbb{N}\}$, then I think (not 100% sure) this set can be uncountably infinite. Anyway, this is a minor technical detail. $\endgroup$ – antkam Dec 25 '19 at 18:45

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