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I am struggling with a process like this: $$X_t=\begin{cases} \frac{\alpha\omega_t}{\alpha\omega_t+\beta(1-\omega_t)} & \text{with prob } p\\ \frac{(1-\alpha)\omega_t}{(1-\alpha)\omega_t+(1-\beta)(1-\omega_t)} & \text{with prob } 1-p \end{cases}$$ where $\alpha\in(0,0.5)$ and $\beta\in(0.5,1)$. Now introduce the process (which is a Bayesian update) $$ \omega_{t+1}=k+hX_t $$ where $h,k\in[0,1]$ Thus, $X_t$ is a simple Bernoulli trial, but it assumes different values at any time step $t$, since it depends on $\omega_t$. My question is then: Is it possible to compute the distribution of $\omega$? I made some numerical simulations: distribution is not always the same and it changes depending on parameters $\alpha, \beta, h, k$. Following I attach two simulations-example of the density of $\omega$ with different combinations of parameters (sorry if I am rude and i didn't rename axis label and title :D).

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    $\begingroup$ This question might better fit in the stats forum stats.stackexchange.com. $\endgroup$
    – Thomas Kojar
    Commented May 17, 2023 at 22:08
  • $\begingroup$ A close relative to the density of $\omega_t$ is finding the limiting/stationary distribution as $t$ grows. If the coin flips are iid, the process $\omega_t$ is Markov since it only depends on the previous step. So there is some hope for finding transition density Matrix (if we restrict to discrete times) and then stationary distribution. $\endgroup$
    – Thomas Kojar
    Commented May 17, 2023 at 22:21
  • $\begingroup$ this is quite a complicated definition also because if the jump step k is too large, then $\omega_t$ quickly heads to $\frac{\beta}{\beta-\alpha}$ and thus the Xt diverges. This boundary effect makes it a bit hard to study. $\endgroup$
    – Thomas Kojar
    Commented May 18, 2023 at 0:46
  • $\begingroup$ Sorry I forgot to mention that $\omega$ lies between 0 and 1, so X cannot diverge. $\endgroup$
    – DreDev
    Commented May 19, 2023 at 20:37

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