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I'd like to find distribution of the almost sure limit $X_\infty$ of

$X_{t+1}= \begin{cases} 1-p+pX_t & \text{w/prob: } X_t \\ pX_t & \text{w/prob: } 1-X_t \end{cases}$

where $X_t$ is a random variable (and also a martingale), and $p\in [0,1]$.

Firstly, to show it converges a.s., i need to show $P(\lim \sup X_n = \lim \inf X_n)=1.$ (Which i guess makes sense intuitively, as sup $X_n$ is evaluated when $X_n=1$ and inf $X_n$ when $X_n=0$, right?)

Then i'm not sure how to take the limit for any $\omega \in \Omega$, i.e. $\underset{n \to \infty}{\lim} X_n(\omega)$.. i guess looking at probabilites since it is discrete and applying Borel Cantelli or something?

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  • $\begingroup$ I believe the result is $P(X_\infty=1)=X_0$ and $P(X_\infty=0)=1-X_0$, as this solves the first-step recurrence. Shall I post that as an answer, or are you mainly interested in a formal proof of this? $\endgroup$
    – joriki
    Apr 28 '20 at 8:28
  • $\begingroup$ @joriki Thanks - ideally looking how to prove that formally, any idea? Also, why do you think it involves $X_0$? $\endgroup$
    – CCZ23
    Apr 28 '20 at 8:44
  • $\begingroup$ Well, for $X_0=0$ the limit is $0$ and for $X_0=1$ the limit is $1$, so the limit must depend on $X_0$ independent of my specific result. You didn't specify an initial value, so I assumed that you wanted the result in dependence on an arbitrary initial value. $\endgroup$
    – joriki
    Apr 28 '20 at 8:47
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Hint:

You should Show $E(X_{\infty}(1-X_{\infty}))= 0$ and since $X_{\infty}(1-X_{\infty})\geq 0$ conclude almost sure $X_{\infty} \sim Bernoulli (x_0)$(Since $E(X_t)=x_0$).

1) $X_t$ is a non-negative martingale, so it converge almost surly.Corollary 2.3. so $ \lim \sup X_n = \lim \inf X_n=X_{\infty}$

2)show $E(X_{t+1}-X_t)^2 \longrightarrow 0$ when $t\longrightarrow \infty$.

3)Show $E(X_{t+1}-X_t)^2=(1-p)^2 E(X_t(1-X_t))$ and hence $E(X_t(1-X_t))\longrightarrow 0$ when $t\longrightarrow \infty$.

Now by $t\longrightarrow \infty$ so $ E(X_{\infty}(1-X_{\infty}))= 0$ almost surly. since $X_t(1-X_t)\geq 0$ so almost sure $X_{\infty}(1-X_{\infty})=0$. So $X_{\infty}=0$ or $X_{\infty}=1$. use the fact $E(X_t)=x_0$ so $X_{\infty} \sim Bernoulli (x_0)$

Proof $E(X_{t+1}-X_t)^2=(1-p)^2 E(X_t(1-X_t))$.

$E\left((X_{t+1}-X_t)^2\mid X_t\right)=(1-p+pX_t-X_t)^2\times X_t+(pX_t-X_t)^2\times (1-X_t)=(1-p)^2(1-X_t)^2\times X_t+(1-p)^2\times X_t^2 (1-X_t) =(1-p)^2(1-X_t)\times X_t\left( (1-X_t) +X_t \right)=(1-p)^2(1-X_t)\times X_t$

So

$E(X_{t+1}-X_t)^2=(1-p)^2 E(X_t(1-X_t))$

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  • $\begingroup$ OK, so is this a supermartingale then? Also, why are we finding $E(X_{t+1}-X_t)^2$ ? $\endgroup$
    – CCZ23
    Apr 28 '20 at 16:42
  • $\begingroup$ Yes. A martingale is also is super-martingale. To show $E(X_t(1-X_t))\longrightarrow 0$. Also check $X_t$ is UI or not? $\endgroup$
    – Masoud
    Apr 28 '20 at 20:01

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