# Find the limit of a random variable

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?

• 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? Apr 28 '20 at 8:28
• @joriki Thanks - ideally looking how to prove that formally, any idea? Also, why do you think it involves $X_0$? Apr 28 '20 at 8:44
• 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. Apr 28 '20 at 8:47

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

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

• OK, so is this a supermartingale then? Also, why are we finding $E(X_{t+1}-X_t)^2$ ? Apr 28 '20 at 16:42
• 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? Apr 28 '20 at 20:01