# How to determine the distribution of the limiting random variable

Given a sequence of random variables $$\{X_n\}_{n \geq 0}$$ defined as follows.

$$X_0 = p, X_{n+1} = qX_n + (1-q)1_{Y_{n+1} \leq X_n}, \forall n \geq 0,$$ where $$p, q \in (0, 1)$$ are constants and $$Y_i \sim U(0, 1)$$ is a sequence of i.i.d random variables with standard uniform distribution for all $$i \geq 0$$. Define $$X_{\infty} = \lim_{n \to \infty}X_n$$.

I am confused about how to determine the distribution of $$X_{\infty}$$.

• You might also check expected values; maybe $\mathbb E[X_n]$, or maybe $\mathbb E[X_{n+1} \mid X_n]$, could be of value in figuring out what this process is doing. May 2 at 15:54
• The sequence $X_n$ is a martingale, so $E[X_{\infty}] = E[X_0] = p$. I suspect it is Bernoulli with mean $p$, but I am not yet seeing how to show whether this is true or not. May 2 at 16:02
• I deleted one of my comments because I think it was more misleading than helpful. Sorry about that! May 2 at 16:06

The limiting distribution is a Bernoulli with mean $$p$$. Assuming that we can show that the convergence $$X_n \to X_\infty$$ holds in $$L^2$$ (this can be proved using martingales), here is a sketch of a proof:

• Show that $$\mathbb{E}[(X_{n+1}-X_n)^2] = (1-q)^2 \mathbb{E}[X_n(1-X_n)].$$
• Using $$L^2$$ convergence, deduce that $$\mathbb{E}[X_\infty (1-X_\infty)] = 0.$$
• Using that $$X_\infty \in [0,1]$$ a.s. deduce that it has Bernoulli distribution. The parameter is the mean which is $$\mathbb{E}[X_\infty] = \mathbb{E}[X_0] = p$$.

An alternative proof which does not require $$L^2$$ convergence. Again the idea is to show that $$\mathbb{E}[X_\infty (1-X_\infty)] = 0$$ and to continue from there. To show this, you can

• Argue that $$X_\infty$$ satisfies the following identity in distribution: $$X_\infty = q X_\infty + (1-q)\mathbf{1}_{U\leq X_\infty},$$ where $$U \sim \mathcal{U}(0,1)$$ is an independent random variable.
• Using this, prove that $$\mathbb{E}[X_\infty (1-X_\infty)] = q(2-q) \mathbb{E}[X_\infty (1-X_\infty)]$$ then conclude as before.