How to determine the distribution of the limiting random variable

Question

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 $X_{\infty}$.

Answer 1

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 $\mathbb{E}[X_\infty] = \mathbb{E}[X_0] = p$.

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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.