Inductively defined random variables

Question

Let $X_0=1$, define $X_n$ inductively by declaring that $X_{n+1}$ is uniformly distributed over $(0,X_n)$. Now I can't understand how does $X_{n}$ gets defined. If someone would just derive the distribution of $X_2$ that would be helpful. I saw this in a problem and I can't really start trying it. Thanks for any help.

Answer 1

You already know the distribution of $X_2$ conditionally on $X_1$, i.e.:

$$ X_2 | X_1 = x_1 \sim U[0,x_1]$$

This means that for $t \in [0,1]$:

$$ F_{X_2|X_1=x_1}(t) = \Pr[X_2 \leq t \vert X_1=x_1] = \min\left(\frac{t}{x_1},1\right)$$

Then the unconditional result follows from $X_1 \sim U[0,1]$:

$$ \begin{aligned} F_{X_2}(t) = \Pr[X_2 \leq t] &= \int_{0}^1 \Pr[X_2 \leq t \vert X_1=x_1]dx_1 \\ &= \int_{0}^1 \min\left(\frac{t}{x_1},1\right) dx_1 \\ &= \int_{0}^t dx_1 + \int_{t}^1 \frac{t}{x_1}dx_1 \\ &= t-t\log(t) \end{aligned} $$

(For the case $t=0$ just use the right limit i.e. $0\log(0)=0$.)

In other words the distribution of $X_2$ has Lebesgue density $f_{x_2}(t) = F_{X_2}'(t) = -\log(t)$ on $[0,1]$.

Answer 2

Here's a way to define those random variables rigorously:

Let $\theta_1, \theta_2, \ldots$ be an i.i.d. sequence of random variables that are uniformly distributed on $[0, 1]$. Now define $X_0 = 1$ and $X_n = X_{n - 1} \theta_n = \prod \limits_{i = 1}^n \theta_i$ for $n > 0$.

Answer 3

The distribution of $X_1$ is $U(0,X_0) =U(0,1) $ which has density \begin{equation} f_{X_1}(x) = \begin{cases} 1 \quad \mbox{for } x \in (0,1) \\ 0 \quad \mbox{otherwise} \end{cases} \end{equation} Now, $X_2 | X_1 = x_1$ follows a $U(0, x_1) $ which has density \begin{equation} f_{X_2 | X_1 = x_1}(x) = \begin{cases} \frac{1}{x_1} \quad \mbox{for } x \in (0,x_1) \\ 0 \quad \mbox{otherwise} \end{cases} \end{equation} The joint density is \begin{equation} f_{X_1, X_2}(x_1,x_2) = f_{X_2 | X_1 = x_1}(x_2) f_{X_1}(x) = \begin{cases} \frac{1}{x_1} \quad \mbox{for } 0 < x_2 < x_1 < 1 \\ 0 \quad \mbox{otherwise} \end{cases} \end{equation} The get the marginal density of $x_2$ we integrate out $x_1$ from the joint density. \begin{equation} \begin{split} f_{X_2}(x_2) &= \int_{x_2}^1 f_{X_1, X_2}(x_1,x_2) d x_1 \\ &= \int_{x_2}^1 \frac{1}{x_1} d x_1 \\ &= \log (x_1) \big|_{x_1 = x_2}^{x_1 =1} \\ &= \log(1) - \log (x_2) \\ &= - \log (x_2) \end{split} \end{equation} The marginal density of $X_2$ is \begin{equation} f_{X_2}(x_2)= \begin{cases} - \log (x_2) \quad \mbox{for } x_2 \in (0,1) \\ 0 \quad \mbox{otherwise} \end{cases} \end{equation}