Sequence of uniformly distribued variables

Question

I have a little problem that looks stupid, but I'm not sure about the solution.

So we have a random variable $X_1$, uniformly distribued on $[0,1]$. We define then $X_2$ conditionally on $\sigma(X_1)$ as uniformly distribued on $[0,X_1]$. By recursion, we define $X_n$, conditionally on $\sigma(X_1,...,X_{n-1})$, as uniformly distribued on $[0,X_{n-1}]$. This means, for $n\geq2$ and for each Borel set $A$, we have that: $$P(X_n\in A\mid \sigma(X_1,...,X_{n-1}))=\frac{\lambda(A\cap [0,X_{n-1}])}{X_{n-1}},$$ where $\lambda$ stands for the Lebesgue measure.

I have to compute the density function of $X_n$ and then prove that $X_n$ tends to $0$ in probability.

So yes for $X_1$ there is no problem at all. For the density (conditionally on the previous variables in the sequence) i had $\frac{1}{X_{n-1}}$, but it seems too easy :/

And then for the limit, i had:

$$P(X_n>\epsilon)=1-P(X_n\leq \epsilon)=1-\frac{\epsilon}{X_{n-1}},$$ and i don't really see how to prove that this goes to $0$.

Thanks for your reading and your answers,

Bests,

Answer 1

If $X_n$ is uniformly distributed in $[0, X_{n-1}]$, that means $E(X_n|X_{n-1})=\dfrac12\,X_{n-1}$, i.e. $EX_n=\dfrac1{2^n}$. Now, you can use the simple estimate $$P(X_n>\epsilon)\le\frac{EX_n}{\epsilon}=\frac1{2^n\epsilon}.$$

Answer 2

With induction it can be shown that for $u\in\left(0,1\right]$: $$P\left(X_{n}\leq u\right)=u\sum_{k=0}^{n-1}\frac{\left(-\ln u\right)^{k}}{k!}$$ having derivative: $$f_{X_{n}}\left(u\right)=\frac{\left(-\ln u\right)^{n-1}}{\left(n-1\right)!}$$

First step:

$$\begin{aligned}P\left(X_{2}\leq u\right) & =1-P\left(X_{2}>u\right)\\ & =1-\int P\left(X_{2}>u\mid X_{1}=x\right)dx\\ & =1-\int_{u}^{1}1-\frac{u}{x}dx\\ & =-u\ln u+u \end{aligned} $$