2
$\begingroup$

Let $\xi_1, \xi_2, \cdots$ be i.i.d with $P(\xi_i=1)=P(\xi_i=0)=\frac{1}{2}$. Let

$X_0 := \frac{1}{2} \; ;\;\; X_n := \frac{1}{2} \xi_1 + \frac{1}{2^2}\xi_2 + \cdots + \frac{1}{2^{n}} \xi_n + \frac{1}{2^{n+1}}, \; \; n \ge 1$

I need to point that the last term is $\frac{1}{2^{n+1}}$, not $\frac{1}{2^{n+1}}\xi_{n+1}$.

I need to show that $X_n \longrightarrow X_{\infty}$ a.s. for some $X_{\infty}$ and also I need to find the distribution of $X_{\infty}$.

I can show that $X_n$ is a martingale and I am trying to derive an upper bound for $\mathbb{E}[\sup\limits_{1 \le i \le n} X_i]$ and then show that the upper bound converges. But that way it's hard for me to see how I can then figure out $X_{\infty}$ and its distribution. I guess the distribution of $X_{\infty}$ is some form of negative binomial, as I can clearly see the geometric and Bernoulli structures involved. I could not come up with a concrete proof though.

I appreciate any insights on this.

$\endgroup$
3
  • $\begingroup$ Yes, thanks for pointing that out. Just corrected it. $\endgroup$ Apr 7, 2016 at 18:41
  • 2
    $\begingroup$ Isn't $|X_n|\leq 1$ for all $n$? $\endgroup$ Apr 7, 2016 at 18:55
  • $\begingroup$ That is right $|X_n| \le 1$ for all $n$. $\endgroup$ Apr 7, 2016 at 19:12

0

You must log in to answer this question.

Browse other questions tagged .