1
$\begingroup$

I am a little confused about the expectation $E(\tau)$ of the stopping time $$\tau=\inf\{k>0 \mid X_k=1\}, $$with the $X_i$ independent random variables with values $-1$ and $1$, where $P(X_1=1)=1/2$ and $P(X_1=-1)=1/2$. I thought this is $$E(\tau)=1\cdot P(\tau=1)+2\cdot P(\tau=2)+\dots=\sum_{i=1}^{\infty}\frac{n}{2^n}.$$ Is this correct and what is the value of the series? Thanks for the help

$\endgroup$
  • $\begingroup$ How do you get $P(\tau=k)=1/2^k$?? $\endgroup$ – user940 Oct 24 '16 at 16:09
  • $\begingroup$ @ByronSchmuland well, $\mathbb{P}(\tau=k) = \mathbb{P}(X_1 = -1, \ldots, X_{k-1}=-1, X_k = 1) = 2^{-k}$ by independence... ? $\endgroup$ – saz Oct 24 '16 at 16:39
  • $\begingroup$ @ByronSchmuland, $\tau$ has geometric distribution with parameter $1/2$. At first I read the problem carelessly and thought that $(X_k)$ is a SRW (so that $\tau$ has heavy tail), but in fact this is simply a coin tossing. $\endgroup$ – Sangchul Lee Oct 24 '16 at 16:43
  • $\begingroup$ @saz Apologizes! I completely misread the question. $\endgroup$ – user940 Oct 24 '16 at 16:48
  • $\begingroup$ The process is not a martingale, and I guess the "martingale" tag threw me off. $\endgroup$ – user940 Oct 24 '16 at 16:49
1
$\begingroup$

Yes, looks good (just as a side remark: it should read $\sum_{i=1}^{\infty}$ instead of $\sum_{n=1}^{\infty}$). In order to calculate $\sum_{i=1}^{\infty} \frac{i}{2^i}$ use that

$$\sum_{i=1}^{\infty} i x^i = \sum_{i=1}^{\infty}(i+1) x^i - \sum_{i=1}^{\infty} x^i = \frac{d}{dx} \left( \sum_{i=1}^{\infty} x^{i+1} \right) - \sum_{i=1}^{\infty} x^i$$

for any $x \in (-1,1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.