-2
$\begingroup$

Suppose $S_n:= \sum^n_{i}X_i$ be asymmetric simple random walk. with $P(X_{1}=+1)=p>1/2$, $P(X_{1}=-1)=q$, $p+q=1$, and $S_{0}=0$. Define $T_1 := \inf\{n: S_n = 1\}$. And we know that for $\theta > 0$, $e^{\theta}E\varphi(\theta)^{-T_1}=1$, where $$\varphi(\theta)= pe^{\theta} +qe^{-\theta}.$$

$\textbf{Question}$: Find the probability $$P(T_1=2k+1), k\ge 0.$$

I am really stuck on this question, could you please give me some details about how to solve this question? Thank you!

$\endgroup$
  • 2
    $\begingroup$ Welcome to Math.SE! Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 2 '18 at 21:55
  • $\begingroup$ Hint: Express $e^\theta$ as a function of $\varphi(\theta)$. $\endgroup$ – Did May 2 '18 at 21:59
  • $\begingroup$ Hi @Did, could you provide more details about that? Thank you! $\endgroup$ – BethNeiers May 3 '18 at 1:49
  • $\begingroup$ Sure - but first, what did you try to apply this hint? Please be specific, or one could be led to believe you did nothing and are just waiting for a full answer to pop up. $\endgroup$ – Did May 3 '18 at 5:52
  • $\begingroup$ Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. $\endgroup$ – robjohn May 6 '18 at 7:41
2
$\begingroup$

Let $T_{x} = \inf\{n \geq 0 : S_{n} = x\}$ for $x \in \mathbb{Z}$. Now you can show that for all $x,y \in \mathbb{Z}$ \begin{equation} \mathbb{P}(T_{y} = n | S_{0}=x) = \mathbb{P}(T_{y-x} = n | S_0 = 0),\qquad n \in \mathbb{N}. \end{equation} So we can assume without loss of generality that $S_{0} = 0$. Now define the interarrival times $\xi_{n} := T_{n} - T_{n-1}$ for $n \geq 1$.

It can be shown that the sequence $\{\xi_{n}\}_{n \in \mathbb{N}}$ is i.i.d.

Because of this you can compute the probability generating function of $T_{x}$ and $T_{-x}$ if you can determine the probability generating function of $T_{1}$ and $T_{-1}$, respectively since for $\varphi(s) = \mathbb{E}(s^{T_{1}} | S_{0} = 0)$ we get \begin{equation} \mathbb{E}(s^{T_{x}} | S_{0} = 0) = (\varphi(s))^{x}. \end{equation}

Now with first step analysis we get

\begin{equation} \varphi(s) = \mathbb{E}(s^{T_{1}} | S_{0} = 0) = \frac{1 - \sqrt{1 - 4pqs^{2}}}{2qs}. \end{equation} Similarly \begin{equation} \psi(s) = \mathbb{E}(s^{T_{-1}} | S_{0} = 0) = \frac{1 - \sqrt{1 - 4pqs^{2}}}{2ps}. \end{equation} So for all $x \in \mathbb{Z}$ we get \begin{equation*} \mathbb{E}(s^{T_{x}}|S_{0} = 0)= \begin{cases} \left(\varphi(s)\right)^{x}, & \mbox{if $x > 0$ } \\ \left(\psi(s)\right)^{|x|}, & \mbox{if $x < 0$}. \end{cases} \end{equation*}

Now $\varphi^{(2k+1)}(0) = (2k+1)! \cdot \mathbb{P}(T_{1} = 2k + 1)$

$\endgroup$
  • $\begingroup$ Since you're new here, let me give you a rough idea about deletion on this site. I appreciate your effort to this PSQ, but questions not meeting the standards of this site are often victims of deletion "by the votes of people", in particular, by some users in the chatroom CRUDE. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 2 '18 at 23:35
  • $\begingroup$ Hi @wayne, I am sorry, but what's the meaning of $\varphi(s)^{(2k+1)}(0)$ and how do you get the last equality? Thank you very much! $\endgroup$ – BethNeiers May 3 '18 at 1:48
  • 3
    $\begingroup$ If you are a second year PHD student in pure mathematics, you have to know what are the meaning of $\varphi^{(2k+1)}(0)$ $\endgroup$ – wayne May 3 '18 at 7:14

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.