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I wonder if someone can please check if I am on the right lines.

I want to show that a random walk with a probability of a right step of $\frac{1}{4}$ is recurrent. I think I might have done something wrong in my deductions.

I have that the probability of being at the origin at the nth step is

$$\left( \begin{array}{c} n\\ \frac{n}{2} \end{array} \right) \left(\frac{1}{4} \cdot \frac{3}{4}\right)^\frac{n}{2}$$

Then because this could be for any $n \ge 1$, I have decided that the probability of being at the origin at step $n$ for some $n$ could be

$$ \sum_{n=1}^\infty \left( \begin{array}{c} n\\ \frac{n}{2} \end{array} \right) \left(\frac{1}{4} \cdot \frac{3}{4}\right)^\frac{n}{2}=\sum_{n=1}^\infty\left( \begin{array}{c} n\\ \frac{n}{2} \end{array} \right) \left(\frac{1}{4}\right)^{n-\frac{n}{2}} \left( \frac{3}{4}\right)^\frac{n}{2}$$

Is this correct? And if it is, this then looks similar to the binomial theorem for $(\frac{1}{4}+\frac{3}{4})^n$ which is of course 1, meaning the random walk would be recurrent. If this is what I need to be doing, can anyone guide me as to how I get this in the correct format to use the binomial theorem.

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  • $\begingroup$ Are you using the gamma function in this summation to define n choose n/2 ? $\endgroup$ – Peter Foreman Feb 2 at 16:44
  • $\begingroup$ "this then looks similar to the binomial theorem for $(\frac{1}{4}+\frac{3}{4})^n$" This might "look similar" to your eyes but this is a completely different object, right? $\endgroup$ – Did Feb 2 at 16:53
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Technical hint: $\binom n {n/2} \sim c 2^n/\sqrt n$. Which you can see by using Stirling's formula or by just knowing that about half of the mass in Pascal's triangle is within $\pm \sqrt n$ of the middle.

Conceptual hint: you are calculating the expected number of returns to $0$ ever. If it is finite, that means ...

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