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Let's say we have a 1D random walk starting at the origin where we go up $1$ with probability $1/5$, down $1$ with probability $1/5$, and stay put with probability $3/5$.

If we walk $n$ steps, what's the expected number of times we reach zero?


Here's a similar question, where instead they assume we move based on a continuous distribution.

I tried turning my discrete distribution into a continuous one using the pdf

$$f(x) = \frac{1}{5}\delta(x+1) + \frac{3}{5}\delta(x) + \frac{1}{5}\delta(x-1),$$

but the approach in that question didn't seem to work out.

Any ideas?

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  • $\begingroup$ Do you count staying at zero as "reaching" zero, or are you interested in the expected number of returns to zero, counting each (possibly extended) stay at zero as one? $\endgroup$ – joriki May 23 '16 at 1:24
  • $\begingroup$ @joriki Yes, I'm counting staying at zero as "reaching" zero. $\endgroup$ – ZapMathigan May 23 '16 at 1:49
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By linearity of expectation, the expected number of visits to the origin is the sum over the expected number of visits to the origin after $0\lt l\le n$ steps. That's the probability of taking $j$ up steps and $j$ down steps and staying put $l-2j$ times, summed for $0\le 2j\le l$, so the total is

$$ \sum_{l=1}^n\left(\frac35\right)^l\sum_{j=0}^{\left\lfloor\frac l2\right\rfloor}\binom l{j,j,l-2j}3^{-2j}\;. $$

I'm not aware that this can be simplified.

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  • $\begingroup$ Thanks for this! Don't we also need to somehow incorporate the 1/5 probabilities into the formula? $\endgroup$ – ZapMathigan May 23 '16 at 2:53
  • $\begingroup$ @ZapMathigan: Yes, certainly, sorry, it should have been $5^{-n}$ instead of $5^{-l}$; fixed. $\endgroup$ – joriki May 23 '16 at 5:23
  • $\begingroup$ Actually I think your first answer was correct. Using your ideas, I had a double sum with summand Binomial[2 j, j] Binomial[l, 2 j] (1/5)^(2 j) (3/5)^(l - 2 j). That simplifies to your first answer. I did some simulations and it seems correct. $\endgroup$ – ZapMathigan May 23 '16 at 5:27
  • $\begingroup$ @ZapMathigan: Right, you had me confused for a moment there :-) You were right that we need to incorporate the $\frac15$ probabilities, but I'd already incorporated them. I reverted the edit. $\endgroup$ – joriki May 23 '16 at 6:56

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