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?
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.