**Downside:**

A particle undergoes Random Stroll on the eight vertices of a dice, by shifting from a given vertex to any of the three adjoining vertices with the identical chance $frac{1}{3}$, independently of the place it has been previously and when. For 2 reverse vertices $x$ and $y$, calculate the anticipated time $mathbb{E}left(sum_{n=0}^{T_x – 1} mathbb{1}_{{X_n = y}}proper)$ spent at $y$ earlier than returning to $x$. Right here $T_x = inf{n in mathbb{N} : X_n = x}$.

**My try**

$(X_n)_{n in mathbb{N}_0}$ is a Markov chain with $X_0 = x$ and state area being the set ${x, y, z}$, the place state $z$ represents the opposite $6$ vertices of the dice. The transition possibilities are:

My plan is to first calculate $p_k$, *the chance of $ok$ visits to $y$ earlier than returning again to $x$*, the place $ok in mathbb{N}$, after which the required expectation needs to be

$$

mathbb{E}left(sum_{n=0}^{T_x – 1} mathbb{1}_{{X_n = y}}proper) = sum_{ok=1}^{infty} ok p_k

$$

I calculate $p_k$ as follows. A typical path ranging from $x$, visiting $y$ $ok$ occasions, after which going again to $x$ appears like:

$$

x zz y z y zzz y cdots zx

$$

Within the path above:

- There are $ok$ occurrences of $y$.
- Between each incidence of $x$ or $y$, there may be no less than one $z$.
- There are $ok+1$ “bins” for $z$‘s. In every field there may be no less than one $z$.

Due to this fact, there should be a minimal of $(ok+1)$ $z$‘s. Suppose there are $m in mathbb{N}_0$ “further” $z$‘s, i.e., there are a complete of $(ok+m+1)$ $z$‘s.

Likelihood of such a path $=frac{1}{3^{ok+m+1}}$

Variety of such paths = Variety of methods to place $m$ indistinguishable balls in $ok+1$ bins = $binom{ok+m}{m}$

Due to this fact,

$$

p_k = sum_{m=0}^{infty} frac{1}{3^{ok+m+1}} binom{ok+m}{m}

$$

**Assist**

I’m unable to simplify the expression for $p_k$ obtained above.

Extra usually, is there a greater technique to resolve this downside?

I do know this query asks one thing very comparable, however I’m unable to calculate the $p$ within the accepted reply, as you’ll be able to see above.