Markov chain: frog jumping on infinite ladder

Question

A frog jumps on an infinite ladder. He goes up one step with probability $(1-p)$, and falls back down to the ground with probability $p$. If the frog has just fallen to the bottom, on average how many jumps will it take before he next reaches step $i$?

I showed that the frog's jumps form a Markov chain, with transition probability $p_{i,i+1}=(1-p)$ from step $i$ to step $i+1$, and probability $p_{i,0}=p$ from step $i$ to the ground. It then follows that the stationary distribution $(\pi_0,\pi_1,\dots)$ is geometric, with $\pi_i=(1-p)^{i}p$. Let $k_{i}^{j}$ denotes the mean time to go from step $i$ to step $j$ and $m_i = k_{i}^{i}$ is the mean return time from step $i$ to itself. Then, using the law of total expectation, we have \begin{align*}&m_i=(1-p)\left(1+k_{i+1}^{i}\right)+p\left(1+k_{0}^{i}\right), \text{ and} \\ &k_{i+1}^{i} = (1-p)\left(1+k_{i+2}^{i}\right)+p\left(1+k_{0}^{i}\right)\end{align*} Solving the latter recurrence relation, we get \begin{equation*}k_{j}^{i} = \frac{1}{p}\left(1+pk_{0}^{i}-\left(\frac{1}{1-p}\right)^j\right)\end{equation*} We also have the well-known result $m_i = \frac{1}{\pi_i}$, so substituting all of this into the first equation gives \begin{equation*}\frac{1}{(1-p)^{i}p} = (1-p)\left(1+\frac{1}{p}\left(1+pk_{0}^{i}-\left(\frac{1}{1-p}\right)^{i+1}\right)\right)+p\left(1+k_{0}^{i}\right)\end{equation*} so we can solve for the desired $k_{0}^{i}$ as \begin{equation*}k_{0}^{i} = \frac{2}{(1-p)^{i}p}-\frac{1}{p}\end{equation*} However, the answer is apparently supposed to be \begin{equation*}\frac{1}{(1-p)^{i}p}-\frac{1}{p}\end{equation*} the explanation being that $m_i = k_{i}^{0}+k_{0}^{i}$, with $m_i = \frac{1}{(1-p)^{i}p}$ as above and $k_{i}^{0} = \frac{1}{p}$ by a property of the geometric distribution. So I can't figure out what I did wrong to get the $2$ in the numerator - any assistance would be much appreciated. Thanks in advance for reading this long post.

Answer 1

If $k_j^i$ is the mean time to go from step $i$ to step $j$, then $k_i^i$ should equal $0$. If the frog is already on step $i$, then it wouldn't have to take any more jumps to reach that step. There's at least two ways to solve this correctly.

Approach $1$: One thing you could do is completely ignore $m_i$ and solve for $k_0^i$ in $k_{j}^{i} = \frac{1}{p}\left(1+pk_{0}^{i}-\left(\frac{1}{1-p}\right)^j\right)$ assuming that $k_i^i = 0$. Plugging in $j = i$ and $k_i^i = 0$ yields that $$k_0^i = \frac{1}{(1-p)^ip}-\frac{1}{p}$$

Approach $2$: You're right that $m_i$ should be the mean return time from $i$ to itself, but the frog has to move first, so it isn't equal to $0$. Instead, it has to fall to the ground $\left(k_i^0 \text{ steps}\right)$ and then go back up to $i$ $\left(k_0^i\text{ steps}\right)$. This is why $m_i$ should equal $k_i^0 + k_0^i$. Since you know $k_i^0 = \frac{1}{p}$ by the geometric distribution and $m_i = \frac{1}{\pi_i} = \frac{1}{(1-p)^{i}p}$, $k_0^i$ would be equal to $\frac{1}{(1-p)^{i}p}-\frac{1}{p}$.