Let $(X_n)$ a Markov chain on $\mathbb Z$ s.t. $$\mathbb P(X_{n+1}=i+1\mid X_n=i)=\mathbb P(X_{n+1}=i-1\mid X_n=i)=\frac{1}{2},$$ for all $n$. We denote $\mathbb P_i$ the probability measure when $X_0=i$, and $\mathbb E_i$ the expectation w.r.t. the measure $\mathbb P_i$. Let $T$ being the first time when $(X_n)$ is in $\{1,4\}$.
Suppose that Let $h_i=\mathbb P_i(T<\infty )$ and $k_i=\mathbb E_i[T]$. For me it's clear from the fact that $$\mathbb P(X_n=j\mid X_1=i)=\mathbb P_i(X_{n-1}=j),$$ and the Markov property that we have that $$h_2=\frac{1}{2}h_1+\frac{1}{2}h_3.$$
However, I don't understand why $$k_2=\frac{1}{2}k_1+\frac{1}{2}k_3+{\color{red}{1}},$$
Where does the $+1$ comes from ? In my lecture it's written : "the $+1$ appear because we count the time for the first step", but I don't really understand the argument. Any explanation ?
Suppose at time $0$ you were in state $1$ or $3$ with equal probability. Then the expectation of the first time when reach your target is $\mathbb P(X_0=1)\mathbb E_1(T)+\mathbb P(X_0=3)\mathbb E_3(T)=\frac12\mathbb E_1(T)+\frac12\mathbb E_3(T)$.
But that's not the case - you were in state $2$ at time $0$, so you are in state $1$ or $3$ with equal probability at time $1$. Since the chain is time-homogeneous, this just means everything happens $1$ time step later than the case above.