Random walks : Hitting and recurrence Times relation

Question

I have trouble understanding that how $$E\left[T_0|X_{0} = 0\right] = 1 + E[H_0|X_0=1] $$ where $T_0 = \inf\{n \geq 1:X_n = 0 \}$ and $H_A =\inf\{ n\geq 0: X_n \in A \}$. In other words $T_0$ is the first recurrent time and $H_0$ is the first hitting time.

Answer 1

Assume that the random walk is defined as $$ X_n = \sum_{k = 1} ^ n \xi_k $$where $(\xi_a)_{a\ge 1}$ are iid and such as $P(\xi_1 = \pm 1) = 1/2$.

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$$1_{T_0 = m,X_0=0} = 1_{T_0 = m,X_0=0, X_1 = 1}+ 1_{T_0 = m,X_0=0, X_1 = -1} $$ Now using the Markov property, and as $T_0$ depends in a deterministic way on $(X_a)_{a\ge 1}$, you get $$ T_0 |(X_0=a, X_1 = b) = T_0 |X_0=a $$ and in particular the following red equality in $$\begin{align} E[T_0|X_0 = 0] &= \sum_{m \ge 1} m P(T_0 = m | X_0 = 0) \\ &= \sum_{m \ge 1} m \sum_{e = \pm 1} P(T_0 = m, X_1 = e | X_0 = 0) \\ &= \sum_{m \ge 1} m \sum_{e = \pm 1} P(T_0 = m| X_1 = e, X_0 = 0)P(X_1 = e)\\ &\color{red}{=} \sum_{m \ge 1} m \sum_{e = \pm 1} P(T_0 = m| X_1 = e)P(X_1 = e)\\ &= \sum_{e = \pm 1} \sum_{m \ge 1} m P(T_0 = m| X_1 = e)P(X_1 = e)\\ &= \sum_{e = \pm 1} E[T_0|X_1 = e]P(X_1 = e) \end{align}$$

Using the symetry you get: $$ = \sum_{e = \pm 1} E[T_0|X_1 = e] \frac 12 = E[T_0|X_1 = 1] $$

Using the definition of $T_A:$ $$ T_0|(X_1 = 1) = (1 + T_0) | (X_0 = 1) $$ and the previous equality becomes: $$ = 1 + E[T_0|X_0 = 1] $$