Hey there, let's break down your two questions about the Markov chain from Ross's book step by step:

问题1：为什么随机变量$N_i$相互独立？ #

First, let's recap the core of the Markov property: given the current state, the future is independent of the past. Now, remember what each $N_i$ represents: the number of transitions it takes to go from the first entry into state $i$ to the first entry into state $i+1$.

Suppose we take two such random variables $N_i$ and $N_j$ where $i < j$. The process for $N_i$ is entirely contained within moving from state $i$ to $i+1$. Once the chain finishes this process and reaches $i+1$, all subsequent behavior (until it hits state $j$ and starts the $N_j$ process) only depends on the current state at each step. By the Markov property, once the chain enters state $j$, every detail of the earlier $N_i$ process is irrelevant to how long it takes to get to $j+1$—the future transitions only care about the current state $j$, not any past history.

The reverse is also true: $N_j$ can't affect $N_i$ because $N_i$ happens entirely before $N_j$ begins, and the Markov property ensures past events don't influence future outcomes given the present state.

More generally, for any sequence of indices $i_0 < i_1 < ... < i_k$, each $N_{i_m}$ only depends on the path from $i_m$ to $i_m+1$. These paths are separated by intermediate state transitions, and each path's evolution is independent of all prior paths thanks to the Markov property. That's why all $N_i$ are independent.

问题2：为什么$\mu_i$的表达式中第一项是1而不是$p$？ #

Your initial conditioning approach is totally valid—let's connect it to the book's equation to see they're equivalent.

First, recall $\mu_i = E[N_i]$. When the chain first enters state $i$, we immediately make one transition (this is where the 1 comes in—we've already used one transition, no matter where we go):

• With probability $p$, we jump directly to $i+1$, so the total number of transitions is just 1.
• With probability $q$, we jump to $i-1$, and now we need additional transitions to get from $i-1$ all the way to $i+1$.

Let's expand your conditioning equation:
$$\mu_i = E[N_i|\text{jump to }i+1]p + E[N_i|\text{jump to }i-1]q$$
We know $E[N_i|\text{jump to }i+1] = 1$. Now, $E[N_i|\text{jump to }i-1]$ is the total transitions needed when we first jump to $i-1$: that's 1 (the initial jump) plus the expected number of transitions to get from $i-1$ to $i+1$.

Substitute that into your equation:
$$\mu_i = 1 \cdot p + \left(1 + E[\text{transitions from }i-1\text{ to }i+1]\right)q$$
Since $p + q = 1$, this simplifies to:
$$\mu_i = 1 + q \cdot E[\text{transitions from }i-1\text{ to }i+1]$$
Which is exactly the equation from the book! The book just rearranged the terms to group the guaranteed 1 transition first, then account for the extra work needed if we jump backward. Your approach and the book's are mathematically identical—they just present the same logic in slightly different forms.

备注：内容来源于stack exchange，提问作者Moon Traveler