Writing $\mathbb{P}(X_n - X_{n-1} = 1 \mid X_{n-1} = x)$ as $\mathbb{P}(X_n - x = 1)$? Suppose we have a Markov chain $\{X_n; n \geq 0\}$ where each variable takes on the states $S_X = \{0, 1, 2\}$.
I am unsure about which of the following is correct:
\begin{align}
    \mathbb{P}(X_n - X_{n-1} = 1 \mid X_{n-1} = x) &= \mathbb{P}(X_n - x = 1 \mid X_{n-1} = x) \tag{1} \\
    \mathbb{P}(X_n - X_{n-1} = 1 \mid X_{n-1} = x) &= \mathbb{P}(X_n - x = 1) \tag{2}
\end{align}
Both seem plausible to me. Even though I suspect $(1)$ is correct, I cannot justify why; intuitively, it seems that replacing $X_{n-1}$ by $x$ makes the condition $X_{n-1} = x$ redundant.
Which equality is correct and how can we give a convincing justification?
 A: The first choice is correct.

Applying the formula for conditional probability
$$
P(A{\,{\large{\mid}}\,} B)=\frac{P(A\land B)}{P(B)}
$$
we get
\begin{align*}
&
P\Bigl(\bigl(X_n-X_{n-1}=1\bigr) {\,{\large{\mid}}\,} \bigl(X_{n-1}=x\bigr)\Bigr)
\\[6pt]
=\;&
\frac{P\Bigl(\bigl(X_n-X_{n-1}=1\bigr) \land \bigl(X_{n-1}=x\bigr)\Bigr)}{P(X_{n-1}=x)}
\\[6pt]
=\;&
\frac{P\Bigl(\bigl(X_n-x=1) \land \bigl(X_{n-1}=x\bigr)\Bigr)}{P(X_{n-1}=x)}
\\[6pt]
=\;&
P\Bigl(\bigl(X_n-x=1) {\,{\large{\mid}}\,} \bigl(X_{n-1}=x\bigr)\Bigr)
\\[6pt]
\end{align*}
which validates the first choice.

As an example to show that the second choice can fail, consider the Markov chain with transition matrix
$$
\begin{array}{c|c|c|c|}
{\vphantom{x_{X_1}}}&0&1&2\\
\hline
0&{\vphantom{\dfrac{x}{X}}}{\large{\frac{1}{2}}}&{\large{\frac{1}{4}}}&{\large{\frac{1}{4}}}\\
\hline
1&{\vphantom{\dfrac{x}{X}}}{\large{\frac{1}{4}}}&{\large{\frac{1}{2}}}&{\large{\frac{1}{4}}}\\
\hline
2&{\vphantom{\dfrac{x}{X}}}{\large{\frac{1}{4}}}&{\large{\frac{1}{4}}}&{\large{\frac{1}{2}}}\\
\hline
\end{array}
$$
where for $i,j\in \{0,1,2\}$, the entry $p(i,j)$ in row $i$, column $j$ is the probability of a $1$-step transition from state $i$ to state $j$.

Now suppose $X_0$ takes values $0,1,2$ with equal likelihood.

Then for $x=0$ we get
\begin{align*}
&
P\Bigl(\bigl(X_1-X_0=1\bigr){\,{\large{\mid}}\,}\bigl(X_0=x\bigr)\Bigr)
\qquad\;\;\;\;\;
\\[4pt]
=\;&
P\Bigl(\bigl(X_1=1\bigr){\,{\large{\mid}}\,}\bigl(X_0=0\bigr)\Bigr)
\\[4pt]
=\;&
p(0,1)
\\[4pt]
=\;&
{\small{\frac{1}{4}}}
\\[4pt]
\end{align*}
whereas
\begin{align*}
&P\bigl(X_1-x=1\bigr)
\\[4pt]
=\;&
P\bigl(X_1=1\bigr)
\\[4pt]
=\;&
P\bigl(X_0=0\bigr)P\Bigl(\bigl(X_1=1\bigr){\,{\large{\mid}}\,}\bigl(X_0=0\bigr)\Bigr)
\\[0pt]
&+
P\bigl(X_0=1\bigr)P\Bigl(\bigl(X_1=1\bigr){\,{\large{\mid}}\,}\bigl(X_0=1\bigr)\Bigr)
\\[0pt]
&+
P\bigl(X_0=2\bigr)P\Bigl(\bigl(X_1=1\bigr){\,{\large{\mid}}\,}\bigl(X_0=2\bigr)\Bigr)
\\[4pt]
=\;&
{\small{\frac{1}{3}}}{\,\cdot\,}p(0,1)
+
{\small{\frac{1}{3}}}{\,\cdot\,}p(1,1)
+
{\small{\frac{1}{3}}}{\,\cdot\,}p(2,1)
\\[4pt]
=\;&
\Bigl({\small{\frac{1}{3}}}\Bigr)
\Bigl(p(0,1)+p(1,1)+p(2,1)\Bigr)
\\[4pt]
=\;&
\Bigl({\small{\frac{1}{3}}}\Bigr)
\Bigl({\small{\frac{1}{4}}}+{\small{\frac{1}{2}}}+{\small{\frac{1}{4}}}\Bigr)
\\[4pt]
=\;&
{\small{\frac{1}{3}}}
\\[4pt]
\end{align*}
A: \begin{align}
    \Pr(X_n - X_{n-1} = 1 \mid X_{n-1} = x) &= \Pr(X_n - x = 1 \mid X_{n-1} = x) \tag{1} \\
    \Pr(X_n - X_{n-1} = 1 \mid X_{n-1} = x) &= \Pr(X_n - x = 1) \tag{2}
\end{align}
That the second one cannot be right is seen by an example: Suppose the transitions $0\to1\to2\to0$ each have probability $0.99$ and the transitions $0\to2\to1\to0$ each have probability $0.01.$
Assuming we start with the stationary distribution, which assigns probability $1/3$ to each of $0,1,2,$ then we have
\begin{align}
& \Pr( X_n-X_{n-1}=1\mid X_{n-1}=0)=0.99 = \Pr(X_n-0=1\mid X_{n-1}=0) \\[6pt]
\text{and } & \Pr( X_n-X_{n-1}=1\mid X_{n-1}=0)=0.99 \ne 1/3 = \Pr(X_n-0=1).
\end{align}
