If $N_t$ is a Poisson process and $Y\in\{-1,1\}$, then $X_t = Y(-1)^{N_t}$ is a Markov process Let $Y$ be a random variable with values in $\{-1,1\}$ independent of a Poisson process $\{N_t\}$ with intensity $\lambda>0$. Set the process $X = \{X_t\}$ by $X_t=Y(-1)^{N_t}$.
Show that $X$ is a Markov process.

My proof:
$P(X_t=j|X_{t-1}=i_{t-1},...,X_0=i_0)=P(i_0(-1)^{N_t}=j)$
and:
$P(X_t=j|X_{t-1}=i_{t-1})=P(X_t=j|X_{t-1}=i_{t-1},Y=i_0)=P(i_0(-1)^{N_t}=j)$.
$X$ is a Markov process.
Is it correct? 
 A: First notice that the state space of $X$ is $\{-1, 1\}$. 
For fixed $t > 0$ and positive integer $k$, and any $0 \leq t_1 \leq t_2 \leq \cdots t_k \leq t$, $(i_1, \ldots, i_k, i) \in \{-1, 1\}^{k + 1}$,
\begin{align*}
& P(X_t = i \mid X_{t_k} = i_k, \ldots, X_{t_1} = i_1) \\
= & P(Y(-1)^{N_t} = i \mid Y(-1)^{N_{t_k}} = i_k, \ldots, Y(-1)^{N_{t_1}} = i_1) \\
= & P((-1)^{N_t} = iY \mid (-1)^{N_{t_k}} = i_k Y, \ldots, (-1)^{N_{t_1}} = i_1 Y) \\
= & \frac{P[(-1)^{N_t} = iY, (-1)^{N_{t_k}} = i_k Y, \ldots, (-1)^{N_{t_1}} = i_1 Y]}{P[(-1)^{N_{t_k}} = i_k Y, \ldots, (-1)^{N_{t_1}} = i_1 Y]} \tag{1}
\end{align*}
Now deal with the numerator and denominator separately, conditioning $Y$,
\begin{align*}
& P[(-1)^{N_t} = iY, (-1)^{N_{t_k}} = i_k Y, \ldots, (-1)^{N_{t_1}} = i_1Y] \\ 
= & P[(-1)^{N_t} = iY, (-1)^{N_{t_k}} = i_k Y, \ldots, (-1)^{N_{t_1}} = i_1 Y \mid Y = 1]P(Y = 1) \\
+ &  P[(-1)^{N_t} = iY, (-1)^{N_{t_k}} = i_k Y, \ldots, (-1)^{N_{t_1}} = i_1 Y \mid Y = -1]P(Y = -1) \tag{2}
\end{align*}
By the independence of $Y$ and $\{N_t\}$ and the fact that $\{N_t\}$ is increment independent, it follows that
\begin{align*}
& P[(-1)^{N_t} = iY, (-1)^{N_{t_k}} = i_k Y, \ldots, (-1)^{N_{t_1}} = i_1 Y \mid Y = 1] \\
= & P[(-1)^{N_t} = i, (-1)^{N_{t_k}} = i_k, \ldots, (-1)^{N_{t_1}} = i_1 \mid Y = 1] \\ 
= & P[(-1)^{N_t} = i, (-1)^{N_{t_k}} = i_k, \ldots, (-1)^{N_{t_1}} = i_1] \\
= & P[(-1)^{N_t - N_{t_k}} = i/i_k, \ldots, (-1)^{N_{t_2} - N_{t_1}} = i_2/i_1, (-1)^{N_{t_1}} = i_1] \\
= & P[(-1)^{N_t - N_{t_k}} = i/i_k]\cdots P[(-1)^{N_{t_2} - N_{t_1}} = i_2/i_1]P[(-1)^{N_{t_1}} = i_1]
\end{align*}
Similarly, 
\begin{align*}
& P[(-1)^{N_t} = iY, (-1)^{N_{t_k}} = i_k Y, \ldots, (-1)^{N_{t_1}} = i_1 Y \mid Y = -1] \\
= & P[(-1)^{N_t - N_{t_k}} = i/i_k]\cdots P[(-1)^{N_{t_2} - N_{t_1}} = i_2/i_1]P[(-1)^{N_{t_1}} = -i_1]
\end{align*}
Therefore, $(2)$ can be rewritten as 
\begin{align*}
& P[(-1)^{N_t} = iY, (-1)^{N_{t_k}} = i_k Y, \ldots, (-1)^{N_{t_1}} = i_1Y] \\
= & P[(-1)^{N_t - N_{t_k}} = i/i_k]\cdots P[(-1)^{N_{t_2} - N_{t_1}} = i_2/i_1] \times \\
& (P[(-1)^{N_{t_1}} = i_1]P(Y = 1) + P[(-1)^{N_{t_1}} = -i_1]P(Y = -1))
\end{align*}
With the same argument as above, the denominator of $(1)$ can be expressed as:
\begin{align*}
& P[(-1)^{N_{t_k}} = i_kY, \ldots, (-1)^{N_{t_1}} = i_1Y] \\
= & P[(-1)^{N_{t_k} - N_{t_{k - 1}}} = i_k/i_{k - 1}]\cdots P[(-1)^{N_{t_2} - N_{t_1}} = i_2/i_1]\times \\
& (P[(-1)^{N_{t_1}} = i_1]P(Y = 1) + P[(-1)^{N_{t_1}} = -i_1]P(Y = -1))
\end{align*}
It then follows that
\begin{align*}
& P(X_t = i \mid X_{t_k} = i_k, \ldots, X_{t_1} = i_1) \\
= & P[(-1)^{N_t - N_{t_k}} = i/i_k] \\
= & P(X_t = i \mid X_{t_k} = i_k),
\end{align*}
showing that $\{X_t\}$ is Markovian. 
