Markov chain and random walk Let $(X_n)$ be independent identically distributed random variables. Let $S_n=\sum\limits_{k=1}^n X_k$. Now we define $Z_n=S_n+S_{n+1}$ and ask if this defines a markov chain. My efforts so far tell me that this is true, but other sources suggest that the $Z_n$ do not define a markov chain. 
Any hints/solutions are appreciated.
 A: We can show that in general this process is not a Markov chain (meaning that unless you fix some particular distribution for $X_n$ you cannot expect the Markov property). In particular we can show that
$$
\mathbb{E}[Z_{n} | Z_{n-1}] \neq \mathbb{E}[Z_{n} | \mathcal{F}_{n-1}]
$$
where $\mathcal{F}_n = \sigma( Z_i \ | \  0 \le i \le n) = \sigma( X_i \ | \  1\le i \le n{+}1)$ where the latter equality follows from the definitions of $Z_i.$ In particular we see that
$$
\mathbb{E}[Z_{n} | \mathcal{F}_{n-1}] = 2S_n +\mathbb{E}[X_{n+1}].
$$
On the other hand we have that
$$
\mathbb{E}[Z_{n} | Z_{n-1}] = Z_{n-1} + \mathbb{E}[ X_{n} +X_{n+1} | Z_{n-1}]
 = 2S_n + \mathbb{E}[X_{n+1}] + \mathbb{E}[X_{n} | Z_{n-1}] - X_n$$
Hence the difference
$$
\mathbb{E}[Z_{n} | Z_{n-1}] - \mathbb{E}[Z_{n} | \mathcal{F}_{n-1}] = X_n-\mathbb{E}[X_{n} | Z_{n-1}] 
$$
and the latter random variable is in general not zero. For example you can take independent Bernoulli trials, than on the set $Z_{n-1} = k$ with $k < 2n -1$ $X_n$ can assume both values $0$ and $1$, while $\mathbb{E}[X_{n} | Z_{n-1}]$ is constant.
REMARK: Note that I added $Z_0$ in the definition of $\mathcal{F_n}.$ This simplifies the argument, but seems a reasonable assumption in my opinion.I would guess the argument still holds if you take $Z_0$ out of the filtration, but it would get more ugly because you have some weird information about your starting point.
A: Disclaimer: not entire confident with my calculations and remarks below, please comment if you find errors.

Let the i.i.d. $X_i$ take values $0, 1, 2$ with probabilities $p_0$, $p_1$, and $p_2$ respectively.
Then one can show
\begin{align}
P(Z_3=4 \mid Z_2 = 2) &= \frac{P(Z_3=4, Z_2=2)}{P(Z_2=2)}
\\
&= \frac{P\left[(X_1,X_2,X_3,X_4)\in \{(0,0,2,0), (0,1,0,2), (1,0,0,2)\}\right]}{P\left[(X_1,X_2,X_3) \in \{(0,0,2), (0,1,0), (1,0,0)\}\right]}
\\
&= \frac{p_0^2 p_2 (p_0 + 2 p_1)}{p_0^2 (p_2 + 2 p_1)}
\\
&= \frac{p_0 + 2 p_1}{p_2 + 2 p_1} \cdot p_2
\end{align}
and $$P(Z_3 = 4 \mid Z_2=2, Z_1=0)= \frac{P[(X_1,X_2,X_3,X_4) = (0,0,2,0)]}{ P[(X_1,X_2,X_3) = (0,0,2)]} = p_0.$$

Remarks:


*

*I don't think one can get a counterexample with coin flips (Bernoulli $X_i$), since one can then determine $(X_1,\ldots,X_{n+1})$ from $(Z_1,\ldots,Z_n)$, which would imply $Z_n$ forms a Markov chain. But I may be wrong about this.

*If instead $Z_n = (S_n, S_{n+1})$ (a pair instead of a sum), then $Z_n$ would form a Markov chain. This can be proved by noting that $Z_{n+1} = 2S_{n+1} + X_{n+2}$.

