Let $\{X_{n}\}$ be a sequence of iid random variables with $P(X_n = j) = a_j > 0 $ for every $j \geq 0$. and let $\{Y_{n}\}$ be a sequence defined by $Y_{n} = X_{n} + X_{n - 1}$ for each $n\geq 1$ and $Y_0 = 0$.
Is $\{Y_{n}\}$ a Markov chain? I think the answer is no because
$$Y_{n + 1} = X_{n + 1} + X_{n} $$
and
$$Y_{n} = X_{n} + X_{n - 1},$$
so $Y_{n + 1}$ does not only depend on $Y_{n}$. For example the $X_{n + 1}$ term is not there. Am I right ?
You are right that $\{Y_n\}$ is in general not a Markov chain, but your reasoning is inproper.
A counter example:
Let's say all $X_n$ takes values $0$ and $1$, with probabilities $1/3$ and $2/3$, respectively. We consider $Y_1, Y_2, Y_3$. If $\{Y_n\}$ were a Markov chain, then we would have $Pr(Y_3 = 1|Y_1 = 0, Y_2 = 1) = Pr(Y_3 = 1|Y_2 = 1)$.
However, the left hand side is equal to \begin{eqnarray*} && Pr(Y_1 = 0, Y_2 = Y_3 = 1)/Pr(Y_1 = 0, Y_2 = 1)\\ &=& Pr(X_0 = 0, X_1 = 0, X_2 = 1, X_3 = 0)/Pr(X_0 = 0, X_1 = 0, X_2 = 1)\\ &=& 1/3, \end{eqnarray*} while a similar calculation shows that the right hand side is equal to $1/2$.
Hence it's not a Markov chain.