# Is first order moving average a Markov process?

Given first order moving average

$$x(n) = e(n) + ce(n-1)$$

where $e(n)$ is a sequence of Gaussian random variables with zero mean and unit variance which are independent of each other, and $c$ is a weighting constant in the interval $\,\, 0 < c \le 1$.

Under these conditions, is $x(n)$ a Markov process?

I tried starting from previous equations

$$\begin{array}{lcl} x(n-2) & = & e(n-2) & + & c\,e(n-3) \\ x(n-1) & = & e(n-1) & + & c\,e(n-2) \\ x(n) & = & e(n) & + & c\,e(n-1) \\ \end{array}$$

Looking at these equations I intuitively write

$$f\Big(x(n) \, | \, x(n-1)\Big) = f\Big(x(n) \, | \, x(n-1), x(n-2)\Big)$$

because $x(n)$ is totally independent of $x(n-2)$. But I can't express this idea in Mathematical language.

-
Got something from an answer below? – Did Mar 29 '13 at 7:55
And your strategy in such cases is to stay silent? Not exactly what I would call a model of good manners nor a fruitful learning disposition. – Did Mar 29 '13 at 8:08
An interesting feature of this site is that it is a counterexample to your (rather sad, I must say) grand principles. Conversely, the strategy you describe is awfully rude to those answering your questions, hence you might want to advertise it clearly so that people who object to it abstain from answering your questions. – Did Mar 29 '13 at 9:51

Assuming $- 1 < c < 1$, and letting $L$ be the backward operator (acting on a sequence, $x_{n - 1} = Lx_n$), you could write \begin{eqnarray*} x_n & = & \left( 1 - cL \right) e_n \end{eqnarray*} Inverting the $\left( 1 - cL \right)$ algebraically \begin{eqnarray*} \left( 1 - cL \right)^{- 1} & := & \sum_{k = 0}^{\infty} c^k L^k \end{eqnarray*} you could write \begin{eqnarray*} \left( 1 - cL \right)^{- 1} x_n & = & x_n + \sum_{k = 1}^{\infty} c^k x_{n - k} \end{eqnarray*} yielding that the process is not markov because of the AR($\infty$) representation \begin{eqnarray*} x_n & = & e_n - \sum_{k = 1}^{\infty} c^k x_{n - k} \end{eqnarray*}

-
Yep. I almost wrote an answer along this line but finally opted for the presentation in mine because the conclusion is not so easy to justify: why this representation of $(x_n)$ implies that $(x_n)$ is not Markov? One should show that there is no miraculous simplification, nor an altogether different representation, making it Markov nevertheless. – Did Mar 23 '13 at 13:30

The process $(x_n)_n$ is not Markov.

To compute the distribution of $x_n$ conditionally on $\mathcal G_{n-1}=\sigma(x_{n-1})$, one proceed as follows. First, note that $x_n=e_n+cx_{n-1}-c^2e_{n-2}$, where $x_{n-1}$ is $\mathcal G_{n-1}$ measurable and $e_n$ is independent of $\mathcal G_{n-1}\vee\sigma(e_{n-2})$. To deal with the $e_{n-2}$ part, note that $(e_{n-2},x_{n-1})$ is gaussian hence $e_{n-2}=\alpha x_{n-1}+\beta y_{n-1}$ for some $\alpha$ and $\beta$ and some gaussian random variable $y_{n-1}$ independent of $\mathcal G_{n-1}$. To identify $\alpha$, $\beta$ and $y_{n-1}$, note that $x_{n-1}=e_{n-1}+ce_{n-2}$ and $y_{n-1}=ce_{n-1}-e_{n-2}$ are independent and that $e_{n-2}=\alpha x_{n-1}+\beta y_{n-1}$ for $\alpha=c/1+c^2$ and $\beta=-1/(1+c^2)$. Thus, conditionally on $\mathcal G_{n-1}$, $e_{n-2}$ is gaussian with mean $\alpha x_{n-1}$ and variance $\beta^2\mathrm{var}(y_{n-1})=1/(1+c^2)$.

Finally, the decomposition $x_n=(c-c^2\alpha)x_{n-1}+(e_n-c^2\beta y_{n-1})$ indicates that $x_n=(c/(1+c^2))x_{n-1}+z_n$ where $z_n$ is gaussian, independent of $x_{n-1}$, centered with variance $\sigma^2=1+c^4\beta^2\mathrm{var}(y_{n-1})=(1+c^2+c^4)/(1+c^2)$. Thus, the distribution of $x_n$ conditionally on $\mathcal G_{n-1}$ is gaussian with mean $(c/(1+c^2))x_{n-1}$ and variance $\sigma^2$.

To compute the distribution of $x_n$ conditionally on $\mathcal H_{n-1}=\sigma(x_{n-1},x_{n-2})$, one proceeds as above. The result is that $x_n=\gamma x_{n-1}+\delta x_{n-2}+t_n$ for some $\gamma$ and $\delta$ and some gaussian random variable $t_n$ independent of $\mathcal H_{n-1}$. Thus, the distribution of $x_n$ conditionally on $\mathcal H_{n-1}$ is gaussian with mean $\gamma x_{n-1}+\delta x_{n-2}$ and variance $\tau^2=1+c^2-\gamma^2-\delta^2-2c\gamma\delta$. Since $\delta\ne0$, the distributions of $x_n$ conditionally on $\mathcal G_{n-1}$ and $\mathcal H_{n-1}$ differ hence $(x_n)_n$ is not a Markov process.

The argument can be adapted to show that, for each $k\geqslant1$, $(x_n)$ is not Markov with memory $k$ either.

-