# How to show that an MA(2) process is strictly stationary?

I have a question about MA(2) model $$X_t = e_t + 0.5e_{t−1} + 0.4e_{t−2}$$ with $$e_t \sim IID(0, σ^2_e$$).

I know by construction, MA process is (weakly) stationary, but can it be strictly stationary given IID?

To show it is strictly stationary, I get the $$E(X_t)$$ and $$E(X_{t-5})$$ and $$V(X_t)$$ and $$V(X_{t-5})$$ and found that both are the same ($$\text{mean}=0$$, and $$\text{var}=1.41\sigma^2$$).

So with that, is the process time-invariant? However, I know that the variance is finite so it implies WSS. I do not get a grab on this concept. I do confuse with IID that assumed the process to be strictly stationary.

• Hi: stricly stationary means that if you look at observations at time t and calculate the joint distribution of say $n$ data points at that time and then look at the data at time $t+h$ and calculate the joint distribution of $n$ data points at that time, then they are the same. So, I think you are doing that sort of but you should check three data points say $x_1$, $x_2$, $x_3$ versus say $x_5$, $x_6$, $x_7$ since you have an MA(2). I say two because the MA(2) process is not correlated after lag 2 so there's really nothing to check except 3 data points. Commented Apr 7, 2021 at 15:25

There's a deeper and far more general result lurking here: a moving window operation on any strictly stationary process produces a strictly stationary process.

The demonstration is just a matter of applying definitions. I'll take you through it step by step, emphasizing the simplicity and--I hope--the obviousness of each one.

Let's begin with this "initial statement" of a trivial proposition. It will be applied repeatedly below.

Let the $$p$$-variate random variables $$\mathbf X = (X_1,\ldots, X_p)$$ and $$\mathbf Y = (Y_1,\ldots, Y_p)$$ have the same distribution and let $$f:\mathbb{R}^p\to \mathbb R^q$$ be any (measurable) function. Then $$f(\mathbf X)$$ and $$f(\mathbf Y)$$ are identically distributed.

Proof: the distributions of the results of applying $$f$$ depend only the the distributions of its arguments, which are identical, QED. Note that the resulting identical distributions are $$q$$-variate and we're talking about the full joint distribution, not just the marginal distributions of their components.

Suppose, then, that $$\mathbf X = (\ldots, X_{-1}, X_0, X_1, \ldots)$$ is a time series process. As a matter of notation, whenever $$T$$ is a finite sequence $$(t_1,\ldots, t_p)$$ of integers, define

• $$\mathbf{X}_T = (X_{t_1}, X_{t_2}, \ldots, X_{t_p})$$ to be the sequence of components of the process picked out by the indexes in $$T$$ and

• For any integer $$h,$$ let $$T + h = (t_1+h, t_2+h, \ldots, t_p+h)$$ be the translate of the indexes in $$T.$$

A process is strictly stationary when, for any finite sequence $$T$$ and any integer $$h,$$ $$\mathbf{X}_T$$ and $$\mathbf{X}_{T+h}$$ have the same distribution.

This is the definition. In words, it says all finite-dimensional marginal distributions of the process (namely, all $$\mathbf{X}_T$$) are time-invariant (namely, they do not change when translated by any lag $$h$$).

Take any $$f$$ as in the initial statement. For any finite sequence $$T$$ and integer $$h,$$ it follows that $$f(\mathbf{X}_T)$$ and $$f(\mathbf{X}_{T+h})$$ are identically distributed.

Let's apply these trivia to the situation in the question. Let $$g:\mathbb{R}^r\to\mathbb{R}$$ be a (measurable) function. Define a new time series process from the process $$\mathbf X$$ via

$$Y_t = g(X_{t-r+1}, X_{t-r+2}, \ldots, X_t).$$

This is the "moving window" application of $$g$$ to windows of length $$r.$$ We need to convince ourselves that when $$\mathbf{X}$$ is strictly stationary, so is $$\mathbf Y.$$

To show this, we must consider an arbitrary finite index set $$S=(s_1,s_2,\ldots, s_q)$$ and arbitrary integer $$h$$ and compare the distributions of $$\mathbf{Y}_S$$ and $$\mathbf{Y}_{S+h}.$$ Notice that, by construction,

$$\mathbf{Y}_S = (Y_{s_1}, Y_{s_2}, \ldots, Y_{s_q}) = (g(X_{s_1-r+1}, \ldots, X_{s_1}), \ldots, g(X_{s_q-r+1}, \ldots, X_{s_q})).$$

The right hand side can be expressed as the image of a function $$f:\mathbb{R}^{rq}\to \mathbb{R}^{q}$$ defined by

$$f(x_1, x_2, \ldots, x_{rq}) = (g(x_1,\ldots, x_q), g(x_{q+1}, \ldots, x_{2q}), \ldots, g(x_{(r-1)q+1}, \ldots, x_{rq})).$$

The initial statement (with $$p=rq$$) applies because this $$f$$ is (obviously) measurable. The relevant index set is

$$T = (s_1-r+1, s_1-r+2, \ldots s_1,\quad s_2-r+1,s_2-r+2,\ldots, s_2,\quad \ldots\\\quad s_q-r+1, s_q-r+2, \ldots, s_q).$$

(Many of these indexes might be the same. I never insisted the indexes must be unique!)

Because $$\mathbf X$$ is strictly stationary, $$\mathbf{X}_T$$ and $$\mathbf{X}_{T+h}$$ have the same $$rq$$-variate distributions. But--applying the initial statement once more--this means $$\mathbf{Y}_S = f(\mathbf{X}_T)$$ and $$\mathbf{Y}_{S+h}= f(\mathbf{X}_{T+h})$$ have the same $$q$$-variate distributions, proving $$\mathbf{Y}$$ is strictly stationary.

The question itself concerns the window function $$g:\mathbb{R}^3\to\mathbb{R}^1,$$ $$g(x_1,x_2,x_3) = 0.4x_1 + 0.5x_2 + x_3,$$ as applied to an iid sequence of variables $$(e_t),$$ which is clearly strictly stationary. Because this (linear) map $$g$$ is measurable, we're done.