Explaining the fit of Correlation and Covariance in AR and MA models 
The primary difference between an AR and MA model is based on the correlation between time series objects at different time points. The correlation between $x(t)$ and $x(t-n)$ for $n > \text{order of MA}$ is always zero. This directly flows from the fact that covariance between $x(t)$ and $x(t-n)$ is zero for MA models (something which we refer from the example taken in the previous section). However, the correlation of $x(t)$ and $x(t-n)$ gradually declines with $n$ becoming larger in the AR model. This difference gets exploited irrespective of having the AR model or MA model.

Source: https://www.analyticsvidhya.com/blog/2015/12/complete-tutorial-time-series-modeling/
I just cannot see how to derive the covariance between $x(t)$ and $x(t-n)$ for MA models.
Moreover, I cannot see how the correlation of $x(t)$ and $x(t-n)$ would declines as $n$ increases in the AR model.
 A: For simplicity, take an $\text{AR}(1)$ and a $\text{MA}(1)$ model, written below (in order):
$$X_t = \frac{1}{2}X_{t - 1} + \epsilon_t$$
$$W_t = \epsilon_t + 0.5\epsilon_{t - 1}$$
For the $\text{AR}(1)$ process we have the representation $X_t = \left(\frac{1}{2}\right)^h X_{t -
 h} + \sum_{j = 0}^{h - 1} \left(\frac{1}{2}\right)^j \epsilon_{t - j}$ (if you are uncomfortable with this, recursively replace $X_{t - j}$ terms in the above definition of the $\text{AR}(1)$ process until you eventually have $\epsilon_{t - n}$ in the expression). From this let's compute $E[X_{t + n}X_t]$ and $E[W_{t + n}W_t]$.
Remember that the only thing that is random in these representations is the process $\{\epsilon_t\}_{t \in \mathbb{Z}}$, and $E[\epsilon_t^2] = 1$, and all are iid. From this we get:
$$E[W_{t + n} W_t] = E[(\epsilon_{t + n} + 0.5 \epsilon_{t + n - 1})(\epsilon_t + 0.5\epsilon_{t - 1}) \\ = E[\epsilon_{t + n}\epsilon_t] + 0.5E[\epsilon_{t + n}\epsilon_{t - 1}] + 0.5E[\epsilon_{t + n - 1}\epsilon_t] + 0.25E[\epsilon_{t + n - 1}\epsilon_{t - 1}] \\
 = 1.25\delta_{0}(n) + 0.5 \left(\delta_{-1}(n) + \delta_{1}(n)\right)$$
($\delta_{j}(n)$ is 1 if $n = j$ and 0 otherwise.) From this you see that for large $n$ the covariance, and thus the correlation, is zero between $W_t$ terms.
Now let's find the covariance function for $X_t$.
$$E[X_{t + n}X_t] = E\left[\left(\left(\frac{1}{2}\right)^n X_{t} + \sum_{j = 0}^{n - 1} \left(\frac{1}{2}\right)^j \epsilon_{t + n - j}\right)X_t\right] \\ = \left(\frac{1}{2}\right)^nE[X_t^2] + \sum_{j = 0}^{n - 1}\left(\frac{1}{2}\right)^j E\left[\epsilon_{t + n - j} X_t\right]$$
If you are willing to believe that $E[X_t^2] = \frac{4}{3}$ and $E\left[\epsilon_{t + n - j} X_t\right] = 0$ (in this situation; basically, the two random variables are independent when $j \neq n$) then the covariance is $\left(\frac{1}{2}\right)^n \frac{4}{3}$, a quantity that goes to zero with large $n$. (The last two claims can be justified.)
What I described works for more general AR and MA models, but those models require more work; this gives you the gist of what is going on.
