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Hot answers tagged moving-average

How does ACF & PACF identify the order of MA and AR terms?

The quotes are from the link in the OP: Identification of an AR model is often best done with the PACF. For an AR model, the theoretical PACF “shuts off” past the order of the model. The phrase “...
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Why do we care if an MA process is invertible?

Invertibility is not really a big deal because almost any Gaussian, non-invertible MA$(q)$ model can be changed to an invertible MA$(q)$ model representing the same process by changing the parameter ...
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Real-life examples of moving average processes

Suppose you are producing some good, stockpiling some of it and selling the rest. Your production in time period $t$ is $x_t=m+\varepsilon_t$ with $\mathbb{E}(\varepsilon_t)=0$ and your stock is $y_t$....
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Under what circumstances is an MA process or AR process appropriate?

I can provide what I think is a compelling answer to the first part of the question ("whence MA?") but am presently pondering an equally compelling answer to the second part of the question ("whence ...
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Moving Average, Exponential Smoothing, and Random Walk for Forecasting

Is it true that a (simple) exponential smoothing model with alpha (smoothing constant) = 1 is the same as MA(1), which is in turn the same as a random walk model? (i.e. using only the most recent ...
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Moving average process - stationarity

First see what definitions say $\{X_t\}$ is strictly stationary if for any $t_1,t_2,...,t_n \in T$ and any $k \in T$ $$P(X_{t_1},...,X_{t_n}) = P(X_{t_1+k},...,X_{t_n+k})$$ that is, we have ...
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Is there a name for a moving average when it is done not across time but some other variable?

Terminology can differ between fields even apparently sharing applications. Based on statistical theory and practice in several fields (time series, spatial series, any application where a response ...
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Do non-invertible MA models imply that the effect of past observations increases with the distance?

Not a big deal - it is strongly stationary and approaches white noise The non-invertible $\text{MA}(1)$ process makes perfect sense, and it does not exhibit any particularly strange behaviour. Taking ...
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Is the MA($\infty$) process with i.i.d. noise strictly stationary?

This process is always strictly stationary by definition. Recall that the process is (strictly) stationary when all $n$-variate distributions formed by selecting any pattern $(s_1,s_2,\ldots,s_{n-1})$ ...
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Any difference between AR(1) model and MA(1) model in practice?

Welcome here! AR(1) and MA(1) use different input values and will not give the same results. AR(1) models $y_t = a_1 \cdot y_{t-1} + \epsilon_t$ indeed use a lagged variable of the outcome. This model ...
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Is the estimation of MA models unique?

Thus, for MA model, does the estimation result not highly depend on what $k$ is selected? For large $k$ we have better statistical properties, but less data for the estimation... is this correct? ...
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Is it good practice to use Linear Least-Squares with SMA?

Of course you can do a fit on a moving average. That is your right. But the statistical diagnostics are not reliable anymore. The reason is that the IID property required in standard OLS are violated ...
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Do non-invertible MA models imply that the effect of past observations increases with the distance?

I don't think it makes sense to ask for an example "from the real world where they [non-invertible MA models] occur". All you observe is $y_1,y_2,\dots,y_n$. As I try to explain in the post you link ...
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What to do if ACF or PACF show significant higher lags?

[I believe this is a duplicate - and while I can find questions with this issue explained in comments, the couple that explain it correctly and fully in answers aren't really answering the same ...
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MA(1) model - prove that correlation of order two equals to zero

Given the covariance of order two, all the terms are uncorrelated ($u_{s}$ is a white process, hence there is no correlation), therefore the covariance, respective correlation, is zero  Cov(Y_{t},Y_{...
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Proposition. Let $y_t$ be a series from $MA(q)$. Then for any $q$, there is $s$ such that $Cov(y_t,y_{t-s})=0$. Obviously, the above proposition is true. But there is no such $s$ for $AR(1)$ process ...