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In Time Series Analysis there is this idea of choosing the distribution of the starting point of a time series in a way such that the time series is stationary. Let for example $\{X_t\}_{t=0}^T%$ be a time series following an AR(1) process ($X_t = \phi X_{t-1} + \varepsilon_t, \ \varepsilon_t \sim wn(0, \sigma^2)$). The Wold(-like) decomposition looks like this: \begin{align} X_t = X_0 + \sum_{i = 0}^{t-1} \phi^{i} \varepsilon_{t-i} \end{align} Relevant for the stationarity is e.g. the variance which is given as \begin{align} Var(X_t) = Var(X_0) + \sigma^2 \sum_{i = 0}^{t-1} \phi^{2i} = Var(X_0) + \sigma^2 \frac{1 - (\phi^2)^{t}}{1 - \phi^2}. \end{align} Now in order for $X_t$ to be stationary, we have to choose $X_0$ in such a way that \begin{align} Var(X_0) = \sigma^2 \sum_{j = t}^{\infty} \phi^{2j} \end{align} Because then we have $Var(X_t) = \frac{\sigma^2}{1-\phi^2}$ which is independent from $t$.
My question is, what is the technical term for choosing $X_0$ in such a way. I only know the german term which is "eingeschwungene Lösung".

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I only know the german term which is "eingeschwungene Lösung".

In English one often uses the stationary distribution.

It may also refer to limiting distribution and possibly in other contexts it could be used as steady state, but that is another (non-statistics) story.

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  • $\begingroup$ Thanks! I think "stationary distribution" comes closest since the AR model can be seen as a Markov chain. $\endgroup$
    – Red
    Mar 29, 2023 at 8:48
  • $\begingroup$ @Red yes stationary distribution comes closest and that's why I had set it appart. Btw. it can also be considered as the 'marginal distribution', in the viewpoint that the time series follows a multivariate distribution. $\endgroup$ Mar 29, 2023 at 8:57

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