According to this question and answer Explosive AR(MA) processes are stationary? the AR(1) process (with $e_t$ white noise):
$$X_{t}=\varphi X_{t-1}+e_{t} \qquad , e_t \sim WN(0,\sigma)$$
is a stationary process if $\varphi>1$ because it can be rewritten as
$$X_t=\sum_{k=0}^\infty {\varphi}^{-k}u_{t+k}$$
But now the variable $X_t$ depends on the future.
I wonder where this representation (which I remember having seen in several places) and the derivation originally comes from.
I am confused about the derivation, and I wonder how it works. When I try to do the derivation myself I am failing.
I can rewrite the process $$X_{t+1}=\varphi X_{t}+e_{t+1}$$ as
$$X_{t}= \varphi^{-1} X_{t+1} -\varphi^{-1} e_{t+1}$$
and replacing $\varphi^{-1} e_{t+1}$ by $u_{t}$ it becomes
$$X_{t}= \varphi^{-1} X_{t+1} + u_{t}$$
such that the expression is 'like' another AR(1) process but in reverse time and now the coefficient is below 1 so it seemingly is stationary (*).
From the above it would follow indeed that $$X_t=\sum_{k=0}^\infty {\varphi}^{-k}u_{t+k}$$
(*) But the $u_t$ is not independent from $X_{t+1}$, because it is actually $e_{t+1}$ times a negative constant.
Best Answer
The question suggests some basic confusion between the equation and the solution
The Equation
Let ${\varphi} > 1$. Consider the following (infinite) system of equations---one equation for each $t\in \mathbb{Z}$: $$ X_{t}=\varphi X_{t-1}+e_{t}, \mbox{ where } e_t \sim WN(0,\sigma), \;\; t \in \mathbb{Z}. \quad (*) $$
Definition Given $e_t \sim WN(0,\sigma)$, a sequence of random variables $\{ X_t \}_{t\in \mathbb{Z}}$ is said to be a solution of $(*)$ if, for each $t$, $$ X_{t}=\varphi X_{t-1}+e_{t}, $$ with probability 1.
The Solution
Define $$ X_t= - \sum_{k=1}^\infty {\varphi}^{-k}e_{t+k}, $$ for each $t$.
$X_t$ is well-defined: The sequence of partial sums $$ X_{t,m} = - \sum_{k=1}^m {\varphi}^{-k}e_{t+k}, \;\; m \geq 1 $$ is a Cauchy sequence in the Hilbert space $L^2$, and therefore converges in $L^2$. $L^2$ convergence implies convergence in probability (although not necessarily almost surely). By definition, for each $t$, $X_t$ is the $L^2$/probability-limit of $(X_{t,m})$ as $m \rightarrow \infty$.
$\{ X_t \}$ is, trivially, weakly stationary. (Any MA$(\infty)$ series with absolutely summable coefficients is weakly stationary.)
$\{ X_t \}_{t\in \mathbb{Z}}$ is a solution of $(*)$, as can be verified directly by substitution into $(*)$.
This is a special case of how one would obtain a solution to an ARMA model: first guess/derive an MA$(\infty)$ expression, show that it is well-defined, then verify it's an actual solution.
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This impression perhaps results from confusing the equation and the solution. Consider the actual solution: $$ \varphi X_{t-1} + e_t = \varphi \cdot \left( - \sum_{k=1}^\infty {\varphi}^{-k}e_{t+k-1} \right) + e_t, $$ the right-hand side is exactly $- \sum_{k=1}^\infty {\varphi}^{-k}e_{t+k}$, which is $X_t$ (we just verified Point #3 above). Notice how $e_t$ cancels and actually doesn't show up in $X_t$.
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I believe Mann and Wald (1943) already considered non-causal AR(1) case, among other examples. Perhaps one can find references even earlier. Certainly by the time of Box and Jenkins this is well-known.
Further Comment
The non-causal solution is typically excluded from the stationary AR(1) model because:
It is un-physical.
Assume that $(e_t)$ is, say, Gaussian white noise. Then, for every non-causal solution, there exists a causal solution that is observationally equivalent, i.e. the two solutions would be equal as probability measures on $\mathbb{R}^{\mathbb{Z}}$. In other words, a stationary AR(1) model that includes both causal and non-causal cases is un-indentified. Even if the non-causal solution is physical, one cannot distinguish it from a causal counterpart from data. For example, if innovation variance $\sigma^2 =1$, then the causal counterpart is causal solution to AR(1) equation with coefficient $\frac{1}{\varphi}$ and $\sigma^2 =\frac{1}{\varphi^2}$.