Why representation of AR process comes up in estimation Let ${X_t}$, $t=...-2,-1,0,1,2...$ be a stochastic process that satisfies:
$X_t=\rho X_{t-1}+\varepsilon_t$
with $|\rho|<1$ and $\varepsilon_t$ is a white noise. In that case, we also know that there is a $\beta$ such that $|\beta|>1$ and another white noise $\eta_t$ such that ${X_t}$ also satisfies:
$X_t=\beta X_{t-1}+\eta_t$
Suppose I have some observations of $X_t$. If I run OLS by data generated by those equivalent process, which one should I expect? Why?
 A: The property you invoke relates to $MA$ processes, not $AR$ processes. To stick with the $MA(1)$ case, for every invertible $MA(1)$ representation of data
$$X_t = \mu + u_t + \theta u_{t-1}, \;\; |\theta|<1,\;\;  E(u_t^2) = \sigma^2$$
there exists an equivalent non-invertible $MA(1)$ representation,
$$X_t = \mu + w_t + \hat \theta w_{t-1}, \;\; |\hat \theta|=\frac 1{|\theta|},\;\;  E(w_t^2) = \theta^2\sigma^2$$
They are equivalent in the sense that they have identical autocovariance-generating functions, and can describe the same data equally well (if the one does, the other does also)
In theory, we could attempt to estimate either model. To estimate the invertible representation, we could turn it into the equivalent $AR(\infty)$ representation and so use for the estimation, past values of $X$. In order to estimate the non-invertible representation, we would need future values of $X$. Moreover, various algorithms for estimation and forecasting  work only if the invertible representation is used.  
Given the above, what do you think you would get?
A: Following what you write, you have $\rho X_{t-1}+\epsilon_t = \beta X_{t-1}+\eta_t$. If you fix a value of $\rho$, then $\beta = \frac{\rho X_{t-1}+\epsilon_t-\eta_t}{X_{t-1}}$. Your $\beta$ dependes on $X_t$ and then for each time $t$, $\beta$ must assumes a different value.
Your relation is true if you have one $\beta$ for each $t$ and $\rho X_{t-1}+\epsilon_t = \beta_t X_{t-1}+\eta_t$. 
