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Let us have an AR(1) model with individual efect $$y_t = \alpha + \theta y_{t-1} + \varepsilon_t$$

with $|\theta|<1$ for stacionarity and $\varepsilon_i$ i.i.d. from distribution with mean $0$ and variance $\sigma^2$.

As $\varepsilon_{t}$ and $y_{t-1}$ are independent and $\mathbb{E}\, \varepsilon_{t}=0$ I would like to show that $$\frac{1}{T} \sum_{t=1}^T \varepsilon_{t}y_{t-1} \xrightarrow[T\rightarrow \infty]{P} 0.$$ For this I cannot use strong law of large numbers as individual terms are dependent (for example with t=5 and t=4 $\varepsilon_5 y_4$ and $\varepsilon_4 y_3$ are not independent).

Is there another way to show this?

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Show that the variance of the mean goes to zero

$$\mathrm{var}\left[\frac{1}{T}\sum_{t=1}^T \epsilon_ty_t{-1} \right]=\frac{1}{T^2}\sum_{s,t=1}^T\mathrm{cov}\left[\epsilon_ty_{t-1},\epsilon_s y_{s-1} \right]$$

The covariances are going decrease exponentially in $|t-s|$ (for large $|t-s|$), so the sum will be $O(T)$ for any fixed $\theta$ and the scaling by $T^2$ makes it go to zero. Convergence in mean square implies convergence in probability by Chebyshev's inequality.

(This is a fairly general approach to the easier LLNs for correlated variables: some of the correlations will be large (but obviously no larger than 1), the rest will be small. Count the number of large ones and bound the sum of the small ones, then invoke Chebyshev.)

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