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To demonstrate consistency we must show, as always, that the sampling error $\hat\beta-\beta$ goes to zero in probability. Recall $$ \hat\beta=\frac{\sum_tx_{t-1}y_t}{\sum_tx_{t-1}^2} $$ Plugging in for $y_t$ gives, after some manipulation, $$ \hat\beta-\beta=\frac{\sum_tx_{t-1}\nu_t}{\sum_tx_{t-1}^2}=\frac{\sum_tx_{t-1}^3v_t}{\sum_tx_{t-1}^2} $$ It shall prove useful to rewrite this as $$ \hat\beta-\beta=\frac{\frac{1}{T^2}\sum_tx_{t-1}^3v_t}{\frac{1}{T^2}\sum_tx_{t-1}^2} $$ From, e.g., Theorem 4.2 in Ibragimov and Phillips (2008), we may deduce that $$ \frac{1}{T^2}\sum_tx_{t-1}^3v_t\to_d\int_0^1W_u(r)^3dW_v(r) $$ Here $W_u(r)$ is the Brownian motion associated with $u_t$ and $W_v(r)$ that associated with $v_t$. That is $1/\sqrt{T}\sum_{t=1}^{[Tr]}u_t\to_dW_u(r)$.

It is a fairly standard result that $$ \frac{1}{T^2}\sum_tx_{t-1}^2\to_d\int_0^1W_u(r)^2dr $$ Hence, $$ \hat\beta-\beta\to_d\frac{\int_0^1W_u(r)^3dW_v(r)}{\int_0^1W_u(r)^2dr} $$ which is some nonstandard random variable, and hence $O_p(1)$. Thus, $\hat\beta-\beta$ is not $o_p(1)$ and hence does not converge to zero in probability.

As for the use of the LIE, recall that with nonstationary time series, expected values often do not exist, in which case the LIE cannot be applied.

To demonstrate consistency we must show, as always, that the sampling error $\hat\beta-\beta$ goes to zero in probability. Recall $$ \hat\beta=\frac{\sum_tx_{t-1}y_t}{\sum_tx_{t-1}^2} $$ Plugging in for $y_t$ gives, after some manipulation, $$ \hat\beta-\beta=\frac{\sum_tx_{t-1}\nu_t}{\sum_tx_{t-1}^2}=\frac{\sum_tx_{t-1}^3v_t}{\sum_tx_{t-1}^2} $$ It shall prove useful to rewrite this as $$ \hat\beta-\beta=\frac{\frac{1}{T^2}\sum_tx_{t-1}^3v_t}{\frac{1}{T^2}\sum_tx_{t-1}^2} $$ From, e.g., Theorem 4.2 in Ibragimov and Phillips (2008), we may deduce that $$ \frac{1}{T^2}\sum_tx_{t-1}^3v_t\to_d\int_0^1W_u(r)^3dW_v(r) $$ Here $W_u(r)$ is the Brownian motion associated with $u_t$ and $W_v(r)$ that associated with $v_t$. That is $1/\sqrt{T}\sum_{t=1}^{[Tr]}u_t\to_dW_u(r)$.

It is a fairly standard result that $$ \frac{1}{T^2}\sum_tx_{t-1}^2\to_d\int_0^1W_u(r)^2dr $$ Hence, $$ \hat\beta-\beta\to_d\frac{\int_0^1W_u(r)^3dW_v(r)}{\int_0^1W_u(r)^2dr} $$ which is some nonstandard random variable, and hence $O_p(1)$. Thus, $\hat\beta-\beta$ is not $o_p(1)$ and hence does not converge to zero in probability.

As for the use of the LIE, recall that with nonstationary time series, expected values often do not exist, in which case the LIE cannot be applied.

To demonstrate consistency we must show, as always, that the sampling error $\hat\beta-\beta$ goes to zero in probability. Recall $$ \hat\beta=\frac{\sum_tx_ty_t}{\sum_tx_t^2} $$ Plugging in for $y_t$ gives, after some manipulation, $$ \hat\beta-\beta=\frac{\sum_tx_t\nu_t}{\sum_tx_t^2}=\frac{\sum_tx_t^3v_t}{\sum_tx_t^2} $$ It shall prove useful to rewrite this as $$ \hat\beta-\beta=\frac{\frac{1}{T^2}\sum_tx_t^3v_t}{\frac{1}{T^2}\sum_tx_t^2} $$ From, e.g., Theorem 4.2 in Ibragimov and Phillips (2008), we may deduce that $$ \frac{1}{T^2}\sum_tx_t^3v_t\to_d\int_0^1W_u(r)^3dW_v(r) $$ Here $W_u(r)$ is the Brownian motion associated with $u_t$ and $W_v(r)$ that associated with $v_t$. That is $1/\sqrt{T}\sum_{t=1}^{[Tr]}u_t\to_dW_u(r)$.

It is a fairly standard result that $$ \frac{1}{T^2}\sum_tx_t^2\to_d\int_0^1W_u(r)^2dr $$ Hence, $$ \hat\beta-\beta\to_d\frac{\int_0^1W_u(r)^3dW_v(r)}{\int_0^1W_u(r)^2dr} $$ which is some nonstandard random variable, and hence $O_p(1)$. Thus, $\hat\beta-\beta$ is not $o_p(1)$ and hence does not converge to zero in probability.

As for the use of the LIE, recall that with nonstationary time series, expected values often do not exist, in which case the LIE cannot be applied.

To demonstrate consistency we must show, as always, that the sampling error $\hat\beta-\beta$ goes to zero in probability. Recall $$ \hat\beta=\frac{\sum_tx_{t-1}y_t}{\sum_tx_{t-1}^2} $$ Plugging in for $y_t$ gives, after some manipulation, $$ \hat\beta-\beta=\frac{\sum_tx_{t-1}\nu_t}{\sum_tx_{t-1}^2}=\frac{\sum_tx_{t-1}^3v_t}{\sum_tx_{t-1}^2} $$ It shall prove useful to rewrite this as $$ \hat\beta-\beta=\frac{\frac{1}{T^2}\sum_tx_{t-1}^3v_t}{\frac{1}{T^2}\sum_tx_{t-1}^2} $$ From, e.g., Theorem 4.2 in Ibragimov and Phillips (2008), we may deduce that $$ \frac{1}{T^2}\sum_tx_{t-1}^3v_t\to_d\int_0^1W_u(r)^3dW_v(r) $$ Here $W_u(r)$ is the Brownian motion associated with $u_t$ and $W_v(r)$ that associated with $v_t$. That is $1/\sqrt{T}\sum_{t=1}^{[Tr]}u_t\to_dW_u(r)$.

It is a fairly standard result that $$ \frac{1}{T^2}\sum_tx_{t-1}^2\to_d\int_0^1W_u(r)^2dr $$ Hence, $$ \hat\beta-\beta\to_d\frac{\int_0^1W_u(r)^3dW_v(r)}{\int_0^1W_u(r)^2dr} $$ which is some nonstandard random variable, and hence $O_p(1)$. Thus, $\hat\beta-\beta$ is not $o_p(1)$ and hence does not converge to zero in probability.

As for the use of the LIE, recall that with nonstationary time series, expected values often do not exist, in which case the LIE cannot be applied.

To demonstrate consistency we must show, as always, that the sampling error $\hat\beta-\beta$ goes to zero in probability. Recall $$ \hat\beta=\frac{\sum_tx_ty_t}{\sum_tx_t^2} $$ Plugging in for $y_t$ gives, after some manipulation, $$ \hat\beta-\beta=\frac{\sum_tx_t\nu_t}{\sum_tx_t^2}=\frac{\sum_tx_t^3v_t}{\sum_tx_t^2} $$ It shall prove useful to rewrite this as $$ \hat\beta-\beta=\frac{\frac{1}{T^2}\sum_tx_t^3v_t}{\frac{1}{T^2}\sum_tx_t^2} $$ From, e.g., Theorem 4.2 in Ibragimov and Phillips (2008), we may deduce that $$ \frac{1}{T^2}\sum_tx_t^3v_t\to_d\int_0^1W_u(r)^3dW_v(r) $$ Here $W_u(r)$ is the Brownian motion associated with $u_t$ and $W_v(r)$ that associated with $v_t$. That is $1/\sqrt{T}\sum_{t=1}^{[Tr]}u_t\to_dW_u(r)$.

It is a fairly standard result that $$ \frac{1}{T^2}\sum_tx_t^2\to_d\int_0^1W_u(r)^2dr $$ Hence, $$ \hat\beta-\beta\to_d\frac{\int_0^1W_u(r)^3dW_v(r)}{\int_0^1W_u(r)^2dr} $$ which is some nonstandard random variable, and hence $O_p(1)$. Thus, $\hat\beta-\beta$ is not $o_p(1)$ and hence does not converge to zero in probability.

To demonstrate consistency we must show, as always, that the sampling error $\hat\beta-\beta$ goes to zero in probability. Recall $$ \hat\beta=\frac{\sum_tx_ty_t}{\sum_tx_t^2} $$ Plugging in for $y_t$ gives, after some manipulation, $$ \hat\beta-\beta=\frac{\sum_tx_t\nu_t}{\sum_tx_t^2}=\frac{\sum_tx_t^3v_t}{\sum_tx_t^2} $$ It shall prove useful to rewrite this as $$ \hat\beta-\beta=\frac{\frac{1}{T^2}\sum_tx_t^3v_t}{\frac{1}{T^2}\sum_tx_t^2} $$ From, e.g., Theorem 4.2 in Ibragimov and Phillips (2008), we may deduce that $$ \frac{1}{T^2}\sum_tx_t^3v_t\to_d\int_0^1W_u(r)^3dW_v(r) $$ Here $W_u(r)$ is the Brownian motion associated with $u_t$ and $W_v(r)$ that associated with $v_t$. That is $1/\sqrt{T}\sum_{t=1}^{[Tr]}u_t\to_dW_u(r)$.

It is a fairly standard result that $$ \frac{1}{T^2}\sum_tx_t^2\to_d\int_0^1W_u(r)^2dr $$ Hence, $$ \hat\beta-\beta\to_d\frac{\int_0^1W_u(r)^3dW_v(r)}{\int_0^1W_u(r)^2dr} $$ which is some nonstandard random variable, and hence $O_p(1)$. Thus, $\hat\beta-\beta$ is not $o_p(1)$ and hence does not converge to zero in probability.

As for the use of the LIE, recall that with nonstationary time series, expected values often do not exist, in which case the LIE cannot be applied.