# A simple question regarding AR(1) process and CDFs

Somewhat a trivial question, but I struggle to get my head around it. Consider we have an AR(1) process, as follows:

$$y_t=\rho y_{t-1}+\varepsilon_t,\quad t=1,...,T$$.

such that $$\varepsilon_t$$ are $$i.i.d$$ and $$\varepsilon_t\sim N(0,1)$$. Evidently, $$y_t$$ and $$y_{t-1}$$ are correlated and as such a joint probability, such as $$P[y_t\leq0,y_{t-1}\leq 0]\neq P[y_{t}\leq 0]P[y_{t-1}\leq0]$$.

But given that $$y_t=\rho y_{t-1}+\varepsilon_t$$ and $$y_{t-1}=\rho y_{t-2}+\varepsilon_{t-1}$$, the aforementioned joint probability can be manipulated and expressed as follows: $$\begin{eqnarray} P[y_t\leq0,y_{t-1}\leq 0]&=&P[\rho y_{t-1}+\varepsilon_t\leq 0,\rho y_{t-2}+\varepsilon_{t-1}\leq 0]\\ &=&P[\varepsilon_t\leq-\rho y_{t-1},\varepsilon_{t-1}\leq-\rho y_{t-2}] \end{eqnarray}$$ But from earlier, we know that $$\varepsilon_1,...,\varepsilon_t$$ are $$i.i.d$$ implying that $$\begin{eqnarray} P[\varepsilon_t\leq-\rho y_{t-1},\varepsilon_{t-1}\leq-\rho y_{t-2}]&=&P[\varepsilon_t\leq-\rho y_{t-1}]P[\varepsilon_{t-1}\leq-\rho y_{t-2}]\\ &=&P[y_t\leq0][y_{t-1}\leq0] \end{eqnarray}$$ which is obviously in contradiction with the earlier results. I know I am missing a step here, so some clarification would be very much appreciated.

• The conclusion after "implying that" appears to assume $y_{t-1}$ and $y_{t-2}$ are independent of each other and of $\varepsilon_t$ and $\varepsilon_{t-1}.$ Is that the case? If not, what is your justification for it? – whuber Nov 15 '19 at 0:17
• I am rather confused. Clearly in and AR(1) process, $y_1$ and $y_2$ are not independent. Hence, as mentioned earlier $P[y_{t-1}\leq 0,y_{t-2}\leq 0]\neq P[y_{t-1}\leq 0]P[y_{t-2}\leq 0]$. However, my point is if the inequalities in the joint probability are expressed in terms of the residuals, and we already know that the residuals are independent, then $P[\varepsilon_{t}\leq -\rho y_{t-1},\varepsilon_{t-1}\leq -\rho y_{t-2}]=P[\varepsilon_{t}\leq -\rho y_{t-1}]P[\varepsilon_{t-1}\leq -\rho y_{t-2}]$. Is that not the case? Or is my question still ambiguous? – Carl Nov 15 '19 at 0:25
• @Carl: in your $\epsilon$ equation, rhs of the inequality is not a constant but a random variable itself, so just because LHS is iid you cannot split it as you have. – Dayne Nov 15 '19 at 2:26
• @Dayne What do you exacly imply by not splitting? So suppose you have expressed $P[y_{t-1}\leq0,y_{t-2}\leq0]$ in terms of residuals as I have. Do you imply that if I were to express the probabilities as bivariate normal CDFs, the covariance matrix should contain non-zero off-diagonal elements (i.e. implying correlation of the residuals)? – Carl Nov 15 '19 at 3:33
• @Carl let me put it this way: let $A, B, C, D$ be four random variables such that $A, B$ are independent and $C, D$ are not. Clearly, $Pr(A<a, B<b)=Pr(A<a)Pr(B<b) \forall a,b \in \mathbb R$. However, do you think the following is also true? $Pr(A<C, B<D)=Pr(A<C)Pr(B<D)$. I think not as now your random variables become $A-C$ and $B-D$ which are not independent. – Dayne Nov 15 '19 at 4:55

Assuming a fixed initial value $$y_0\in\mathbb R$$, we have $$y_t = \rho^t y_0 + \sum_{i=1}^t\rho^{i-1}\varepsilon_t$$ for $$t=1,2,\ldots,T$$. Note that the linear combination of independent normal random variables is again normal, so $$\sum_{i=1}^t\rho^{i-1} \varepsilon_t$$ has $$N\left(0,\frac{1-\rho^{2(t-1)}}{1-\rho^2}\right)$$ distribution. For any $$t\in\{1,2,\ldots,T\}$$ we may compute the joint distribution of $$(y_t,y_{t-1})$$ by \begin{align} \mathbb P(y_t\leqslant u, y_{t-1}\leqslant v) &= \mathbb P\left(\rho^ty_0 + \sum_{i=1}^t \rho^{t-i}\varepsilon_i\leqslant u, \rho^ty_0 + \sum_{i=1}^{t-1} \rho^{t-i}\varepsilon_i \leqslant v\right)\\ &= \mathbb P\left(\sum_{i=1}^t \rho^{t-i}\varepsilon_i\leqslant u-\rho^ty_0, \sum_{i=1}^{t-1} \rho^{t-i}\varepsilon_i\leqslant v-\rho^ty_0, \right)\\ &= \mathbb P\left(\varepsilon_t+ \sum_{i=1}^{t-1} \rho^{t-i}\varepsilon_i\leqslant s-\rho^ty_0, \sum_{i=1}^{t-1} \rho^{t-i}\varepsilon_i\leqslant v-\rho^ty_0 \right). \end{align}