Closed-form expression for h-step forecast for AR(2) process

Question

I'm trying to derive $\mathbb E[y_{t+h} \mid y_t]$ where $y_t$ is an AR(2) process $$ y_t = \phi_1 y_{t-1} + \phi_2y_{t-2} + c + \varepsilon_t $$ As per usual I'm recursively applying $\mathbb E[y_{t+h} \mid y_t] = \mathbb E[\mathbb E[y_{t+h} \mid y_{t + 1}] \mid y_t]$ for $h= 2, 3, \dots$ to see if some pattern emerges, but I'm getting $$ \mathbb E[y_{t+2} \mid y_t] = (\phi_1^2 + \phi_2)y_t + \phi_1\phi_2y_{t-1} + (\phi_1 + 1)c\\ \mathbb E[y_{t+3} \mid y_t] = (\phi_1^3 + 2 \phi_1 \phi_2)y_t + (\phi_1^2 + \phi_2^2)y_{t-1} + (\phi_1^2 + \phi_1 + \phi_2 + 1)c \\ \mathbb E[y_{t+4} \mid y_t] = (\phi^4_1 + 2\phi^2_1 \phi_2 + \phi_1^2 + \phi_2)y_t + (\phi_1^3 + 2 \phi_1 \phi_2^2)y_{t-1} + (\phi_1^3 + 2 \phi_1 \phi_2 + \phi^2_1 + \phi_1 + \phi_2 + 1)c\\ \mathbb E[y_{t+5} \mid y_t] = (\phi^5_1 + 2 \phi^3_1 + 2 \phi^3_1\phi_2 + 2 \phi_1 \phi^2_2 + \phi_1 \phi_2)y_t + \cdots $$ from which it's hard to see any pattern. I expected to perhaps see some binomial-expansion type pattern but I cannot see any. Searches online also yield surprisingly little results. Does anybody have a reference to such a formula if it exists?

Answer 1

I strongly doubt that it's possible to obtain a formula for $\ \mathbb{E}\big[y_{t+h}\,\big|\,y_t\big]\ $ without making some further assumptions. Even if you assume that $\ \epsilon_t\ $ are independent normal variates with mean $\ 0\ $ and common variance $\ \sigma^2\ $ (which I will take for granted in the rest of this answer), that's still not enough to pin down a unique distribution for $\ \big(y_{t+h}\,, y_t\,\big)\ $. If some initial values $\ (y_1,y_0)\ $, say, are assumed to be bivariate normal, and independent of $\ \epsilon_t\ $, then $\ \big(z_{t+h}\,, z_t\,\big)\ $ will be bivariate normally distributed, so you can then use the result that for such variables $$ \mathbb{E}\big[y_{t+h}\,\big|\,y_t\,\big]=\frac{\mathbb{Cov}\big[y_{t+h}\,,y_t\big]\big(y_t-\mathbb{E}\big[y_t\big]\big)}{\mathbb{Var}\big[y_t\big]}+\mathbb{E}\big[y_{t+h}\,\big]\ . $$

To simplify matters, let $\ e_t=\mathbb{E}\big[y_t\big]\ $ and $\ z_t=y_t-e_t\ $. Then $\ \mathbb{E}\big[y_s\,\big|\,y_t\big]=\mathbb{E}\big[z_s\,\big|\,z_t\big]+e_s\ $, $$ e_s=\phi_1e_{s-1}+\phi_2e_{s-2}+c\ , $$ and $$ z_t=\phi_1z_{t-1}+\phi_2z_{t-2}+\epsilon_t $$ Write this as $$ \pmatrix{z_t\\z_{t-1}}=A\pmatrix{z_{t-1}\\z_{t-2}}+\pmatrix{\epsilon_t\\\epsilon_{t-1}} $$ where $\ A=\pmatrix{\phi_1&\phi_2\\1&0}\ $ and $\ \epsilon_1=0\ $. Assume (as will likely be the case) that the polynomial $\ x^2-\phi_1x-\phi_2\ $ has two distinct roots $\ \lambda_1,\lambda_2\ $ (the eigenvalues of $\ A\ $) neither of which is $1$. Then $$ \big(\lambda_1-\lambda_2)^{-1}\pmatrix{1&-\lambda_2\\-1&\lambda_1}A\pmatrix{\lambda_1&\lambda_2\\1&1}=\pmatrix{\lambda_1&0\\0&\lambda_2}\ . $$ Let \begin{align} \pmatrix{v_t\\w_t}&=\big(\lambda_1-\lambda_2)^{-1}\pmatrix{1&-\lambda_2\\-1&\lambda_1}\pmatrix{z_t\\z_{t-1}}\\ \pmatrix{\delta_t\\\gamma_t}&=\big(\lambda_1-\lambda_2)^{-1}\pmatrix{1&-\lambda_2\\-1&\lambda_1}\pmatrix{\epsilon_t\\\epsilon_{t-1}}\ . \end{align} Then \begin{align} v_t&=\lambda_1v_{t-1}+\delta_t\\ w_t&=\lambda_2w_{t-1}+\gamma_t\\ z_t&=\lambda_1v_t+\lambda_2w_t\ , \end{align} from which we obtain \begin{align} v_{t+1}&=\lambda_1^tv_1+\sum_{i=0}^{t-1}\lambda_1^i\delta_{t+1-i}\\ w_{t+1}&=\lambda_2^tw_1+\sum_{i=0}^{t-1}\lambda_2^i\gamma_{t+1-i}\\ z_{t+1}&=\lambda_1v_{t+1}+\lambda_2w_{t+1}\\ &=\lambda_1^{t+1}v_1+\lambda_2^{t+1}w_1+\sum_{i=0}^{t-1}\big(\lambda_1^{i+1}\delta_{t+1-i}+\lambda_2^{i+1}\gamma_{t+1-i}\big)\\ &=\frac{\big(\lambda_1^{t+1}-\lambda_2^{t+1}\big)z_1+\lambda_1\lambda_2\big(\lambda_2^t-\lambda_1^t\big)z_0}{\lambda_1-\lambda_2}\\ &\hspace{1em}+\sum_{j=2}^{t+1}\big(\lambda_1^{t+1-j}+\lambda_2^{t+1-j}\,\big)\epsilon_j\\ &=\frac{\big(\lambda_1^{t+1}-\lambda_2^{t+1}\big)z_1+\lambda_1\lambda_2\big(\lambda_2^t-\lambda_1^t\big)z_0}{\lambda_1-\lambda_2}\\ &\hspace{1em}+\sum_{j=0}^{t-1}\big(\lambda_1^j+\lambda_2^j\,\big)\epsilon_{t+1-j}\ . \end{align} Similarly, we have $$ e_s=\frac{\big(\lambda_1^s-\lambda_2^s\big)\big(e_1-\kappa)+\lambda_1\lambda_2\big(\lambda_2^{s-1}-\lambda_1^{s-1}\big)\big(e_0-\kappa\big)}{\lambda_1-\lambda_2}+\kappa\ , $$ where $\ \kappa=\frac{c}{1-\phi_1-\phi_2}\ $.

To simplify matters further, I'll now assume that $\ y_1\ $ and $\ y_0\ $ are (non-random) constants. Then $\ z_0=z_1=0\ $, $\ z_{t+h}\,, z_t\ $ will be bivariate normally distributed with zero means, and $\ \mathbb{E}\big[z_{t+h}\,\big|\,z_t\,\big]=\frac{\mathbb{E}\big[z_{t+h}\,z_t\big]z_t}{\mathbb{E}\big[z_t^2\big]}\ $. Calculations like those given below could also be carried through under the weaker assumption that $\ (y_1,y_0)\ $ are bivariate normal, and independent of $\ \epsilon_t\ $, but the formulae will end up being even messier than those appearing below, and I couldn't be bothered trying to wade through their derivation.

From above we have \begin{align} \mathbb{E}\big[z_{t+h}\,z_t\big]&=\sum_{j=2}^{t+h}\sum_{k=2}^t\big(\lambda_1^{t+h-j}+\lambda_2^{t+h-j}\ \ \big)\big(\lambda_1^{t-k}+\lambda_2^{t-k}\ \ \big)\mathbb{E}\big[\epsilon_j\epsilon_k\big]\\ &=\sigma^2\sum_{j=2}^{t+h}\sum_{k=2}^t\big(\lambda_1^{t+h-j}+\lambda_2^{t+h-j}\ \ \big)\big(\lambda_1^{t-k}+\lambda_2^{t-k}\ \ \big)\delta_{jk}\\ &=\sigma^2\sum_{j=2}^t\big(\lambda_1^{t+h-j}+\lambda_2^{t+h-j}\ \ \big)\big(\lambda_1^{t-j}+\lambda_2^{t-j}\ \ \big)\\ &=\sigma^2\Bigg(\lambda_1^h\Bigg(\frac{1-\lambda_1^{2t-2}}{1-\lambda_1^2}\Bigg)+\lambda_2^h\Bigg(\frac{1-\lambda_2^{2t-2}}{1-\lambda_2^2}\Bigg)\\ &\hspace{4em}+\big(\lambda_1^h+\lambda_2^h\big)\Bigg(\frac{1-\big(\lambda_1\lambda_2\big)^{t-1}}{1-\lambda_1\lambda_2}\Bigg)\Bigg)\ . \end{align} Putting $\ h=0\ $ in this formula, we get $$ \mathbb{E}\big[z_t^2\big]=\sigma^2\Bigg(\frac{1-\lambda_1^{2t-2}}{1-\lambda_1^2}+\frac{1-\lambda_2^{2t-2}}{1-\lambda_2^2}+2\Bigg(\frac{1-\big(\lambda_1\lambda_2\big)^{t-1}}{1-\lambda_1\lambda_2}\Bigg)\Bigg)\ . $$ Now setting $$ g\big(t,h,\lambda_1,\lambda_2\big)=\lambda_1^h\Bigg(\frac{1-\lambda_1^{2t-2}}{1-\lambda_1^2}\Bigg)+\lambda_2^h\Bigg(\frac{1-\lambda_2^{2t-2}}{1-\lambda_2^2}\Bigg)\\ \hspace{7em}+\big(\lambda_1^h+\lambda_2^h\big)\Bigg(\frac{1-\big(\lambda_1\lambda_2\big)^{t-1}}{1-\lambda_1\lambda_2}\Bigg)\ , $$ we get \begin{align} \mathbb{E}\big[z_{t+h}\,\big|\,z_t\,\big]&=\frac{g\big(t,h,\lambda_1,\lambda_2\big)z_t}{g\big(t,0,\lambda_1,\lambda_2\big)}\\ \mathbb{E}\big[y_{t+h}\,\big|\,y_t\,\big]&=\frac{g\big(t,h,\lambda_1,\lambda_2\big)\big(y_t-e_t\big)}{g\big(t,0,\lambda_1,\lambda_2\big)}+e_{t+h}\ . \end{align} If $\ \big|\lambda_1\big|<1\ $ and $\ \big|\lambda_2\big|<1\ $ then for large $\ t\ $ we have \begin{align} \mathbb{E}\big[z_{t+h}\,z_t\big]&\approx2\sigma^2\big(\lambda_1^h+\lambda_2^h\big)\\ \mathbb{E}\big[z_t^2\big]&\approx4\sigma_2\\ \mathbb{E}\big[z_{t+h}\,\big|\,z_t\,\big]&\approx\frac{\big(\lambda_1^h+\lambda_2^h\big)z_t}{2}\\ e_t&\approx\kappa\\ \mathbb{E}\big[y_{t+h}\,\big|\,y_t\,\big]&\approx\frac{\big(\lambda_1^h+\lambda_2^h\big)\big(y_t-\kappa\big)}{2}+\kappa\ . \end{align} Note that here \begin{align} \mathbb{E}\big[y_{t+h}\,\big|\,y_{t+1}\,\big]&\approx\frac{\big(\lambda_1^{h-1}+\lambda_2^{h-1}\big)\big(y_{t+1}-\kappa\big)}{2}+\kappa\\ \mathbb{E}\big[\mathbb{E}\big[y_{t+h}\,\big|\,y_{t+1}\,\big]\,\big|\,y_t\big]&\approx\frac{\big(\lambda_1^{h-1}+\lambda_2^{h-1}\big)\mathbb{E}\big[y_{t+1}-\kappa\,\big|\,y_t\big]}{2}+\kappa\\ &\approx\frac{\big(\lambda_1^{h-1}+\lambda_2^{h-1}\big)\big(\lambda_1+\lambda_2\big)\big(y_t-\kappa\big)}{4}+\kappa\\ &\not\approx\mathbb{E}\big[y_{t+h}\,\big|\,y_t\,\big] \end{align} except for some very special values of $\ h,\lambda_1\ $ and $\ \lambda_2\ $.