K-step ahead forecast for an AR(2)

Question

Given an AR(2) process $y_{t}=\phi_{1}y_{t-1}+\phi_2 y_{t-2}$, I understand that the two-step ahead forecast (that is, $\mathbb{E}[y_{t+2}|y_t]$ is given by $(\phi_{1}^2+\phi_{2})y_{t}+\phi_{1}\phi_{2}y_{t-1}$. However, I am unable to derive a general expression for the k-step ahead forecast (that is, for $\mathbb{E}[y_{t+k}|y_{t}]$. How might I go about doing this? Thank you.

Answer 1

Charles: I still haven't had time to look carefully at tdm's soluton but I did make my own so, assuming mine is correct, you can have two choices for what to do. I also still plan on going over tdm's solution when I have more time.

The original model is an AR(2): $y_{t} = \phi_1 y_{t-1} + \phi_2 y_{t-2}$.

Using the backshift operator, this can be re-written as

$y_t(1 - \phi_{1} L) = \phi_2 y_{t-2} $

$y_{t} = \frac{\phi_{2}}{(1 - \phi_{1} L)} y_{t-2}$

Assuming that $|\phi_{1}| < 1$, the denominator in the equation immediately above can be written as an infinite sum so that the expression for $y_{t}$ becomes:

`$y_{t} = \phi_{2} \sum_{i=0}^{\infty} \phi_{1}^{i} y_{t-2-i}$

So, given that one is at time $t$, an expression for the prediction of $y_{t+1}$ can be written that is made up of the past $y_{t}$. Specifically, if one wants to calculate the prediction of $y_{t+1}$, then using the expression derives above but increasing the index by 1, results in:

$\hat{y}_{t+1} = \phi_{2} \sum_{i=0}^{\infty} \phi_{1}^{i} y_{t - 2 + 1 - i}$

Note that the first term in the infinite sum above, is $y_{t-1}$ so everything in the sum is known at time $t$.

Similarly, if one is at time $t$ and wants to calculate the prediction of $y_{t+2}$, then using the same expression but increasing the index by 2, results in:

$\hat{y}_{t+2} = \phi_{2} \sum_{i=0}^{\infty} \phi_{1}^{i} y_{t - 2 + 2 -i}$

Again note that the first term in the infinite sum above, is $y_{t}$ so everything in the sum is known at time $t$.

Similarly, if one is at time $t$ and wants to calculate the prediction of $y_{t+3}$, then using the same expression but increasing the index by 3, results in:

$\hat{y}_{t+3} = \phi_{2} \sum_{i=0}^{\infty} \phi_{1}^{i} y_{t - 2 + 3 -i}$

Now notice that the first term in the infinite sum is $y_{t+1}$ which is not available at time $t$. Therefore, for the first term in the sum, the prediction at time $t+1$, $\hat{y}_{t+1}$, needs to be used in place than the actual value, $y_{t+1}$, since it is not known at time $t$.

Similarly, for $\hat{y}_{t+4}$, the predictions $\hat{y}_{t+1}$ and $\hat{y}_{t+2}$ will need to be used in the summation, in place of $y_{t+1}$ and $y_{t+2}$ since they are also not available at time $t$.

Using the same argument, the same forecasting procedure can to be used all the way upto and including $\hat{y}_{t+k}$.

$\hat{y}_{t+k} = \phi_{2} \sum_{i=0}^{\infty} \phi_{1}^{i} y_{t - 2 + k - i}$

Note though that the previous predictions need to be used in place of $y_{t+1},y_{t+2},\ldots, y_{t+k-1}$ in the expression for the infinite sum. I hope this helps. It seems like there should be an easier way but I could not discern a pattern when I played around so this seemed like the only way.

Answer 2

Let's define $C_k$ such that: $$ \mathbb{E}(y_{t+k}|y_t) = C_k y_t. $$ Then: $$ \begin{align*} \mathbb{E}(y_{t+k+1}|y_t) &= \phi_1 \mathbb{E}(y_{t+k}|y_t) + \phi_2 \mathbb{E}(y_{t+k-1}|y_t),\\ &= \phi_1 C_k y_t + \phi_2 C_{k-1} y_{t},\\ &= (\phi_1 C_k + \phi_2 C_{k-1}) y_t \end{align*} $$ This gives us the following recursive second order difference equation: $$ C_{k+1} = \phi_1 C_k + \phi_2 C_{k-1}. $$ Or, updated one period: $$ C_{k+2}- \phi_1 C_{k+1} - \phi_2 C_k = 0. $$ This is a second order homogeneous difference equation, which is of the form $C_k = \alpha a_1^k + \beta a_2^k$. Computing $a_1$ and $a_2$ can be done by solving the following characteristic function: $$ (a^2- \phi_1 a - \phi_2) = 0 $$ This gives the following two roots: $$ a_1 = \frac{\phi_1 + \sqrt{\phi_1^2 + 4 \phi_2}}{2} \text{ and } a_2 = \frac{\phi_1 - \sqrt{\phi_1^2 + 4 \phi_2}}{2}. $$ So in general we have: $$ C_k = \alpha a_1^k + \beta a_2^k $$ To find $\alpha$ and $\beta$, we need two starting values, say $C_0 = 1$ and $C_1 = \phi_1$. In other words, we assume that $\mathbb{E}(y_t|y_t) = y_t$ and $\mathbb{E}(y_{t+1}|y_t) = \phi_1 y_t$.

This gives the following two conditions: $$ \alpha + \beta = 1\\ \alpha \left(\frac{\phi_1 + \sqrt{\phi_1^2 + 4 \phi_2}}{2}\right) + \beta \left(\frac{\phi_1 - \sqrt{\phi_1^2 + 4 \phi_2}}{2}\right) = \phi_1.\\ $$

Solving this system in terms of $\alpha$ and $\beta$ gives: $$ \beta = \frac{\sqrt{\phi_1^2 + 4 \phi_2} - \phi_1}{2 \sqrt{\phi_1^2 + 4 \phi_2}}\\ \alpha = \frac{\phi_1 + \sqrt{\phi_1^2 + 4 \phi_2} }{2 \sqrt{\phi_1^2 + 4 \phi_2}} $$

This gives the following closed form expression condition: $$ \begin{align*} \mathbb{E}(y_{t+k}|y_t) &= C_k y_t,\\ &=\left[\begin{array}{ll} &\frac{\phi_1 + \sqrt{\phi_1^2 + 4 \phi_2} }{2 \sqrt{\phi_1^2 + 4 \phi_2}}\left(\frac{\phi_1 + \sqrt{\phi_1^2 + 4 \phi_2}}{2}\right)^k\\ + &\frac{\sqrt{\phi_1^2 + 4 \phi_2} - \phi_1}{2 \sqrt{\phi_1^2 + 4 \phi_2}} \left(\frac{\phi_1 - \sqrt{\phi_1^2 + 4 \phi_2}}{2}\right)^k\end{array}\right] y_t \end{align*} $$