K-step ahead forecast for an AR(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: $$ begin{align*} mathbb{E}(y_{t+k}|y_t) &= C_k y_t, &=left[begin{array}{ll} &frac{phi_1 + sqrt{phi_1^2 + 4 phi_2} }{2 sqrt{phi_1^2 + 4 phi_2}}left(frac{phi_1 + sqrt{phi_1^2 + 4 phi_2}}{2}right)^k + &frac{sqrt{phi_1^2 + 4 phi_2} - phi_1}{2 sqrt{phi_1^2 + 4 phi_2}} left(frac{phi_1 - sqrt{phi_1^2 + 4 phi_2}}{2}right)^kend{array}right] y_t end{align*} $$

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.