I have some trouble solving the following problem:
Given is the causal AR(1) model $X$:
$X_t = qX_{t-1}+Y_t$
Where $Y_t$ is distibuted $IID(0,\sigma^2)$
Now I'm trying to figure out how to derive:
$E[(P_nX_{n+l}-X_{n+l})^2]$$=\frac{1 -q^{2l} }{1 -q^{2}}\sigma^2$
Extra information given: $l\ge1$ and $P_nX_{n+l}$ is the $l$-step ahead forecast.
Looking at the final equation of the derivation, it looks like the following rule will be used:
if $|\beta|\lt1$ then the summation of '$\beta^i$' from $i=0$ to $i=n-1$ is equal to:
$\frac{1 -\beta^{n} }{1 -\beta}$
How to get to such an equation with a summation is not clear to me. Maybe it's because I am not very familiar with $l$-step ahead forcasting. Help is appreciated.
What you're looking at is the MSE. It can be rewritten as:
$$ \begin{align} E[(\hat{X}_{n+l}-X_{n+l})^2]&=V(X_{n+l})+[X_{n+l}-E(\hat{X}_{n+l})]^2\\ &=V(X_{n+l}) \end{align} $$ where the last equality is true because $E(\hat{X}_{n+l})=X_{n+l}$. Furthermore, this notation is a bit sloppy because actually we are doing this conditional on all $X$ up to $X_n$. So what we really want to find is:
$$ V(X_{n+l}|X_n, X_{n-1}, \dots) $$ where we for simplicity will let $\mathcal{I}_n=\{X_t : t\leq n, \, t\in\mathbb{Z}\}$
$$ \begin{align} V(X_{n+l}|\mathcal{I}_n)&=V(qX_{n+l-1}+Y_{n+l}|\mathcal{I}_n)=V(Y_{n+l}|\mathcal{I}_n)+q^2V(X_{n+l-1}|\mathcal{I}_n)\\ &=\sigma^2+q^2V(qX_{n+l-2}+Y_{n+l-1}|\mathcal{I}_n)\\ &=\sigma^2+q^2\sigma^2+q^4V(qX_{n+l-3}+Y_{n+l-2}|\mathcal{I}_n)\\ &\vdots\\ &=\sigma^2+q^2\sigma^2+q^4\sigma^2+\cdots+q^{2(l-1)}V(qX_{n}+Y_{n+1}|\mathcal{I}_n)\\ &=\sigma^2+q^2\sigma^2+q^4\sigma^2+\cdots+q^{2(l-1)}V(Y_{n+1}|\mathcal{I}_n)\\ &=\sigma^2+q^2\sigma^2+q^4\sigma^2+\cdots+q^{2(l-1)}\sigma^2\\ &=\sigma^2\sum_{k=0}^{l-1}\left(q^2\right)^k\\ &=\sigma^2\frac{1-\left(q^2\right)^l}{1-q^2}\\ &=\sigma^2\frac{1-q^{2l}}{1-q^2}. \end{align} $$ Some remarks: