I'm trying to derive the result that $\mathbb{E}(X_{n+h} - m(X_n))^2 = \sigma^2(1-\rho(h)^2)$ where $m(X_n) = \mu + \rho(h)(X_n - \mu)$ but i can't get the same result. I will post my derivation below. I'm wondering if what they intended to write was $\mathbb{E}((X_{n+h} - m(X_n))^2 | X_n)$ instead? This is because if $X_{n+h}|X_{n}$ is normally distributed with mean $m(X_n)$ and variance $\sigma^2(1-\rho(h)^2)$ then it's clear $\mathrm{Var}(X_{n+h}|X_{n}) = \mathbb{E}(X_{n+h}-m(X_n))^2|X_{n}) = \sigma^2(1-\rho(h)^2)$. Then again maybe it is correct and there is something wrong in my derivation?
So $\mathbb{E}(X_{n+h} - m(X_n))^2) = \mathbb{E}(X_{n+h}^2) - 2\mathbb{E}(X_{n+h}m(X_n))+\mathbb{E}(m(X_n)^2)$
I will caculate the following components seperately
- $\mathbb{E}(X_{n+h}^2)$
- $2\mathbb{E}(X_{n+h}m(X_n))$
- $\mathbb{E}(m(X_n)^2)$
For 1. Clearly as the time series is stationary $\mathbb{E}(X_{n+h}^2) = \mu^2+\sigma^2$
For 3. We have $\mathbb{E}(\mu^2+2\mu\rho(h)(X_n - \mu)+\rho(h)^2(X_{n} - \mu)^2) = \mu^2+\rho(h)^2\sigma^2$
For 2. We have
\begin{eqnarray} 2\mathbb{E}(X_{n+h}\mu-\rho(h)X_{n+h}(X_n - \mu)) &=& 2(\mu^2 - \rho(h)\mathbb{E}(X_{n+h}X_n)-\rho(h)\mu^2) \end{eqnarray}
Now we need to calculate $\rho(h)\mathbb{E}(X_{n+h}X_{n})$ now by definition $\rho(h) = \frac{\mathbb{E}(X_{n+h}X_{n}) - \mu^2}{\sigma^2}$ and so $\mathbb{E}(X_{n+h}X_{n}) = \rho(h)\sigma^2+\mu^2$ and so $\rho(h)^2\mathbb{E}(X_{n+h}X_n) = \rho(h)^2\sigma^2+ \rho(h)\mu^2$. Substituting this in we get for
- $$2\mathbb{E}(X_{n+h}m(X_n)) = 2(\mu^2(1-\rho(h)^2) + \rho^2(h)\sigma^2+\rho(h)\mu^2)$$
I now get for 1.+3.-2.
$$ 2\mu^2 + (1+\rho(h)^2)\sigma^2 - 2\mu^2 - 2\rho(h)^2\sigma^2 - 2\rho(h)^2\mu^2 - 2\rho(h)\mu^2 = (1-\rho(h)^2)\sigma^2 - 2\rho(h)^2\mu^2 - 2\rho(h)\mu^2$$
Which is not what they have. Does anyone know where i may have gone wrong, or if they intended a conditional on $X_{n}$ in their equation?
EDIT: I think i have found the error in my calculation. I put for 2.
\begin{eqnarray} 2\mathbb{E}(X_{n+h}\mu-\rho(h)X_{n+h}(X_n - \mu)) &=& 2(\mu^2 - \rho(h)\mathbb{E}(X_{n+h}X_n)-\rho(h)\mu^2) \end{eqnarray}
but the formula for $m(x) = \mu + \rho(h)(X_n - \mu)$ and hence it should've been $\mathbb{E}(X_{n+h}\mu+\rho(h)X_{n+h}(X_n - \mu))$ in which case we get $(\mu^2 - \rho(h)\mu^2 + \mathbb{E}(\rho(h) X_{n}X_{n+h}) = \mu^2-\rho(h)\mu^2 + \rho(h)(\rho(h)\sigma^2+\mu^2) = \mu^2 +\rho(h)^2\sigma^2$
then for 1.+3. - 2. we get $(\mu^2+\sigma^2)+(\mu^2+\rho(h)^2\sigma^2) - 2(\mu^2+\rho(h)\sigma^2) = \sigma^2-\rho(h)^2\sigma^2 $.
By law of iterated expectations $\mathbb{E}(X_{n+h}-m(X_n))^2)=\mathbb{E}\mathbb{E}(X_{n+h}-m(X_n))^2 | X_n)=\mathbb{E}(\sigma^2(1-\rho(h)^2))=\sigma^2(1-\rho(h)^2)$. Leaving out the conditional expectation is not a mistake as here the two happen to be equal.