Let $r_{t+1}$ be a random variable and suppose that $\mathbb{E}_t(r_{t+1}) = r+ x_t$ where $r$ is a constant, $x_t$ is a zero-mean random variable, and $\mathbb{E}_t$ is the expectation conditional on time $t$. Further assume that $x_t$ follows an AR(1) process:
$$x_{t+1} = \phi x_t + \xi_{t+1} $$ where $-1 < \phi < 1$. Show that: $$\mathbb{E}_t\left[\sum_{j=0}^{\infty} \rho^j r_{t+1+j} \right] = \frac{r}{1-\rho} + \frac{x_t}{1-\rho \phi} $$ where $\rho$ is just a constant.
I know that the variance of $\xi_t$ is $\sigma_{\xi_t}^2 = (1-\phi^2) \sigma_{x_t}^2$ but that's as far as I got. Any help would be appreciated.
I may be confused about the question, but assuming that $|\rho| < 1$ is known I think you can just solve it without worrying about $\xi$ too much. We have that $E[r_{t+n}] = r + x_{t+n-1}$, if we can map the Expectation across the infinite sum, this gives us:
\begin{equation} \begin{split} E[\sum_{j=0}^\infty \rho^jr_{t+j+1}] & = \sum_{j=0}^\infty\rho^jE[r_{r+j+1}]\\ & = \sum_{j=0}^\infty\rho^jE[r + x_{t+j}]\\ & = \sum_{j=0}^\infty\rho^jr + \sum_{j=0}^\infty\rho^j E[x_{t+j}]\\ \end{split} \end{equation}
Note that we also have that $E[x_{t+j}] = \phi^jx_t$, this follows from repeatedly applying the expectation, but conditioning on the known time t. This gives us:
\begin{equation} \begin{split} \sum_{j=0}^\infty\rho^jr + \sum_{j=0}^\infty\rho^j E[x_{t+j}] & = \sum_{j=0}^\infty\rho^jr + \sum_{j=0}^\infty\rho^j\phi^jx_t\\ & = \frac{r}{1-\rho} + \frac{x_t}{1-\rho\phi} \end{split} \end{equation}