Vtyjkyuk

Let $r_t+1$ be a random variable and suppose that $mathbbE_t(r_t+1) = r+ x_t$ where $r$ is a constant, $x_t$ is a zero-mean random variable, and $mathbbE_t$ 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:
$$mathbbE_tleft[sum_j=0^infty rho^j r_t+1+j right] = fracr1-rho + fracx_t1-rho phi $$ 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:

beginequation
beginsplit
E[sum_j=0^infty rho^jr_t+j+1] & = sum_j=0^inftyrho^jE[r_r+j+1]\
& = sum_j=0^inftyrho^jE[r + x_t+j]\
& = sum_j=0^inftyrho^jr + sum_j=0^inftyrho^j E[x_t+j]\
endsplit
endequation

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:

beginequation
beginsplit
sum_j=0^inftyrho^jr + sum_j=0^inftyrho^j E[x_t+j] & = sum_j=0^inftyrho^jr + sum_j=0^inftyrho^jphi^jx_t\
& = fracr1-rho + fracx_t1-rhophi
endsplit
endequation