Conditional expectation of a random walk process with noise For each $t \in \mathbb{N}$, define a process $e_t$ by $e_t = z_t + v_t$, where $v_t$ follows normal distribution $N(0, \sigma_v^2)$ independently and $z_t$ is a random walk with normal error:
$$
z_t = z_{t-1} + u_t = \sum_{s=1}^t u_s, \quad u_s \sim N(0,\sigma_u^2).
$$
$z_t$ and $v_t$ are independent.
I want to derive the conditional expectation of future $e$ at time $t$, i.e.,  $E[e_{t+k} \mid e_t, e_{t-1},\dots], k>1$.
Here is what I understand thus far:
The relevant moments are
$$
E[e_t] = 0, Var(e_t) = t\sigma_u^2 + \sigma_v^2., Cov(e_{t+k},e_t) = t \sigma_u^2
$$
For example, for a simple case such as $E[e_2 \mid e_1]$, from the result concerning the conditional expectation of normal variables, I can derive that
$$
E[e_2 \mid e_1] = \frac{Cov(e_1,e_2 )}{Var(e_1)} e_1 = \frac{\sigma_u^2}{\sigma_u^2 + \sigma_v^2} e_1.
$$
However, I am not sure how to tractably compute $E[e_{t+k} \mid e_{t}, e_{t-1}, \dots]$ (I know the formula for conditional expectation of $n$ normal variables).
 A: \begin{align}
\mathsf{E}[e_{t+k}\mid e_t,e_{t-1},\ldots,e_1]&=\mathsf{E}\!\!\left[z_t+\sum_{i=t+1}^{t+k}u_t+v_{t+k}\mid e_t,e_{t-1},\ldots,e_1\right] \\
&=\mathsf{E}[z_t\mid e_t,e_{t-1},\ldots,e_1]
\end{align}
Since $z_t, e_t,\ldots,e_1$ are jointly normal, the last conditional expectation is the linear projection of $z_t$ on $\mathbf{e}_t:=[e_t,\ldots,e_1]^{\top}$, that is,
$$
\mathsf{E}[z_t\mid e_t,e_{t-1},\ldots,e_1]=\mathbf{e}_t^{\top}\beta,
$$
where
$$
\beta=\left(\mathsf{E}[\mathbf{e}_t\mathbf{e}_t^{\top}]\right)^{-1}\mathsf{E}[\mathbf{e}_tz_t].
$$
Letting $\mathbf{z}_t:=[z_t,\ldots,z_1]^{\top}$ and $\mathbf{v}_t:=[v_t,\ldots,v_1]^{\top}$
\begin{align}
\mathsf{E}[\mathbf{e}_t\mathbf{e}_t^{\top}]&=\mathsf{E}[(\mathbf{z}_t+\mathbf{v}_t)(\mathbf{z}_t+\mathbf{v}_t)^{\top}]=\mathsf{E}[\mathbf{z}_t\mathbf{z}_t^{\top}]+\mathsf{E}[\mathbf{v}_t\mathbf{v}_t^{\top}] \\
&=\sigma_u^2U_tU_t^{\top}+\sigma_v^2I_t,
\end{align}
where $U_t$ is a $t\times t$ upper triangular matrix whose nonzero entries are ones, and
$$
\mathsf{E}[\mathbf{e}_tz_t]=\mathsf{E}[\mathbf{z}_tz_t]=\sigma_u^2U_t\mathbf{i}_t,
$$
where $\mathbf{i}_t$ is a vector of ones.

Combining these results, one gets
$$
\mathsf{E}[e_{t+k}\mid \mathbf{e}_t]=\mathbf{e}_t^{\top}(U_tU_t^{\top}+I_t\sigma_v^2/\sigma_u^2)^{-1}U_t\mathbf{i}_t.
$$
