not following a step in a paper on rational expectations
Presumably $I_t$ stands for the information set at time $t$. So note that $E[\epsilon_j|I_t] = \epsilon_j$ $\forall j \leq t$. This will come into play when working out the LHS.

We wish to verify Equation (4), which is 

$$
E(p_{t+1}|I_t) = \frac{\Psi(B) - \psi_0}{B} \epsilon_t,
$$

which is equivalent to

$$
$$

$$
\begin{align*}
B(E(p_{t+1}|I_t)) = \Psi(B) \epsilon_t - \psi_0 \epsilon_t. &amp;&amp;(*)
\end{align*}
$$

$$
$$



We are given that

$$
\Psi(B) \epsilon_t = \psi_0 \epsilon_t + \psi_1 \epsilon_{t-1} + \psi_2 \epsilon_{t-2} + \dots + \psi_k \epsilon_{t-k} + \dots
$$

so the RHS of $(*)$ is

$$
$$

$$
\boldsymbol{\Psi(B) \epsilon_t - \psi_0 \epsilon_t = \psi_1 \epsilon_{t-1} + \psi_2 \epsilon_{t-2} + \dots + \psi_{k} \epsilon_{t-k} + \dots}.
$$

$$
$$



Now, the LHS. Since $p_t = \psi_0 \epsilon_t + \psi_1 \epsilon_{t-1} + \psi_2 \epsilon_{t-2} + \dots + \psi_k \epsilon_{t-k} + \dots,$

$$
$$

$$
\begin{align*}
E(p_{t+1}|I_t) &amp;= \psi_0 E(\epsilon_{t+1}|I_t) + \psi_1 E(\epsilon_t|I_t) + \psi_2 E(\epsilon_{t-1}|I_t) + \dots  + \psi_k E(\epsilon_{t-k+1}|I_t) + \dots \\
&amp;= \psi_1 \epsilon_{t} + \psi_2 \epsilon_{t-1} + \dots + \psi_{k} \epsilon_{t-k+1} + \dots
\end{align*}
$$

$$
$$

since $E(\epsilon_{t+1}|I_t) = 0$ because $\epsilon$ is an independent white noise process.

Given that $B(\cdot)$ is the lag operator, the LHS of $(*)$ is thus

$$
$$

$$
\begin{align*}
\boldsymbol{B(E(p_{t+1}|I_t))} &amp;= B(\psi_1 \epsilon_{t} + \psi_2 \epsilon_{t-1} + \dots + \psi_{k} \epsilon_{t-k+1} + \dots) \\
&amp;= \psi_1 B(\epsilon_{t}) + \psi_2 B(\epsilon_{t-1}) + \dots + \psi_{k} B(\epsilon_{t-k+1}) + \dots \\
&amp;= \boldsymbol{\psi_1 \epsilon_{t-1} + \psi_2 \epsilon_{t-2} + \dots + \psi_{k} \epsilon_{t-k} + \dots}.
\end{align*}
$$

$$
$$



Thus the LHS = RHS; that is,

$$
\begin{align*}
E(p_{t+1}|I_t) = \frac{\Psi(B) - \psi_0}{B} \epsilon_t. &amp;&amp;\blacksquare
\end{align*}
$$