# not following a step in a paper on rational expectations

Hi: I'm reading a paper by Broze and Szafarz titled "On Econometric Models with Rational Expectations and I don't follow a step in section 3.2 which discusses the method of undetermined coefficients. So, I was hoping that someone could explain the derivation. I will write down the equations and then explain the step that I don't follow.

1) $$p_{t} = a \times E(p_{t+1} | I_{t}) + u_{t}$$ This equation is given. It comes out of the Cagan Hyperinflation model. $$u_{t}$$ is an error term.

2) $$u_{t} = \epsilon_{t} + \alpha \epsilon_{t-1} + \ldots + \alpha_{k} \epsilon_{t-k} + \ldots = A(B) \epsilon_{t}$$

3) $$p_{t} = \psi_{0} \epsilon_{t} + \ldots \psi_{k} \epsilon_{t-k} + \ldots = \Psi(B) \epsilon_{t}$$

4) $$E(p_{t+1}| I_{t} ) = \frac{\Psi(B) - \psi_{0}}{B} \epsilon_{t}$$ So, $$B$$ denotes the lag operator which is often denoted as $$L$$ in the literature. The explanations of the equations follow.

1) As stated above, this is given. This is clear.

2) This is the standard infinite MA representation that follows from Wold's decomposition. This is clear.

3) This is the guess that is made in the method of undetermined coefficients. This is clear.

4) This is not clear to me. They say "it follows". I see that it's kind of coming by subtracting $$\psi_{0} \epsilon_{t}$$ from both sides of 3) but $$B$$ is the infinite lag polynomial and they are dividing by it which confuses me.

Thanks in advance for the help. Some discussion has been omitted so, if anyone needs to see the actual pages, I can scan and send them. My email is my name with a 2 on the end of it and the mail server is that famous one that starts with a g.

                                        Mark

• Equation (4) should be $E(p_{t+1}|I_t) = \frac{\Psi(B) - \psi_0}{B} \epsilon_t$ instead right? In other words, $\Psi$ is a vector $\psi_k$ for $k = 0, \dots, \infty$? Oct 18, 2018 at 2:39
• thanks. I edited a lot because I figured out part of it. help is still appreciated. Oct 18, 2018 at 2:42

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. &&(*) \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) &= \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 \\ &= \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))} &= B(\psi_1 \epsilon_{t} + \psi_2 \epsilon_{t-1} + \dots + \psi_{k} \epsilon_{t-k+1} + \dots) \\ &= \psi_1 B(\epsilon_{t}) + \psi_2 B(\epsilon_{t-1}) + \dots + \psi_{k} B(\epsilon_{t-k+1}) + \dots \\ &= \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. &&\blacksquare \end{align*}

• Hi Kenneth. I didn't go over your solution yet ( too late and brain is closed for the night ) but I'm checking the answer anyway and will print out and read carefully. Thanks a lot. Oct 18, 2018 at 8:03