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## A Appendix

### A.1 The maximization problem and the optimality conditions

Our formal description of the consumer’s maximization problem follows Mankiw and Reis (2007). Suppose that a consumer has the information set available in period t. We consider an inattentive consumer who receives the newest information only with the probability $ 1 - \omega $ every period. Given the infrequent information updating and the information set in period t, he chooses a consumption plan to solve the following maximization problem:

$$ \begin{align*}V\left(A_{t},X_{t}\right)\ =&\ \max_{\left\{C_{t+j,j}\right\}_{i=0}^{\infty}}\sum_{j=0}^{\infty}\left(\beta\omega\right)^{j}u\left(C_{t+j,j}\right)\\&+\beta\left(1-\omega\right)\mathbb{E}_{t}\sum_{j=0}^{\infty}\left(\beta\omega\right)^{j}V\left(A_{t+j+1},X_{t+j+1}\right),\\ subject to\ A_{t+j+1}\ =&\ \left(1+r\right)A_{t+j}+X_{t+j}-C_{t+j,j},\ for j=0,1,2,...,\\\end{align*} $$

where $ V(A_{t}, X_{t}) $ is the value function of the consumer in period t. Using the information available in period t, he chooses the consumption plan $ \{ C_{t+j,j} \}_{j=0}^{\infty} $ , where $ C_{t+j,j} $ represents the amount of goods consumed in period $ t+j $ but predetermined j periods in advance. The value function is somewhat complicated since the plan is specified as consumption from period t and revised once he receives the new information. Other variables, parameters, and the function $ u(C) $ are defined in the main text.

The first-order condition and the envelope condition are

$$ u^{\prime}\left(C_{t+j,j}\right)\ =\ \beta\left(1-\omega\right)\mathbb{E}_{t}\sum_{k=0}^{\infty}\left[\beta\omega\left(1+r\right)\right]^{k}V^{\prime}\left(A_{t+j+k+1},X_{t+j+k+1}\right), $$

$$ \begin{array}{r c l}{V^{\prime}\left(A_{t},X_{t}\right)}&{=}&{\beta\left(1-\omega\right)\left(1+r\right)\mathbb{E}_{t}\displaystyle\sum_{k=0}^{\infty}\left[\beta\omega\left(1+r\right)\right]^{k}V^{\prime}\left(A_{t+k+1},X_{t+k+1}\right),}\\ \end{array} $$

for $ j = 0, 1, 2, \ldots $ , respectively. Here, $ u'(C) = du(C) / dC $ and $ V'(A_t, X_t) $ is the partial derivative with respect to the foreign asset. That is, $ V'(A, X) = \partial V(A, X) / \partial A $ .

Evaluating (26) at j = 0 yields $ u' \left( C_{t,0} \right) = \beta \left( 1 - \omega \right) \mathbb{E}_{t} \sum_{k=0}^{\infty} \left[ \beta \omega \left( 1 + r \right) \right]^{k} V' \left( A_{t+k+1}, X_{t+k+1} \right) $ . This result implies

$$ V^{\prime}\left(A_{t},X_{t}\right)=\left(1+r\right)u^{\prime}\left(C_{t,0}\right). $$
Rewriting (27) recursively, we have

$$ \begin{align*}V^{\prime}\left(A_{t},X_{t}\right)\ =&\ \beta\left(1-\omega\right)\left(1+r\right)\mathbb{E}_{t}\sum_{k=0}^{\infty}\left[\beta\omega\left(1+r\right)\right]^{k}V^{\prime}\left(A_{t+k+1},X_{t+k+1}\right)\\ =\ \beta\left(1-\omega\right)\left(1+r\right)\mathbb{E}_{t}V^{\prime}\left(A_{t+1},X_{t+1}\right)\\ +\beta\omega\left(1+r\right)\mathbb{E}_{t}\left\{\beta\left(1-\omega\right)\left(1+r\right)\sum_{k=0}^{\infty}\left[\beta\omega\left(1+r\right)\right]^{k}\mathbb{E}_{t+1}V\left(A_{t+k+2},X_{t+k+2}\right)\right\}\\ =\ \beta\left(1-\omega\right)\left(1+r\right)\mathbb{E}_{t}V^{\prime}\left(A_{t+1},X_{t+1}\right)+\beta\omega\left(1+r\right)\mathbb{E}_{t}V^{\prime}\left(A_{t+1},X_{t+1}\right)\\ =\ \beta\left(1+r\right)\mathbb{E}_{t}V^{\prime}\left(A_{t+1},X_{t+1}\right).\end{align*} $$

Combining (28) and (29), we have the Euler equation that is standard in the RE model:

$$ u^{\prime}\left(C_{t,0}\right)=\beta\left(1+r\right)\mathbb{E}_{t}u^{\prime}\left(C_{t+1,0}\right). $$

We next consider the expected marginal utility $ \mathbb{E}_{t}u^{\prime}\left(C_{t+1,0}\right) $ . Update (26) by one period and set j=0. Then, we have $ u^{\prime}\left(C_{t+1,0}\right)=\beta\left(1-\omega\right)\mathbb{E}_{t+1}\sum_{k=0}^{\infty}\left[\beta\omega\left(1+r\right)\right]^{k}V^{\prime}\left(A_{t+k+2},X_{t+k+2}\right) $ . By the law of iterated expectations,

$$ \begin{align*}\mathbb{E}_{t}u^{\prime}\left(C_{t+1,0}\right)&=\beta\left(1-\omega\right)\mathbb{E}_{t}\sum_{k=0}^{\infty}\left[\beta\omega\left(1+r\right)\right]^{k}V^{\prime}\left(A_{t+k+2},X_{t+k+2}\right)\\&=\ u^{\prime}\left(C_{t+1,1}\right).\end{align*} $$

Therefore, along with the assumption of $\beta(1+r)=1$, (30) implies

$$ C_{t+j,j}=C_{t,0}for j=1,2,\ldots $$

We note that the optimal consumption in period t under the SI and RE models must be the same because the information used to determine the optimal period-t consumption is identical. Therefore, (2) and (31) imply (6) in the main text: $ C_{t+j,j} = rA_{t} + X_{t}^{p} $ , for j = 0, 1, 2, ....

### A.2 Inattentive consumer’s saving

Using (6), it is straightforward to derive (9) in the main text:

$$ \begin{align*}S_{t+j,j}\quad&=\quad A_{t+j+1}-A_{t+j}=rA_{t+j}+X_{t+j}-C_{t+j,j}\\&=\quad r\left(A_{t+j}-A_{t}\right)+X_{t+j}-X_{t}^{p}\\&=\quad r\underbrace{(A_{t+j}-A_{t+j-1})}_{s_{t+j-1,j-1}}+X_{t+j}-X_{t+j-1}+\underbrace{r\left(A_{t+j-1}-A_{t}\right)+X_{t+j-1}-X_{t}^{p}}_{s_{t+j-1,j-1}}\\&=\quad(1+r)S_{t+j-1,j-1}+\Delta X_{t+j}for j=1,2,3,\ldots\end{align*} $$
### A.3 Proof of Proposition 1

#### A.3.1 Derivation of (13)

To obtain the current account in the economy, we aggregate all individual savings over information differing across households. Notice that $ S_{t,k} $ is the period-t saving of the inattentive consumers who updated their information k periods ago. Using (32), we have an individual's saving in period t:

$$ \begin{align*}S_{t,k}\ &=\left(1+r\right)S_{t-1,k-1}+\Delta X_{t}=\left(1+r\right)\left[\left(1+r\right)S_{t-2,k-2}+\Delta X_{t-1}\right]+\Delta X_{t}\\&=\left(1+r\right)^{2}S_{t-2,k-2}+\Delta X_{t}+\left(1+r\right)\Delta X_{t-1}\\&=\ldots\\&=\left(1+r\right)^{k}S_{t-k,0}+\Delta X_{t}+\left(1+r\right)\Delta X_{t-1}+\ldots+\left(1+r\right)^{k-1}\Delta X_{t-k+1},\end{align*} $$

where $ S_{t-k.0} = X_{t-k} - X_{t-k}^{p} $ for $ k = 1, 2, 3, \ldots $ . The distribution of $ S_{t,k} $ follows (10) in the main text so that the current account is aggregated by

$$ \begin{aligned}CA_{t}\ =&\ \left(1-\omega\right)S_{t,0}\\&+\left(1-\omega\right)\omega\left(1+r\right)S_{t-1,0}+\left(1-\omega\right)\omega\Delta X_{t}\\&+\left(1-\omega\right)\omega^{2}\left(1+r\right)^{2}S_{t-2,0}+\left(1-\omega\right)\omega^{2}\Delta X_{t}+\left(1-\omega\right)\omega^{2}\left(1+r\right)\Delta X_{t-1}\\&+\ldots\end{aligned} $$

Using the definition of $ \omega^{*}=\omega(1+r) $ , this equation can be rearranged as

$$ \begin{align*}CA_{t}\quad=\quad&(1-\omega)\sum_{k=0}^{\infty}(\omega^{*})^{k}S_{t-k,0}\\&+\omega\left(1-\omega\right)\sum_{k=0}^{\infty}\omega^{k}\Delta X_{t}+\omega\omega^{*}\left(1-\omega\right)\sum_{k=0}^{\infty}\omega^{k}\Delta X_{t-1}+\omega(\omega^{*})^{2}\left(1-\omega\right)\sum_{k=0}^{\infty}\omega^{k}\Delta X_{t-2}+\ldots\\=\quad&(1-\omega)\sum_{k=0}^{\infty}(\omega^{*})^{k}S_{t-k,0}+\omega\left[\Delta X_{t}+\omega^{*}\Delta X_{t-1}+(\omega^{*})^{2}\Delta X_{t-2}+\ldots\right]\\=\quad&\sum_{k=0}^{\infty}(\omega^{*})^{k}\left[\left(1-\omega\right)S_{t-k,0}+\omega\Delta X_{t-k}\right].\end{align*} $$

Assume that $ \omega^{*}=\omega(1+r)<1 $ . Then, using the lag operator L, $ CA_{t}=(1-\omega^{*}L)^{-1}\left[(1-\omega)S_{t,0}+\omega\Delta X_{t}\right] $ . This equation is equivalent to (13) in Proposition 1.

#### A.3.2 Derivation of (14)

In the proof, we derive the stochastic process of $ c a_{t} $ under the generalized case of non-zero $ \mu $ . It is convenient because the general case allows us to derive (22) and (23) in Proposition 3. To prove
the second part of Proposition 1, we divide both sides of (13) by $ X_{t-1} $ :

$$ \exp\left(g_{t}\right)c a_{t}=\omega^{*}c a_{t-1}+\left(1-\omega\right)\exp\left(g_{t}\right)s_{t,0}+\omega\left[\exp\left(g_{t}\right)-1\right], $$

where $ s_{t,0} \equiv S_{t,0}/X_{t} $ . We note that the steady state value of $ s_{t,0} $ equals the steady state value of $ ca_{t} $ because $ S_{t,0} $ is identical to the current account under the RE model. Let ca be the steady state value of $ ca_{t} $ . In the steady state, (33) becomes

$$ \exp\left(\mu\right)\cdot ca=\omega^{*}ca+\left(1-\omega\right)\exp\left(\mu\right)ca+\omega\left[\exp\left(\mu\right)-1\right]. $$

Noting that $ \exp(\mu)=1+\mu,\omega^{*}=\omega(1+r) $ , we can solve the above equation for ca as follows:

$$ c a=-\frac{\mu}{r-\mu}. $$

That is, ca is zero if and only if $ \mu = 0 $ . In addressing the case of non-zero $ \mu $ , it is convenient to introduce the deviations of a variable from the mean:

$$ c\hat{a}_{t}=c a_{t}-c a,\hat{s}_{t,0}=s_{t,0}-c a,\mathrm{a n d}\hat{g}_{t}=g_{t}-\mu. $$

The first-order Taylor expansion of both sides of (33) around the steady state is

$$ \exp\left(\mu\right)\cdot c\hat{a}_{t}+\left[\exp\left(\mu\right)c a\right]\hat{g}_{t}=\omega^{*}c\hat{a}_{t-1}+\left(1-\omega\right)\exp\left(\mu\right)\hat{s}_{t,0}+\left(1-\omega\right)\left[\exp\left(\mu\right)c a\right]\hat{g}_{t}+\omega\exp\left(\mu\right)\hat{g}_{t}. $$

Arranging terms yields

$$ c\hat{a}_{t}=\omega_{\mu}^{*}c\hat{a}_{t-1}+\left(1-\omega\right)\hat{s}_{t,0}+\omega\left(1-c a\right)\hat{g}_{t}, $$

where $ \omega_{\mu}^{*}=\omega(1+r)/\exp(\mu)=\omega(1+r)/(1+\mu). $

We would like to rewrite the last two terms of (35). As a preparation, note that $ s_{t,0} = 1 - X_{t}^{p} / X_{t} $ since $ S_{t,0} = X_{t} - X_{t}^{p} $ . Here, we have

$$ \frac{X_{t}^{p}}{X_{t}}=\frac{r}{1+r}\sum_{j=0}^{\infty}\left(\frac{1}{1+r}\right)^{j}\mathbb{E}_{t}\left(\frac{X_{t+j}}{X_{t}}\right). $$

Focusing on each $ X_{t+j}/X_{t} $ inside the summation, we have the first-order Taylor expansions of $ X_{t+j}/X_{t} $ :
$$ \begin{array}{rcl}\displaystyle\frac{X_{t}}{X_{t}}&=&\left(1+\mu\right)^{0}\times1\\\displaystyle\frac{X_{t+1}}{X_{t}}&\simeq&\left(1+\mu\right)^{1}\times\left(1+\hat{g}_{t+1}\right)\\\displaystyle\frac{X_{t+2}}{X_{t}}&\simeq&\left(1+\mu\right)^{2}\times\left(1+\hat{g}_{t+1}+\hat{g}_{t+2}\right)\\&&\cdots\\\displaystyle\frac{X_{t+j}}{X_{t}}&\simeq&\left(1+\mu\right)^{j}\times\left(1+\hat{g}_{t+1}+\hat{g}_{t+2}+\ldots+\hat{g}_{t+j}\right).\end{array} $$

Substituting these approximants into $ X_{t}^{p}/X_{t} $ in (36), we have

$$ \begin{align*}\frac{X_{t}^{p}}{X_{t}}\quad&\simeq\quad\frac{r}{1+r}\sum_{j=0}^{\infty}\left(\frac{1+\mu}{1+r}\right)^{j}+\frac{r}{1+r}\sum_{j=1}^{\infty}\left(\frac{1+\mu}{1+r}\right)^{j}\mathbb{E}_{t}\hat{g}_{t+1}+\frac{r}{1+r}\sum_{j=2}^{\infty}\left(\frac{1+\mu}{1+r}\right)^{j}\mathbb{E}_{t}\hat{g}_{t+2}+\ldots\\&=\quad\frac{r}{r-\mu}+\frac{r}{r-\mu}\left(\frac{1+\mu}{1+r}\right)\mathbb{E}_{t}\hat{g}_{t+1}+\frac{r}{r-\mu}\left(\frac{1+\mu}{1+r}\right)^{2}\mathbb{E}_{t}\hat{g}_{t+2}+\ldots\\&=\quad\frac{r}{r-\mu}+\frac{r}{r-\mu}\sum_{j=1}^{\infty}\left(\frac{1+\mu}{1+r}\right)^{j}\mathbb{E}_{t}\hat{g}_{t+j}\\&=\quad\frac{r}{r-\mu}+\frac{r}{r-\mu}\frac{\phi}{(1+r)/\left(1+\mu\right)-\phi}\hat{g}_{t},\end{align*} $$

where the last line is due to AR(1) process of $ \hat{g}_{t} $ .

Equation (37) provides the expression of the second term of (35). Since $ \hat{s}_{t,0}=1-X_{t}^{p}/X_{t}-ca $ , the second term of (35) is given by

$$ \left(1-\omega\right)\hat{s}_{t,0}=-\left(1-\omega\right)\frac{r}{r-\mu}\frac{\phi}{\left(1+r\right)/\left(1+\mu\right)-\phi}\hat{g}_{t}. $$

Plugging in (34) and (38) into (35) yields

$$ \begin{align*}c\hat{a}_{t}\quad=&\quad\omega_{\mu}^{*}c\hat{a}_{t-1}-\left(1-\omega\right)\frac{r}{r-\mu}\frac{\phi}{\left(1+r\right)/\left(1+\mu\right)-\phi}\hat{g}_{t}+\omega\frac{r}{r-\mu}\hat{g}_{t}\\=&\quad\omega_{\mu}^{*}c\hat{a}_{t-1}+\frac{r}{r-\mu}\left[\omega-\left(1-\omega\right)\frac{\phi}{\left(1+r\right)/\left(1+\mu\right)-\phi}\right]\hat{g}_{t}\\=&\quad\omega_{\mu}^{*}c\hat{a}_{t-1}+\frac{r}{r-\mu}\frac{\omega_{\mu}^{*}-\phi}{\left(1+r\right)/\left(1+\mu\right)-\phi}\hat{g}_{t}.\end{align*} $$

Using $ \hat{g}_{t}=\phi\hat{g}_{t-1}+\varepsilon_{t}=(1-\phi L)^{-1}\varepsilon_{t} $ , we can further rewrite the above equation as

$$ c\hat{a}_{t}=\left(\omega_{\mu}^{*}+\phi\right)c\hat{a}_{t-1}-\left(\omega_{\mu}^{*}\phi\right)c\hat{a}_{t-2}+\frac{r}{r-\mu}\frac{\omega_{\mu}^{*}-\phi}{\left(1+r\right)/\left(1+\mu\right)-\phi}\varepsilon_{t}, $$

which proves (22) in the main text. Imposing $ \mu = 0 $ also yields (14) in the main text.
### A.4 Proof of Proposition 2

#### A.4.1 Derivation of (17)

To obtain changes in the aggregate consumption $ \Delta C_{t+1} $ , we aggregate all individual changes in consumption between periods t and $ t+1 $ . Notice that consumers do not change their individual consumption if they do not realize shocks between period t and $ t+1 $ (i.e., $ C_{t+1,k+1} - C_{t,k} = 0 $ for any $ k \geq 0 $ ). Therefore, for aggregation, it suffices to consider only consumers who receive the newest information in period $ t+1 $ . When aggregating changes in their consumption, however, we need to take into account the history of information updating. For this reason, we first consider a sequence of changes in consumption $ \{C_{t+j+1,0} - C_{t+j,j}\}_{j=0}^{\infty} $ , given the foreign asset holdings $ A_{t} $ . This sequence consists of changes in consumption for an individual consumer, conditional on the fact that he receives no additional information between t and $ t+j+1 $ . We derive the expression for the changes in individual consumption as accumulated forecast errors. We then cross-sectionally aggregate changes in individual consumption differing across histories of individual information updating. The aggregation is made based on the distribution of information updating.

Consider a sequence of changes in consumption $ \{C_{t+j+1,0}-C_{t+j,j}\}_{j=0}^{\infty} $ . The simplest is the case of j=0 (i.e., $ C_{t+1,0}-C_{t,0} $ ). If a consumer with information in period t updates the information in period $ t+1 $ , the changes in consumption are the same as that in the RE model:

$$ \begin{align*}C_{t+1,0}-C_{t,0}\ =&r\left(A_{t+1}-A_{t}\right)+\left(X_{t+1}^{p}-X_{t}^{p}\right)\\=&rS_{t,0}+\left(X_{t+1}^{p}-X_{t}^{p}\right)\\=&rX_{t}+\left(X_{t+1}^{p}-\left(1+r\right)X_{t}^{p}\right)\\=&\ \Delta\mathbb{E}_{t+1}X_{t+1}^{p},\end{align*} $$

where the first equality uses the optimal consumption given by $ C_{t,0} = rA_{t} + X_{t}^{p} $ (See (6) in the main text); the second equality arises from $ A_{t+1} - A_{t} = S_{t,0} $ ; the third equality is from $ S_{t,0} = X_{t} - X_{t}^{p} $ ; and the final equality can be obtained from the definition of the non-financial permanent income $ X_{t}^{p} = r / (1 + r) \sum_{j=0}^{\infty} (1 + r)^{-j} \mathbb{E}_{t} X_{t+j} $ . Equation (40) indicates that changes in consumption in period $ t + 1 $ equal unrecognized changes in permanent income.

Next, consider the case of j = 1. Suppose that a consumer with information in period t does not update his information set in period $ t + 1 $ , but updates the information in period $ t + 2 $ . Changes in consumption between periods $ t + 1 $ and $ t + 2 $ can analogously be calculated as

$$ \begin{align*}C_{t+2,0}-C_{t+1,1}\ &=\ C_{t+2,0}-C_{t,0}\\&=\ r\left(A_{t+2}-A_{t+1}+A_{t+1}-A_{t}\right)+\left(X_{t+2}^{p}-X_{t+1}^{p}\right)+\left(X_{t+1}^{p}-X_{t}^{p}\right)\\&=\ r S_{t+1,1}+\left(X_{t+2}^{p}-X_{t+1}^{p}\right)+r S_{t,0}+\left(X_{t+1}^{p}-X_{t}^{p}\right)\\&=\ r S_{t+1,1}+\left(X_{t+2}^{p}-X_{t+1}^{p}\right)+\Delta\mathbb{E}_{t+1}X_{t+1}^{p}.\end{align*} $$
Here, the third equality is based on the fact that $ A_{t+2} - A_{t+1} = S_{t+1,1} $ because the consumer does not update information in period $ t + 1 $ . The last equality can be derived from (40). Furthermore, the first two terms of the right-hand side can be rewritten as

$$ \begin{align*}rS_{t+1,1}+\left(X^{p}_{t+2}-X^{p}_{t+1}\right)\ &=\ r\left[\left(1+r\right)S_{t,0}+\Delta X_{t+1}\right]+\left(X^{p}_{t+2}-X^{p}_{t+1}\right)\\&=\ r\left[rX_{t}+X^{p}_{t+1}-\left(1+r\right)X^{p}_{t}\right]+\left[rX_{t+1}+\left(X^{p}_{t+2}-\left(1+r\right)X^{p}_{t+1}\right)\right]\\&=\ r\Delta\mathbb{E}_{t+1}X^{p}_{t+1}+\Delta\mathbb{E}_{t+2}X^{p}_{t+2},\end{align*} $$

where the first equality is due to (32); the second equality is obtained from arranging terms with $ S_{t,0}=X_{t}-X_{t}^{p} $ ; and the third equality is due to (40). Substituting the above equation into (41), we obtain changes in consumption when the consumer with information in period t updates information in period $ t+2 $ :

$$ C_{t+2,0}-C_{t+1,1}=\Delta\mathbb{E}_{t+2}X_{t+2}^{p}+(1+r)\Delta\mathbb{E}_{t+1}X_{t+1}^{p}. $$

This equation indicates that, if he updates the information in period $ t+2 $ but did not update it in period $ t+1 $ , the changes in consumption include the unrecognized changes in permanent income in both periods $ t+1 $ and $ t+2 $ .

For the general case of j > 1, we repeat the above calculation. The general expression of changes in individual consumption is

$$ C_{t+j,0}-C_{t+j-1,j-1}=\Delta\mathbb{E}_{t+j}X_{t+j}^{p}+(1+r)\Delta\mathbb{E}_{t+j-1}X_{t+j-1}^{p}+\ldots+(1+r)^{j-1}\Delta\mathbb{E}_{t+1}X_{t+1}^{p},\mathrm{f o r}j=1,2,. $$

We now aggregate the consumption change between t and $ t+1 $ across all consumers, using the distribution of inattentive consumers: $ C_{t} = (1 - \omega) \sum_{k=0}^{\infty} \omega^{k} C_{t,k} $ . Recall that it is sufficient to consider only consumers who receive the newest information in period $ t+1 $ . The fraction of information-updating consumers is $ 1 - \omega $ . Therefore, changes in the aggregate consumption become

$$ \begin{align*}\Delta C_{t+1}\ =&\ \left(1-\omega\right)^{2}\left\{\Delta\mathbb{E}_{t+1}X_{t+1}^{p}\right.\ \left.\quad(44)\right.\\&\ +\omega\left[\Delta\mathbb{E}_{t+1}X_{t+1}^{p}+\left(1+r\right)\Delta\mathbb{E}_{t}X_{t}^{p}\right]\\&\ +\omega^{2}\left[\Delta\mathbb{E}_{t+1}X_{t+1}^{p}+\left(1+r\right)\Delta\mathbb{E}_{t}X_{t}^{p}+\left(1+r\right)^{2}\Delta\mathbb{E}_{t-1}X_{t-1}^{p}\right]\\&\ +\omega^{3}\left[\Delta\mathbb{E}_{t+1}X_{t+1}^{p}+\left(1+r\right)\Delta\mathbb{E}_{t}X_{t}^{p}+\left(1+r\right)^{2}\Delta\mathbb{E}_{t-1}X_{t-1}^{p}+\left(1+r\right)^{3}\Delta\mathbb{E}_{t-2}X_{t-2}^{p}\right]+\ldots\Bigg\}.\end{align*} $$

In the above equation, the first line in the curly brackets represents changes in consumption for consumers whose period-t consumption was $ C_{t,0} $ . The second line points to that for consumers whose period-t consumption was $ C_{t,1} (= C_{t-1,0}) $ . We can understand the remaining lines analogously. This
equation can further be simplified to

$$ \begin{align*}\Delta C_{t+1}\ =&\ \left(1-\omega\right)\left[\left(1-\omega\right)\sum_{k=0}^{\infty}\omega^{k}\Delta\mathbb{E}_{t+1}X_{t+1}^{p}\right.\\&\left.+\left(1+r\right)\omega\left(1-\omega\right)\sum_{k=0}^{\infty}\omega^{k}\Delta\mathbb{E}_{t}X_{t}^{p}\right.\\&\left.+\left(1+r\right)^{2}\omega^{2}\left(1-\omega\right)\sum_{k=0}^{\infty}\omega^{k}\Delta\mathbb{E}_{t-1}X_{t-1}^{p}+\ldots\right]\\=&\ \left(1-\omega\right)\left[\Delta\mathbb{E}_{t+1}X_{t+1}^{p}+\omega^{*}\Delta\mathbb{E}_{t}X_{t}^{p}+(\omega^{*})^{2}\Delta\mathbb{E}_{t-1}X_{t-1}^{p}+\ldots\right]\\=&\ \frac{1-\omega}{1-\omega^{*}L}\Delta\mathbb{E}_{t+1}X_{t+1}^{p},\end{align*} $$

where the second equality comes from $ (1-\omega)\sum_{k=0}^{\infty}\omega^{k}=1 $ . By multiplying $ 1-\omega^{*}L $ by both sides and arranging terms, we obtain (17) in the main text:

$$ \Delta C_{t+1}=\omega^{*}\Delta C_{t}+\left(1-\omega\right)\Delta\mathbb{E}_{t+1}X_{t+1}^{p}. $$

#### A.4.2 Derivation of (18)

Turning to the derivation of (18), divide both sides of (17) by $ X_{t} $ :

$$ \Delta c_{t+1}=\omega^{*}\exp\left(-g_{t}\right)\Delta c_{t}+\left(1-\omega\right)\Delta\mathbb{E}_{t+1}\left(\frac{X_{t+1}^{p}}{X_{t}}\right). $$

We approximate each term in the right-hand side of (46). First, the first-order Taylor approximation of the first term is

$$ \omega^{*}\exp\left(-g_{t}\right)\Delta c_{t}\simeq\omega^{*}\exp\left(-\mu\right)\left(\Delta c\right)+\omega^{*}\exp\left(-\mu\right)\left(\Delta c_{t}-\Delta c\right)-\frac{\omega^{*}}{1+\mu}\left(\Delta c\right)\hat{g}_{t}, $$

where $ \Delta c $ is the steady state value of changes in consumption (divided by net output). In the steady state, however, $ \Delta c = 0 $ . Therefore, noting that $ \exp(-\mu) = (1 + \mu)^{-1} $ and $ \omega_{\mu}^{\star} \equiv \omega(1 + r) / (1 + \mu) $ , we can simplify the above equation as

$$ \omega^{*}\exp\left(-g_{t}\right)\Delta c_{t}\simeq\omega_{\mu}^{*}\Delta c_{t}. $$