<p>No, we don't ignore Jensen's Inequality. You made a mistake in taking the derivative with respect to <span class="math-container">$C_{t+1}$</span>. The household needs to decide at time <span class="math-container">$t+i$</span>. Start with</p>
<p><span class="math-container">$$\mathcal{L} = E_t \sum_{i=0}^{\infty} \left[\beta^i u(C_{t+i}, N_{t+i}) + \lambda_{t+i} (B_{t-1+i} + W_{t+i}N_{t+i}-T_{t+i}-P_{t+i}C_{t+i}-Q_{t+i}B_{t+i}) \right]$$</span></p>
<p>This Lagrangian yields the first-order conditions:</p>
<p><span class="math-container">\begin{gather}
\frac{\partial\mathcal{L}} {\partial C_{t+i}} = E_{t+i}u'_{C_{t+i}}-E_{t+i}P_{t+i}\lambda_{t+i}=0 \tag{1} \: \forall \: i\\
\frac{\partial\mathcal{L}} {B_{t-1+i}} = -E_{t+i}\lambda_{t+i} Q_{t+i} + E_{t+i}[\lambda_{t+i+1}]=0 \rightarrow Q_{t+i} = E_{t+i}\left[\frac{\lambda_{t+i+1}} {\lambda_{t+i}}\right] \tag{3}
\end{gather}</span></p>
<p>Now you use (1) to substitute in (3)
<span class="math-container">$$
Q_{t+i} = E_{t+i}\left[\frac{\lambda_{t+i+1}} {\lambda_{t+i}}\right] =E_{t+i}\left[\frac{E_{t+1+i}u'_{C_{t+1+i}}} {E_{t+i}u'_{C_{t+i}}}\right]=E_{t+i}\left[\frac{u'_{C_{t+1+i}}} {u'_{C_{t+i}}}\right]
$$</span>
where the last step used the law of iterated expectations <span class="math-container">$E_t[E_{t+1}[X_{t+1}]]=E_t[X_{t+1}]$</span>.</p>

CC BY-SA 4.0
2024-12-02T14:41:48.540Z
Firstly, you formulated your Lagrangian to consist of expected values with information set available in period t (i.e. with E_t). Why you switch to the E_{t + i} in (1) and (2)? Secondly, in (3) you use the fact that E_{t+i} (\lambda_{t + i + 1}) / E_{t+i} (\lambda_{t + i}) = E_{t + i} [\lambda_{t + i + 1} / \lambda_{t + i}]. Why does it hold?
John Stock
CC BY-SA 4.0
2024-12-02T17:34:55.777Z
Firstly, because the actual choice for consumption is made at each point in time t+i. That's why (1) has an $\forall i$. At time t, the household when the household chooses $C_t$, it needs to forecast the decisions it is going to make later on, which is when these FOCs are needed. Secondly, because $\lambda_{t+i}$ is contained in the information set at time $t+i$. It is not a random variable anymore and can be treated as a constant.
jpfeifer
CC BY-SA 4.0
2024-12-10T12:15:59.783Z
Thanks, I was unable to find such explanation in textbooks
John Stock
CC BY-SA 4.0
2024-11-24T18:06:52.710Z
2024-11-24T18:06:52.710Z
jpfeifer
<p>There is a standard Euler equation derivation for household utility maximization problem:</p>
<p><span class="math-container">$$max_{\{C_i,N_i, B_i\}_{i=0}^{\infty}} \quad U=E_0\sum_{t=0}^{\infty} \beta^t u(C_t, N_t)$$</span></p>
<p><span class="math-container">$$s.t. \quad P_tC_t+Q_tB_t=B_{t-1}+W_tN_t-T_t$$</span></p>
<p>If we change our initial point of time from zero to <span class="math-container">$t$</span>, then Lagrangian for the problem is:</p>
<p><span class="math-container">$$\mathcal{L} = E_t \sum_{i=0}^{\infty} \left[\beta^i u(C_{t+i}, N_{t+i}) + \lambda_{t+i} (B_{t-1+i} + W_{t+i}N_{t+i}-T_{t+i}-P_{t+i}C_{t+i}-Q_{t+i}B_{t+i}) \right]$$</span></p>
<p>This Lagrangian yields first order conditions:</p>
<p><span class="math-container">\begin{gather}
\frac{\partial\mathcal{L}} {\partial C_t} = u'_{C_t}-P_t\lambda_t=0 \tag{1} \\
\frac{\partial\mathcal{L}} {\partial C_{t+1}} = \beta u'_{C_{t+1}} - E_t(P_{t+1}\lambda_{t+1})=0 \tag{2} \\
\frac{\partial\mathcal{L}} {B_t} = -\lambda_t Q_t + E_t[\lambda_{t+1}]=0 \rightarrow Q = \frac {E_t[\lambda_{t+1}]} {\lambda_t} \tag{3}
\end{gather}</span></p>
<p>In (1-2) we can move term with <span class="math-container">$P_i$</span> to the RHS and then divide and we get to the:</p>
<p><span class="math-container">\begin{gather}
\frac {u'_{C_t}} {\beta E_t u'_{C_{t+1}}} = \frac {\lambda_t P_t} {E_t(\lambda_{t+1} P_{t+1})} \tag {4}
\end{gather}</span></p>
<p>It is usual to substitute expression for Q from (3) in (4) to get the Euler equation.</p>
<p><span class="math-container">$$1 = \beta Q^{-1} E_t \{\frac {u'_{C_{t+1}}} {u'_{C_t}} \frac {P_t} {P_{t+1}} \}$$</span></p>
<p>However it supposes that <span class="math-container">$E_t(\lambda_{t+1} P_t) = E_t\lambda_{t+1} E_t P_{t+1}$</span>, which means that <span class="math-container">$\lambda_{i}$</span> and <span class="math-container">$P_i$</span> are uncorrelated. Is this a standart premise in DSGE literature or I am missing something?</p>

<utility><dsge>
Expectations operator in Euler equation
John Stock
# Expectations operator in Euler equation

(Asked by: username_0 on 2024-11-23)

There is a standard Euler equation derivation for household utility maximization problem:

$$max\_{\{C\_i,N\_i, B\_i\}\_{i=0}^{\infty}} \quad U=E\_0\sum\_{t=0}^{\infty} \beta^t u(C\_t, N\_t)$$

$$s.t. \quad P\_tC\_t+Q\_tB\_t=B\_{t-1}+W\_tN\_t-T\_t$$

If we change our initial point of time from zero to $t$, then Lagrangian for the problem is:

$$\mathcal{L} = E\_t \sum\_{i=0}^{\infty} \left[\beta^i u(C\_{t+i}, N\_{t+i}) + \lambda\_{t+i} (B\_{t-1+i} + W\_{t+i}N\_{t+i}-T\_{t+i}-P\_{t+i}C\_{t+i}-Q\_{t+i}B\_{t+i}) \right]$$

This Lagrangian yields first order conditions:

\begin{gather}
\frac{\partial\mathcal{L}} {\partial C\_t} = u'\_{C\_t}-P\_t\lambda\_t=0 \tag{1} \\
\frac{\partial\mathcal{L}} {\partial C\_{t+1}} = \beta u'\_{C\_{t+1}} - E\_t(P\_{t+1}\lambda\_{t+1})=0 \tag{2} \\
\frac{\partial\mathcal{L}} {B\_t} = -\lambda\_t Q\_t + E\_t[\lambda\_{t+1}]=0 \rightarrow Q = \frac {E\_t[\lambda\_{t+1}]} {\lambda\_t} \tag{3}
\end{gather}

In (1-2) we can move term with $P\_i$ to the RHS and then divide and we get to the:

\begin{gather}
\frac {u'\_{C\_t}} {\beta E\_t u'\_{C\_{t+1}}} = \frac {\lambda\_t P\_t} {E\_t(\lambda\_{t+1} P\_{t+1})} \tag {4}
\end{gather}

It is usual to substitute expression for Q from (3) in (4) to get the Euler equation.

$$1 = \beta Q^{-1} E\_t \{\frac {u'\_{C\_{t+1}}} {u'\_{C\_t}} \frac {P\_t} {P\_{t+1}} \}$$

However it supposes that $E\_t(\lambda\_{t+1} P\_t) = E\_t\lambda\_{t+1} E\_t P\_{t+1}$, which means that $\lambda\_{i}$ and $P\_i$ are uncorrelated. Is this a standart premise in DSGE literature or I am missing something?

**Tags:** utility, dsge

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## 1 answers

### Answer by username_1 on 2024-11-24. Score: 1 | Accepted answer

No, we don't ignore Jensen's Inequality. You made a mistake in taking the derivative with respect to $C\_{t+1}$. The household needs to decide at time $t+i$. Start with

$$\mathcal{L} = E\_t \sum\_{i=0}^{\infty} \left[\beta^i u(C\_{t+i}, N\_{t+i}) + \lambda\_{t+i} (B\_{t-1+i} + W\_{t+i}N\_{t+i}-T\_{t+i}-P\_{t+i}C\_{t+i}-Q\_{t+i}B\_{t+i}) \right]$$

This Lagrangian yields the first-order conditions:

\begin{gather}
\frac{\partial\mathcal{L}} {\partial C\_{t+i}} = E\_{t+i}u'\_{C\_{t+i}}-E\_{t+i}P\_{t+i}\lambda\_{t+i}=0 \tag{1} \: \forall \: i\\
\frac{\partial\mathcal{L}} {B\_{t-1+i}} = -E\_{t+i}\lambda\_{t+i} Q\_{t+i} + E\_{t+i}[\lambda\_{t+i+1}]=0 \rightarrow Q\_{t+i} = E\_{t+i}\left[\frac{\lambda\_{t+i+1}} {\lambda\_{t+i}}\right] \tag{3}
\end{gather}

Now you use (1) to substitute in (3)
$$
Q\_{t+i} = E\_{t+i}\left[\frac{\lambda\_{t+i+1}} {\lambda\_{t+i}}\right] =E\_{t+i}\left[\frac{E\_{t+1+i}u'\_{C\_{t+1+i}}} {E\_{t+i}u'\_{C\_{t+i}}}\right]=E\_{t+i}\left[\frac{u'\_{C\_{t+1+i}}} {u'\_{C\_{t+i}}}\right]
$$
where the last step used the law of iterated expectations $E\_t[E\_{t+1}[X\_{t+1}]]=E\_t[X\_{t+1}]$.

> **Comments:**
> **username_0**: Firstly, you formulated your Lagrangian to consist of expected values with information set available in period t (i.e. with E_t). Why you switch to the E_{t + i} in (1) and (2)? Secondly, in (3) you use the fact that E_{t+i} (\lambda_{t + i + 1}) / E_{t+i} (\lambda_{t + i}) = E_{t + i} [\lambda_{t + i + 1} / \lambda_{t + i}]. Why does it hold?  
> **username_1**: Firstly, because the actual choice for consumption is made at each point in time t+i. That's why (1) has an $\forall i$. At time t, the household when the household chooses $C_t$, it needs to forecast the decisions it is going to make later on, which is when these FOCs are needed. Secondly, because $\lambda_{t+i}$ is contained in the information set at time $t+i$. It is not a random variable anymore and can be treated as a constant.  
> **username_0**: Thanks, I was unable to find such explanation in textbooks  

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