Expectations operator in Euler equation

Question

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?

Answer 1

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 $E_t[E_{t+1}[X_{t+1}]]=E_t[X_{t+1}]$.