Relationship between the Lagrangian and consumption euler equation?

Question

I think that the best way for me to ask my question would be to start with an example.

Suppose that consumers have a two-period horizon and their instantaneous utility is: $\:$

$U(C_{t})=ln \: C_{t}$

$\:$

Where $C_{t}>0$$\:$ denotes consumption. Assume that agents supply a fixed amount of labour $L$ and have no initial bonds $B$ or capital $K$. Agents discount utility of the second period with the discount factor $1>\beta>0$. $\:$

The budget constraint is: $\:$

$C_{1}+\frac{C_{2}}{1+r_{1}}$ =$\:$ $(w/P)_{1}$$L$ $\:$ + $\frac{(w/P)_{2}L}{1+r_{1}}$ $\:$

The Lagrangian is: $\:$

$\mathcal L = \ln C_1 + \beta \ln C_2 - \mu \left[ C_1 + \frac{C_2}{1+r_1} -(w / P)_{1} L - \frac{(w/P)_{2}L}{1+r_{1}} \right]$

I now have two questions:

1. Assuming that I compute the correct FOCs, how do I go about deriving the consumption euler equation? $\:$

2. Could you guys please help me with the formatting of my lagrangian? I want to make the square parentheses large enough to encapsulate everything. $\:$

Thanks.

Answer 1

In this simple problem, using a Lagrangean is an overkill, direct substitution of the constraint is perhaps better.

Anyway, treating the two consumption levels as two distinct decision variables under the budget constraint, the Euler equation emerges from the combination of the two first-order conditions:

$FOC$'s

$$\frac {\partial \mathcal L}{\partial C_1} = 0 \Rightarrow \frac 1{C_1} = \mu$$

$$\frac {\partial \mathcal L}{\partial C_2} = 0 \Rightarrow \frac {\beta}{C_2} = \mu\frac {1}{1+r_1}$$

Substitute the first into the second

$$\frac {\beta}{C_2} = \frac 1{C_1}\frac {1}{1+r_1}$$

re-arrange and you 're done.

I would suggest to also do that with a general utility function, where in the Euler equation the first derivatives of the utility will appear.

Answer 2

$\mathcal L = \ln C_1 + \beta \ln C_2 - \mu \left[ C_1 + \frac{C_2}{1+r_1} -(w / P)_{1} L - \frac{(w/P)_{2}L}{1+r_{1}} \right]$

$\frac {\partial \mathcal L}{\partial C_1}=\frac 1{C_1} - \mu = 0$

$\frac {\partial \mathcal L}{\partial C_2}= \frac{\beta}{C_2} - \mu\frac {1}{1+r_1}=0$

$\frac {\partial \mathcal L}{\partial \mu}= C_{1}+\frac{C_{2}}{1+r_{1}}-(w/P)_{1}L-\frac{(w/P)_{2}L}{1+r_{1}}=0$

These conditions characterise a maximum since the second-order condition $\frac {\partial \mathcal (ln \ C_{1} + \beta \ ln \ C_{2})}{\partial^{2} C_{1}}$ is negative.

Substitute the first FOC into the second and you get:

$C_{2} = \beta(1+r_{1})C_{1}$