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I was wondering if someone could help me clarify a result from my lecture notes. I have put them as a picture. It concerns the result on the last slide (the other three slides are included as well because it may be useful).

enter image description here

The problem is, I do not see how it follows from the Euler equation that $MRS=1+r$.

So from the lecture notes, the Euler equation is

\begin{equation*} u'(C_1) = \beta(1+r)u'(C_2) \end{equation*} This is the first order condition for an optimisation problem with respect to $C_1$ so really we have \begin{equation*} u_{C_1}(C_1) = \beta(1+r)u_{C_1}(C_2) \end{equation*} To get the marginal rate of substitution, we need $u_{C_2}(C_2)$. So somehow we must go from $u_{C_1}(C_2)$ to $u_{C_2}(C_2)$.

From the budget constraint \begin{equation*} C_1 + \frac{C_2}{1+r} = Y_1 + \frac{Y_2}{1+r} \implies C_1 = \frac{-C_2}{1+r} + Y_1 + \frac{Y_2}{1+r} \end{equation*} We expressed $C_2$ in terms of $C_1$, i.e. as a function of $C_1$, i.e. $C_2=C_2(C_1)$. Then by the chainrule \begin{equation*} \frac{\partial u}{\partial C_2} =\frac{\partial u}{\partial C_1}\frac{\partial C_1}{\partial C_2} = \frac{\partial u}{\partial C_1}\frac{-1}{1+r} \end{equation*} This is where I am stuck. I cannot get the result from the Euler equation as the lecture notes do.

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The OP's confusion comes from the fact that he or she does not apply the chain rule correctly and interprets the Euler condition

$u'(c_1) = \beta (1+r) u'(c_2)$

as

$ u_{C_1}(C_1) = \beta(1+r) u_{\boldsymbol{C_1}}(C_2) $

This is not the correct way to read the Euler condition and one should really read

$ u_{C_1}(C_1) = \beta(1+r) u_{\boldsymbol{C_2}}(C_2) $

Any doubt can be cleared by rederiving the FOC. The maximisation problem is

$\max_{C_1,C_2} u(C_1) + \beta u(C_2) \qquad s.t. \qquad $

\begin{equation*} C_1 + \frac{C_2}{1+r} = Y_1 + \frac{Y_2}{1+r} \implies C_2 = {-C_1}({1+r}) + Y_1(1+r) + {Y_2} \end{equation*}

So substituting the constraint, the problem can be written as

\begin{align}\max_{C_1} u\Big(C_1\Big) + \beta u\Big(\underbrace{{-C_1}({1+r}) + Y_1(1+r) + {Y_2}}_{=C_2}\Big)\end{align}

To find the FOC, one needs to apply the chain rule to the second period's utility function. Applying the chain rule carefully we have

\begin{align} \frac{\partial \beta u(C_2)}{\partial C_1} = \beta \frac{\partial u(C_2)}{\partial C_2} \frac{\partial C_2}{\partial C_1}\end{align}

In our case that is

\begin{align} \frac{\partial \beta u(C_2)}{\partial C_1} = \beta \frac{\partial u(C_2)}{\partial C_2} (-1)(1+r)\end{align}

Or to put it in yet another way

\begin{align}\frac{\partial \beta u(C_2)}{\partial C_1} & = \frac{\partial \big[\beta u\big({-C_1}({1+r}) + Y_1(1+r) + {Y_2}\big)\big]}{\partial C_1}\\ & = \beta \frac{\partial \big[\ u\big({-C_1}({1+r}) + Y_1(1+r) + {Y_2}\big)\big]}{\partial \big({-C_1}({1+r}) + Y_1(1+r) + {Y_2}\big)} \frac{\partial \big({-C_1}({1+r}) + Y_1(1+r) + {Y_2}\big)}{\partial C_1} \\ & = \beta \frac{\partial \big[\ u\big({-C_1}({1+r}) + Y_1(1+r) + {Y_2}\big)\big]}{\partial \big({-C_1}({1+r}) + Y_1(1+r) + {Y_2}\big)} (-1) (1+r) \\ &= \beta \frac{\partial u(C_2)}{\partial C_2} (-1)(1+r) \end{align}

Then, by definition, given a utility function $U(c_1,c_2)$, the MRS of good one with respect to good two is

$MRS := \frac{{\partial U}/{\partial c_1}}{{\partial U}/{\partial c_2}}$

In your case, as stated in your notes, you get

$MRS = \frac{{\partial u}/{\partial c_1}}{\beta{\partial u}/{\partial c_2}} = \frac{u'(c_1)}{\beta{u'(c_2)}}$

At an equilibrium, when consumers maximize utility, the Euler condition must be satisfied, that is

$u'(c_1) = \beta (1+r) u'(c_2)$

So replace this specific value for $u'(c_1)$ in the formula for the MRS and you get

$MRS = \frac{u'(c_1)}{\beta{u'(c_2)}} = \frac{\beta (1+r) u'(c_2)}{\beta{u'(c_2)}} = (1+r)$.

As a good exercise, try to convince yourself that if the MRS takes another value, then the consumer can benefit from reallocating resources between the first and the second period.

  • Thank you for the answer. But the problem is in the Euler equation ' denotes taking the derivative with respect to $C_1$. In the other equation ' denotes taking the derivative with respect to $C_2$ – user133993 Apr 21 '14 at 04:54
  • For example if we take $u(x)=x^2$ then $u(C_1)=C_1^2$ and $u(C_2)=C_2^2$ then it obviously matters whether we take the partial derivative with respect to $C_1$ or $C_2$ And they don't simply cancel with each other since C1 and C2 are not the same but are related thorough the budget constraint in the first slide – user133993 Apr 21 '14 at 05:06
  • Ok, I am re-reading your question and now I understand why you are puzzle. I'll try to help you in an edit of my answer. – Martin Van der Linden Apr 21 '14 at 13:11
  • Thank you very much for your help. This cleared up a lot of confusion for me. – user133993 Apr 21 '14 at 22:17
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    You're welcome. I checked you profile and saw that you never accepted an answer before. If you were unsatisfied by the answers you received, then you did right and you should not change anything. If on the contraty, some answer solved you problem and you have no more question about it, think of accepting the answer by checking the "v" sign below the upvote/downvote buttons. This shows to the community that the question does not need attention anymore and helps keep the site organize (as well as the site's statistics accurate). – Martin Van der Linden Apr 22 '14 at 02:24
  • I also see from you profile that you never casted a vote. Again, you are never obliged to upvote/downvote question, and you may have good reasons for never casting votes. However, if you find that some answers or questions are particularily good, feel free to upvote them (everyone can vote on any question/answer). This also helps MathSE operating smoothly. – Martin Van der Linden Apr 22 '14 at 02:27