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Currently I am working out advanced micro problems that deal with household's optimal savings behavior. Although I think I understand the material quite well, I am not sure about the specific detail of the solution of the following (shortened) problem.

Let $u: \mathbb{R}_+ \to \mathbb{R}_+, c_i \mapsto c_i^2$ be the (consumption) utility function for the two time periods $i \in \{1,2\}$. Assume that the income $y_2$ is risky with riskiness represented as $\beta$ and $\beta$ being a mean-preserving spread. What can you say about the change in optimal savings if $\beta$ increases?

My thoughts about this are as follows. What we are looking for here is the sign of $$\frac{ds^*}{d\beta}$$

where $s^*$ denotes optimal savings. In the lecture we showed that its sign only depends on the sign of $$K = \frac{d\mathbb{E}[u']}{d\beta}$$ Since $\beta$ is a meaning-preserving spread we then said that the sign of $K$ only depends on the sign of $u'''$, i.e. that $K < 0$ if $u''' < 0$ and vice versa but we omitted the case $K = 0$. But since $u(c_i) = c_i^2$ it is $u''' = 0$.

My questions are thus

  1. What is the conclusion if $u''' = 0$? Is it the same as the positive (> 0) case?

  2. Why exactly do we, if $\beta$ is a MPS, only need to care about the argument's second derivative in order to determine the sign of the expectation?

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  • $\begingroup$ Are you saying that $y_2$ has a distribution $F$, which is a mean preserving spread of some other distribution $G$? If so, what is $\beta$? Is it the variance of $F$? Or is it the discount factor for period 2's utility? $\endgroup$ – Herr K. May 31 '17 at 20:01
  • $\begingroup$ Yes, in the lecture we introduced the density for the variable $y_2$ as $f(\beta, y_2)$ where $\beta$ is a parameter of the function which may serve as a mean-preserving spread. $\endgroup$ – Taufi May 31 '17 at 20:24
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Your question is essentially about precautionary savings, i.e. the response of savings to risk.

What determines precautionary savings is the coefficient of relative prudence: $CRP=-\frac{U'''}{U''}C$, where $U$ is the utility function, $C$ is consumption and primes denote derivatives. As we always have decreasing marginal utility in economics, i.e. $U''<0$, then the sign of the savings will depend on the sign of the third derivative, as you mentioned.

Think of this coefficient as similar to the risk aversion coefficients. The difference is that prudence means that agents act in anticipation of risk by saving more, whereas risk aversion reflects how agents react when presently faced with risk.

With quadratic utility, the coefficient of relative prudence is zero and there are no precautionary savings. Hence there is no effect of risk on savings. With such utility we therefore have certainty equivalence, which is quite convenient and hence often used. Mean preserving spreads don't matter then. So no, it is not the same as in the positive case, where savings would go up in resonse to risk.

What is crucial here, as you noted, is the third derivative. The coefficient really only uses the second derivative as a normalization.

A mean-preserving spread keeps the mean/expected value of consumption the same. However, it does not neccssarily keep the derivatives (marginal utility) the same, which are what matter for (savings) decisions. If the marginal utility is linear (quadratic utility), then a mean-preserving spread keeps the expected marginal utilities the same as well. Hence, there are no precautionary savings (no reaction) in that case.

If the marginal utilities are convex, then the mean-preserving spread changes things in a way that matters, i.e. it becomes beneficial to save more. This follows from Jensen's inequality. If it is concave, the opposite is true. So the relevant function to look at is $U'$, as this is how the agent makes decisions. When we want to see whether a function is convex or concave we look at the second derivative. So the second derivative of $U'$ is $U'''$, which is why the third derivative matters here.

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  • $\begingroup$ Thanks so much for the time you took to write this answer up. It really cleared things for me. $\endgroup$ – Taufi Jun 1 '17 at 16:24

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