I working a lot with Epstein-Zin utility (standard in asset pricing models). But I am having some issues wrapping my head around some intuition for how this utility function works.

Let's think about a standard problem of maximization of utility of a consumer who lives two periods. The agent is endowed with wealth of $w_1$ in period 1, and receives an income of $y_2$ in period. His savings for period 2 are given by $w_2$. The risk-free rate is $r_f$.

$$max_{c_t} u(c_t)$$

subject to: $w_2 = (w_1 - c_1)(1+r_f) + y_1$.

Now assume two utility functions:

  1. the value function with CRRA utility: $$ V_t = \frac{C_t^{1-\gamma}}{1-\gamma} + \beta E_t[V_{t+1}]$$

  2. Epstein-zin utility: $$ V_t = \bigg \{(1-\beta) C_t^{1-1/\psi}+\beta E_t[V_{t+1}^{1-\gamma}]^{\frac{1-1/\psi}{1-\gamma}} \bigg \}^{\frac{1}{1-1/\psi}} $$

Now here's my confusion. Let say I start solving this model by backward induction as is standard in the literature. On the last period of life ($t=2$) agents just consume all wealth available to them ($w^2$).

  1. The utility function under the CRRA case takes the shape (on the last period): $$V_2 = \frac{w_2^{1-\gamma}}{1-\gamma}$$

  2. The utility function under the EZ case takes the shape (on the last period): $$V_2 = (1-\beta)^{\frac{1}{1-1/\psi}} w_2$$

So in some sense in the first case the utility function is concave in wealth whereas on the latter case it is linear in wealth. So they have dramatically different shapes. This is even more puzzling if one things that if we set $\psi$ (the elasticity of intertemporal substitution) to be $\frac{1}{\gamma}$, the EZ utility should collapse to the CRRA case.

What am I missing?


1 Answer 1


While the concavity of a felicity function $u$ affects choices a value function is not unique up to monotone transformations. It is an ordinal and not a cardinal concept. Consider your CRRA equation


$$\Leftrightarrow V_t^{\frac{1}{1-\gamma}}=\left(\frac{C_t^{1-\gamma}}{1-\gamma}+\beta\mathbb{E}_t[V_{t+1}]\right)^{\frac{1}{1-\gamma}}$$

Setting $\tilde{V}_t=V_t^{\frac{1}{1-\gamma}}$


Now $\tilde{V}_2$ is also linear but the equation defines the same ranking towards different policies than the original equation.

As you mentioned, EZ corresponds to CRRA in the special case $\gamma=\frac{1}{\psi}$. Plugging in

$$ V_t = \bigg \{(1-\beta) C_t^{1-\gamma}+\beta E_t[V_{t+1}^{1-\gamma}] \bigg \}^{\frac{1}{1-\gamma}} $$

Now at first sight this might look like different preferences than your standard CRRA specification. But let us rewrite the problem:

$$ V_t^{1-\gamma} = \bigg \{(1-\beta) C_t^{1-\gamma}+\beta E_t[V_{t+1}^{1-\gamma}] \bigg \}$$


$$ V_t^{1-\gamma}(1-\beta)^{-1}(1-\gamma)^{-1} = \bigg \{ \frac{C_t^{1-\gamma}}{1-\gamma}+\beta E_t[V_{t+1}^{1-\gamma}(1-\beta)^{-1}(1-\gamma)^{-1}] \bigg \}$$

Define $\hat{V}_t= V_t^{1-\gamma}(1-\beta)^{-1}(1-\gamma)^{-1}$. You get

$$ \hat{V}_t = \bigg \{ \frac{C_t^{1-\gamma}}{1-\gamma}+\beta E_t[\hat{V}_{t+1}] \bigg \}$$

Which has exactly the same mathematical form as your CRRA problem. From this you should realize that this is just your original CRRA problem but written in a slightly different way.

  • $\begingroup$ I understand that it keeps the same ranking. However, marginal utility of wealth is linear in the second case and concave in the first ... (just take derivatives with respect to wealth on my last two equations ... ) $\endgroup$
    – phdstudent
    Commented Feb 15, 2022 at 20:46
  • $\begingroup$ @phdstudent Concavity of a felicity function matters for choices but value functions are not unique up to monotone transformations. Also calling the derivative of a value function the "marginal utility of wealth" does not make sense. You don't really care about numerical values of a value function. $\endgroup$
    – fes
    Commented Feb 16, 2022 at 6:49
  • $\begingroup$ The only property you can really deduce from your value functions is that the agent prefers more wealth to less wealth. Because you could write the CRRA problem in a way that the value function is linear, concavity of this value function does not mean anything. $\endgroup$
    – fes
    Commented Feb 16, 2022 at 7:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.