# How to derive the standard euler equation from the Expanding Variety Model

In the standard expanding variety model where the representative household faces the following CRRA utility function

$$\int_{0}^{\infty} \exp(-\rho t) \frac{C(t)^{1-\theta}-1}{1-\theta}dt$$

and final goods that is produced competitively with the following production $$Y(t) = \frac{1}{1-\beta} \left( \int_{0}^{N(t)}x(v,t)^{1-\beta}dv\right)L^{\beta}$$

with the following resource constraint

$$C(t) + X(t) + Z(t) \leq Y(t)$$

where $$X(t)$$ is the total spending on machines, and $$Z(t)$$ is the total expenditure on research and development.

The innovation possibilities frontier takes the form

$$\dot{N}(t) = \eta Z(t)$$

The intermediary goods firm faces the following problem where

$$r(t) V(v,t) - \dot{V}(v,t) = \pi(v,t)$$

The thing I don't understand about the model is how the Euler Equation is derived. In almost everywhere I look the Euler Equation is the standard

$$\frac{\dot{C}(t)}{C(t)} = \frac{1}{\theta}(r(t)-\rho)$$

I don't understand how you would derive this equation.

• Hint: set up the Hamiltonian Jul 5, 2020 at 17:19

## 1 Answer

The way you would go about solving this problem is as the ChinG said is by setting up the Hamiltonian. In this case this is:

$$\mathcal{H}:e^{-\rho t} \frac{C(t)^{1-\theta}-1}{1-\theta}+\mu(t)\left[Y(t)-C(t)-X(t)-Z(t)\right]$$

Taking the first order condition for this problem we get: $$\frac{\partial \mathcal{H}}{\partial C(t)}:e^{-\rho t}C(t)^{-\theta}-\mu(t)=0$$ or $$e^{-\rho t}C(t)^{-\theta}=\mu(t)$$

Taking the natural logs of this equation and differentiating with respect to time we get: $$\ln\left[\frac{\partial \mathcal{H}}{\partial C(t)}\right]:-\rho t -\theta \ln [C(t)]=\ln[\mu(t)]$$ $$\frac{\partial\ln\left[\mathcal{H}_{C(t)}\right]}{\partial t}:-\rho-\theta\frac{\dot{C}(t)}{C(t)}=-\frac{\dot{\mu}(t)}{\mu(t)}$$ Rearranging the above and noting that $$\frac{\dot{\mu}(t)}{\mu(t)}=r(t)$$ we get: $$\frac{\dot{C}(t)}{C(t)}=\frac{1}{\theta}(r(t)-\rho)$$ Which is the Euler equation.

Hope this helps.

• Hi, thanks for the response! On thing I don't quite understand is why $\dot{\mu}/\mu = r(t)$? Jul 8, 2020 at 14:55
• @Finalblue Theres a ton of math to get that result so I tried to be sneaky and handwave a bit. I could tell you that intuitively we think of $\mu$ as the marginal utility from output over time. We interpret the growth in that utility from output over time $\dot{\mu}/\mu$ as the rate of return on the market $r(t)$.
– EconJohn
Jul 9, 2020 at 1:37