How can I calculate with the average of the consumption in an RBC framework?

In the following open-economy RBC model consumption level ($C_t$) is not the most important for the representative households, but the difference of the consumption level and the earlier level of the average consumption ($\overline{C_{t-1}}$): $$H_t=C_t-b \overline{C_{t-1}}$$ $$0<b<1$$

The consumer's utility function: $$E_0 \sum\limits_{t=1}^{\infty}\beta^{t-1}\left( \frac{H_t^{1-\sigma}}{1-\sigma}+\frac{L_t^{1+\eta}}{1+\eta}\right)$$

After solving the Utility Maximization Problem, deriving the First Order Conditions, here's the Euler-equation: $$\beta E_t \left(\frac{C_{t+1}-b \overline{C_{t}}}{C_t-b \overline{C_{t-1}}}\right)^{-\sigma}(1+r_t)=1$$

Question: I've got no idea what I should do with the average of the consumption ($\overline{C_{t}}$) term. How can I calculate with this? I need to find the steady state, and do the log-linearization, but I don't understand the substance of this variable.

• If this is a representative consumer model, what is the meaning of "average" consumption? How does it differ from actual consumption? Oct 31 '16 at 16:57
• I have no idea. Do you think it's just a trick with the notation, Alecos? Oct 31 '16 at 17:02
• I think it makes no sense to talk about average consumption. The whole formulation is a familiar "habit formation" variant of the usual model, but using past actual consumption, not past average consumption. Oct 31 '16 at 17:07

Consider the risk free steady state. In that situation the Euler equation becomes:

$$\left(\frac{C_{t+1}-b \overline{C_{t}}}{C_t-b \overline{C_{t-1}}}\right)^{-\sigma}=\frac{1}{\beta \cdot (1+r)}$$

Which implies

$$\Rightarrow \ln[C_{t+1}-b \overline{C_{t}}] -\ln[C_t-b \overline{C_{t-1}}] = \frac{\ln[\beta] + \ln[1+r]}{\sigma}$$

Assume that in the non-stochastic setting that growth is constant of the form: $$C_{t+1} = \gamma \cdot C_{t}$$

Note that by definition $$\overline{C_{t}} \equiv \frac{1}{t}\sum^t_{\tau=1} C_{\tau}$$ Under a constant growth rate $$\Rightarrow \overline{C_{t}}=\frac{1}{t}\sum^t_{\tau=1} \gamma^{\tau-t}C_{t} = \frac{C_t}{t} (1+\frac{1}{\gamma-1}) = \frac{C_t}{t} (\frac{\gamma}{\gamma-1})$$ $$\Rightarrow \overline{C_{t+1}}= \frac{C_{t+1}}{t} (\frac{\gamma}{\gamma-1}) = \gamma \frac{C_{t}}{t} (\frac{\gamma}{\gamma-1}) = \gamma \overline{C_{t}}$$

$$\Rightarrow \frac{\ln[\beta] + \ln[1+r]}{\sigma} = \ln[C_{t+1}-b \overline{C_{t}}] -\ln[C_t-b \overline{C_{t-1}}]$$ $$= \ln[\gamma C_{t}-b \gamma \overline{C_{t-1}}] -\ln[C_t-b \overline{C_{t-1}}] = \ln[\gamma]$$

Therefore the one period growth rate in consumption ($\gamma$) in the non-stochastic model is $$\gamma = e^{\frac{\ln[\beta] + \ln[1+r]}{\sigma}} = [\beta(1+r)]^\sigma$$