Log linearising EUler equation

Question

I am trying to solve a problem that asks to log linearise following Euler equation of the New Keynesian model:

$$C^{-\sigma}_t=\beta E_tC^{-\sigma}_{t+1}(1+i_t)/(1+\pi_{t+1}).$$

The solution is given as: $$\tilde{C}_t=\beta E_t\tilde{C}-\frac{1}{\sigma}(i_t-E_t\tilde{C}_{t+1}-\rho).$$

And also there are following relationships in the steady state given $1=\beta(1+r)$, $\beta=(1+\rho)^{-1}$ and $ln1=ln \beta+r$ what leads to $r=-ln\beta=\rho$.

Can anyone explain to me how the log linearization is done? I know that the first step would be to take the logarithm of everything (and subtract the logarithm of the constant steady state parameter) or to use Taylor approximation. However I have a problem with the $(1+i_t)$ and $(1+\pi_{t+1})$ terms. Furthermore why does $1=\beta(1+r)$ hold in steady state? Note that $i$ denotes nominal interest rate.

Thank you.

UPDATE: So below I will give my solution path. I am not sure though whether it is appropriate to do it like this: $$C^{-\sigma}_t=\beta E_tC^{-\sigma}_{t+1}(1+i_t)/(1+\pi_{t+1})$$ $$lnC^{-\sigma}_t=ln\beta+lnE_t(C^{-\sigma}_{t+1})+ln(1+i_t)-ln(1+\pi_{t+1})\qquad (1)$$ Subtracting (1) with $$lnC^{-\sigma}=ln\beta+ln(C^{-\sigma})+ln(1+i)-ln(1+\pi)$$yields $$\tilde{C}_t=E_t(\tilde{C}_{t+1})-\frac{1}{\sigma}(\tilde{(1+i_t)}-\tilde{(1+\pi_{t+1})})$$ and since $\tilde{1+i_t}=ln(1+i_t)-ln(1+i)=i_t-i$ and $\tilde{1+\pi_{t+1}}=\pi_{t+1}-\pi$: $$\tilde{C}_t=E_t(\tilde{C}_{t+1})-\frac{1}{\sigma}(i_t-i+E\pi_{t+1}-\pi)\qquad(2)$$ $i$ is the nominal interest rate in steady state and can be defined as $i=r+\pi$ and $r=\rho$: $$\tilde{C}_t=E_t(\tilde{C}_{t+1})-\frac{1}{\sigma}(i_t-\rho+\pi-E\pi_{t+1}-\pi)\qquad(3)$$

Leading us to the final equation: $$\tilde{C}_t=\beta E_t\tilde{C}-\frac{1}{\sigma}(i_t-E_t\tilde{C}_{t+1}-\rho).$$

Answer 1

As you say the first step is to take log of both sides after that you are just applying the rules for logarithms and rearrange.

For example: $$\ln (XZ)=\ln X + \ln Z$$ $$\ln X/Z= \ln X - \ln Z$$ $$\ln X^a = a \ln X$$ $$\ln 1 = 0$$

Also an important approximations that hold close to zero are applied here as well these are:

$\ln(1+x) \approx x $ for $x$ close to zero (which for interest rates and inflation which are usually just couple of percents applies).

Also Taylor approximation is actually a different way how to linearize relationship so although it’s an example of linearization it’s not necessary log-linearization. In fact the result $\ln(1+x)$ is based on Taylor approximation but it’s not log linearization because just applying logs there won’t produce loglinear expression.

Using these rules you can prove all the above solutions. I will leave the first equation for you as an exercise, for the other equations you can see that:

Log linearizing $1=\beta(1+r)$ gives: $ \ln 1= \ln (\beta(1+r))$ which after simplification gives us $0= \ln \beta + \ln (1+r)$ or $\ln \beta = -r $

From the second equation $\beta=(1+\rho)^{-1}$ log linearizing gives us $\ln \beta =-\ln(1+\rho) \implies \ln \beta = -\rho$. Hence you get the equality that $-r=\ln \beta = -\rho$ then you can multiply all sides by -1 to move the minus in the middle of the equality.

The $1=\beta (1+r)$ comes from the fact that rational person would want the marginal utility of consumption today and in future to be equal so actually the equation properly reads:

$$u_t^{\prime} = \beta (1+r) u_{t+1}^{\prime}$$

Which can be rewritten as: $u_t^{\prime} / u_{t+1}^{\prime} = \beta (1+r) $ and if $u_{t}^{\prime}= u_{t+1}^{\prime}$ you get the result that $1=\beta (1+r)$. Again this is because in steady state you want marginal utility of consumption to be equal in each and every period.