<p>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.</p>

<p>For example:
<span class="math-container">$$\ln (XZ)=\ln X + \ln Z$$</span>
<span class="math-container">$$\ln X/Z= \ln X - \ln Z$$</span>
<span class="math-container">$$\ln X^a = a \ln X$$</span>
<span class="math-container">$$\ln 1 = 0$$</span></p>

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

<p><span class="math-container">$\ln(1+x) \approx x $</span> for <span class="math-container">$x$</span> close to zero (which for interest rates and inflation which are usually just couple of percents applies).</p>

<p>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 <span class="math-container">$\ln(1+x)$</span> is based on Taylor approximation but it’s not log linearization because just applying logs there won’t produce loglinear expression.</p>

<p>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:</p>

<p>Log linearizing  <span class="math-container">$1=\beta(1+r)$</span> gives: <span class="math-container">$ \ln 1= \ln (\beta(1+r))$</span> which after simplification gives us <span class="math-container">$0= \ln \beta + \ln (1+r)$</span> or <span class="math-container">$\ln \beta = -r $</span></p>

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

<p>The <span class="math-container">$1=\beta (1+r)$</span> 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:</p>

<p><span class="math-container">$$u_t^{\prime} = \beta (1+r) u_{t+1}^{\prime}$$</span></p>

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

CC BY-SA 4.0
2020-04-10T15:59:53.623Z
Hey, thank you for your explanation. Especially the intuitive explanation in the last part was helpful. I have added the solution path to the log-linearized equation, may you take a look? Could you try to explain the intuition of replacing $i=r+\pi$ and not $i_t=r_t+\pi_{t+1}$, also why do we replace the steady state real interest rate $r$ by $\rho$ and not just leave it as it is. I know that the log linearization should finally lead us to the IS curve equation. But tbh I don't really get the intuition.
randomname
CC BY-SA 4.0
2020-04-10T22:16:59.137Z
@randomname you are welcome if you found the answer helpful consider accepting it. Also your calculation appears correct. If you are using it in homework/assignment you can make the steps look more elegant if do the approximation of ln(1+i)=i before subtraction as tilde over sum does not look good but otherwise it appears correct. You use \pi t because the final solution is expressed in form of changes from now whereas the original equation is expressed in form of levels. Because of the equality you could use rho it’s just matter of preference.
1muflon1
CC BY-SA 4.0
2020-04-10T22:25:17.033Z
@randomname you could derive in principle IS even without log linearization - but solving non-linear set of equations is a nightmare. So there is no intuitive connection between log linearization and IS it’s just done for convenience. It is also less taxing on your computer when doing estimations/symulations if you use linearized model.
1muflon1
CC BY-SA 4.0
2020-04-10T13:34:08.770Z
2020-04-10T13:42:15.150Z
2020-04-10T13:42:15.150Z
1muflon1
<p>I am trying to solve a problem that asks to log linearise following Euler equation of the New Keynesian model:</p>

<p><span class="math-container">$$C^{-\sigma}_t=\beta E_tC^{-\sigma}_{t+1}(1+i_t)/(1+\pi_{t+1}).$$</span></p>

<p>The solution is given as: <span class="math-container">$$\tilde{C}_t=\beta E_t\tilde{C}-\frac{1}{\sigma}(i_t-E_t\tilde{C}_{t+1}-\rho).$$</span></p>

<p>And also there are following relationships in the steady state given <span class="math-container">$1=\beta(1+r)$</span>, <span class="math-container">$\beta=(1+\rho)^{-1}$</span> and <span class="math-container">$ln1=ln \beta+r$</span> what leads to <span class="math-container">$r=-ln\beta=\rho$</span>.</p>

<p>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 <span class="math-container">$(1+i_t)$</span> and <span class="math-container">$(1+\pi_{t+1})$</span> terms. Furthermore why does <span class="math-container">$1=\beta(1+r)$</span> hold in steady state? Note that <span class="math-container">$i$</span> denotes <strong>nominal</strong> interest rate.</p>

<p>Thank you.</p>

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

<p>Leading us to the final equation:
<span class="math-container">$$\tilde{C}_t=\beta E_t\tilde{C}-\frac{1}{\sigma}(i_t-E_t\tilde{C}_{t+1}-\rho).$$</span></p>

<macroeconomics><economic-growth><new-keynesian-economics><steady-state><log-linearization>
Log linearising EUler equation
randomname
# Log linearising EUler equation

(Asked by: username_0 on 2020-04-10)

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:
$$\tilde{C}\_t=\beta E\_t\tilde{C}-\frac{1}{\sigma}(i\_t-E\_t\tilde{C}\_{t+1}-\rho).$$

**Tags:** macroeconomics, economic-growth, new-keynesian-economics, steady-state, log-linearization


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## 1 answers

### Answer by username_1 on 2020-04-10. Score: 3 | Accepted answer

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.

> **Comments:**
> **username_0**: Hey, thank you for your explanation. Especially the intuitive explanation in the last part was helpful. I have added the solution path to the log-linearized equation, may you take a look? Could you try to explain the intuition of replacing $i=r+\pi$ and not $i_t=r_t+\pi_{t+1}$, also why do we replace the steady state real interest rate $r$ by $\rho$ and not just leave it as it is. I know that the log linearization should finally lead us to the IS curve equation. But tbh I don't really get the intuition.  
> **username_1**: @username_0 you are welcome if you found the answer helpful consider accepting it. Also your calculation appears correct. If you are using it in homework/assignment you can make the steps look more elegant if do the approximation of ln(1+i)=i before subtraction as tilde over sum does not look good but otherwise it appears correct. You use \pi t because the final solution is expressed in form of changes from now whereas the original equation is expressed in form of levels. Because of the equality you could use rho it’s just matter of preference.  
> **username_1**: @username_0 you could derive in principle IS even without log linearization - but solving non-linear set of equations is a nightmare. So there is no intuitive connection between log linearization and IS it’s just done for convenience. It is also less taxing on your computer when doing estimations/symulations if you use linearized model.  

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