<p>Thanks for the hint and the link! I think I now managed to find the solution.</p>

<p>Putting the exponent on the LHS and replacing </p>

<p>$C_t^{1-\gamma}$ with $C+(1-\gamma) C^{1-\gamma}C \tilde{c}$ </p>

<p>and</p>

<p>$C_t(i)^{1-\gamma}$ with $C(i)+(1-\gamma)C(i)^{1-\gamma}\tilde{c}_t(i)$</p>

<p>i get (subtracting Steady State values):</p>

<p>$(1-\gamma) C^{1-\gamma}C \tilde{c}=\int_0^1 (1-\gamma)C(i)^{1-\gamma}\tilde{c}_t(i) \ di$</p>

<p>Assuming steady state values of $C(i)$ are constant across $i$, i can take all values out of the integral and simplify to get:</p>

<p>$\tilde{c}=\int_0^1 \tilde{c}_t(i) \ di$</p>

CC BY-SA 3.0
2016-09-23T14:48:01.263Z
2016-09-23T14:48:01.263Z
Chris tie
<p>In a New-Keynesian Model, the Consumption index</p>

<p>$C_t=\left(\int_0^1{C_t(i)^{1-\gamma} \ di}\right)^{\frac{1}{1-\gamma}}$</p>

<p>is log-linearized to</p>

<p>$\tilde{c_t}=\int_0^1{\tilde{c_t}(i)  \ di}$</p>

<p>where variables with tilde are log deviations from steady state and $i=[0,1]$ are varieties of the consumption good. </p>

<p>Even though I feel comfortable log linearizing other equations, I have no idea how to deal with the integral. How do you get to the result? And what would a second order approximation look like?</p>

CC BY-SA 3.0
2016-09-21T02:06:35.043Z
Hint: bring the exponent over to the LHS. Then try the method used here http://math.stackexchange.com/questions/1519583/log-linearizing-y-t-int-01-fx-it-di
VCG
CC BY-SA 3.0
2016-09-21T06:10:30.603Z
@VCG I think this is a nice and non-trivial technical trick. If you would type out a detailed answer (perhaps with some explanation about why the approximation works) I would upvote it. Also, if the method works, why do you not vote it up on math?
Giskard
<macroeconomics><new-keynesian-economics><monopolistic-competition>
Log-Linearizing a Dixit-Stiglitz function
Chris tie
# Log-Linearizing a Dixit-Stiglitz function

(Asked by: username_0 on 2016-09-20)

In a New-Keynesian Model, the Consumption index

$C\_t=\left(\int\_0^1{C\_t(i)^{1-\gamma} \ di}\right)^{\frac{1}{1-\gamma}}$

is log-linearized to

$\tilde{c\_t}=\int\_0^1{\tilde{c\_t}(i) \ di}$

where variables with tilde are log deviations from steady state and $i=[0,1]$ are varieties of the consumption good.

Even though I feel comfortable log linearizing other equations, I have no idea how to deal with the integral. How do you get to the result? And what would a second order approximation look like?

**Tags:** macroeconomics, new-keynesian-economics, monopolistic-competition
> **Comments:**
> **username_1**: Hint: bring the exponent over to the LHS. Then try the method used here http://math.stackexchange.com/questions/1519583/log-linearizing-y-t-int-01-fx-it-di  
> **username_2**: @VCG I think this is a nice and non-trivial technical trick. If you would type out a detailed answer (perhaps with some explanation about why the approximation works) I would upvote it. Also, if the method works, why do you not vote it up on math?  

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## 1 answers

### Answer by username_0 on 2016-09-23. Score: 4 | Accepted answer

Thanks for the hint and the link! I think I now managed to find the solution.

Putting the exponent on the LHS and replacing

$C\_t^{1-\gamma}$ with $C+(1-\gamma) C^{1-\gamma}C \tilde{c}$

and

$C\_t(i)^{1-\gamma}$ with $C(i)+(1-\gamma)C(i)^{1-\gamma}\tilde{c}\_t(i)$

i get (subtracting Steady State values):

$(1-\gamma) C^{1-\gamma}C \tilde{c}=\int\_0^1 (1-\gamma)C(i)^{1-\gamma}\tilde{c}\_t(i) \ di$

Assuming steady state values of $C(i)$ are constant across $i$, i can take all values out of the integral and simplify to get:

$\tilde{c}=\int\_0^1 \tilde{c}\_t(i) \ di$


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