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?

  • 2
    $\begingroup$ Hint: bring the exponent over to the LHS. Then try the method used here math.stackexchange.com/questions/1519583/… $\endgroup$
    – VCG
    Commented Sep 21, 2016 at 2:06
  • 2
    $\begingroup$ @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? $\endgroup$
    – Giskard
    Commented Sep 21, 2016 at 6:10

1 Answer 1


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


$C_t(i)^{1-\gamma}$ 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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