Log-linearizing $Y_t=\int_0^1 F(X_{it}) di$

Question

I want to prove that log-linearizing the expression $Y_t=\int_0^1 F(X_{it}) di$ yields:

$$Yy_t \approx F'(X)X\int_0^1 x_{it} di$$

Where: $\{X_{it}\}_{i \in (0,1)}$ is a continuum of strictly positive and stationary variables,

Usually my approach to loglinearize is to define $y_t=ln(\frac{Y_{t}}{Y})$,$x_{it}=ln(\frac{X_{it}}{X})$, where $X$, $Y$ are the steady state (stationary) solutions for each variable, and then use a Taylor expasion of order one.

However I haven't been able to find the desired result in this case. Any recomendations?

Answer 1

To see why this holds, we first log-linearize $Y_t$ and $F(X_{it})$ around $Y$ and $X$ respectively and thus obtain:

$$Y_t \approx Y+Y y_{t}$$ $$F(X_{it}) \approx F(X)+ F'(X)X x_{it}$$

Using this to conditions in our original equation we get:

$$ Y+Y y_{t} \approx \int_0^1 F(X)+ F'(X)X x_{it} di =F(X)+ F'(X)X \int_0^1 x_{it} di $$

But since on the steady state we have that $Y=\int_0^1 F(X) di=F(X)$, then we get:

$$Y y_t \approx F'(X)X\int_0^1 x_{it} di$$

which completes our proof.