Aligning Observations to model variables

Estimation ML/Bayesian estimation
1

Hi Prof,

I have read your paper “Specifying Observation Equations” but just wanted to clarify some stuff. If my Model is RBC and variables are in levels.

Since the data I have for Consumption exhibit a trend, then de-trending the variable to align it with model variables follows the rational in King/Rebelo (1999) where the deterministic trend is defined as X_{t+1}=(1+g_y)X_t) where g_y is the growth rate of Y and is equivalent to (1+g) (1+n). Then, Y_{model_t}=Y_{observed_t}/X_t.

  1. This rationale is what Dynare does by only declaring g_Y as the growth rate for C without specifying anything about X_t ? That is:

observation_trends;
C (g_y);
end;

  1. g_y should be calculated from data as a long term average of growth rate of Y ? That is, g_y = mean(log(Y_t) - log(Y_{t-1})) ? right ?
2 
observation_trends;
C (g_y);
end;

means that C has a linear trend with growth rate g_y. In most cases, you actually want

observation_trends;
log_C (g_y);
end;

because you want the logarithm to have an linear trend, which means the level has an exponential trend.
Yes, you many want to compute the average growth rate over the sample.

3

Thanks Prof.,

One more clarrification if you may. My understanding that DSGE models are stationarized growth models by dividing by X_t where X_{t+1}=(1+g_y) X_t following King & Rebelo (1999).

  1. If X_0 = 1, then X_t=(1+g_y)^t and C_t/X_t is detrended given that my variables in the RBC model are in level. Therefore, why I need to impose an exponential trend form and what is the rationale behind imposing an exponential trend ?
  2. In case we opt for an exponential trend, then all we need to add in the model bloc is log_C=log(C) ?
4

I don’t understand. (1+g_y)^t is clearly an exponential function.

5

I got it. I thought you are referring to some sort of e^{X_t} exponential form. Thanks a lot.

6

(1+g_t)^t\approx e^{g_tt}

7

You are right Prof., Taylor first order approximation of ln(1+g) = g and therefore e^{g*t} \approx e^{ln(1+g)^t} = (1+g)^t . It’s crystal clear now .