correlación espuria

Question

He leído que si dos series de tiempo,$Y_t$ y$X_t$, son tendencia estacionaria, entonces la regresión$Y_t$ on$X_t$ da como resultado una regresión espuria debido a una tendencia temporal omitida Variable . Permitir$Y_t = \delta_0 + \delta_1t + u_t$ y$X_t = \gamma_0 + \gamma_1t + v_t$. Quiero mostrar que$Y_t$ es una función lineal de$X_t$, una tendencia temporal determinista y un término de error. ¿Puede alguien proporcionar una prueba matemática de esto?

Answer 1

La regresión no sería espuria. Si$Y_t=\delta_0+\delta_1 t+u_t$ y$X_t=\gamma_0+\gamma_1t+v_t$ entonces

Alfa_0$$t=\frac{1}{\gamma_1}X_t-\frac{\gamma_0}{\gamma_1}-\frac{1}{\gamma_1}v_t$ \ hat \ alpha_1$

and

$ \ delta_0$Y_t=\delta_0-\frac{\delta_1\gamma_0}{\gamma_1}+\frac{\delta_1}{\gamma_1}X_t+u_t-\frac{1}{\gamma_1}v_t$$

Now this is simply a regression

$$Y_t=\alpha_0+\alpha_1X_t+\varepsilon_t$% {\ Gamma_1} {\ gamma_1} {\ gamma_1} {\ gamma_1} {\ gamma_1}$

and it is possible to show that OLS estimates $ \ gamma_1$ and $ \ gamma_1 = % \ Delta_0 = 3$ are consistent and assymptoticaly normal with means $ \ delta_1 = 4$ and $ \ alpha_0 = 1$ respectively, albeit with non-standard normalizing constants. The mathematical details can be found in this answer.

The consistency can be illustrated by the following code:

gend <- function(n) { 
    data.frame(x=1+2*1:n+rnorm(n),y=3+4*1:n+rnorm(n))
}

> set.seed(13)
> coef(lm(y~x,data=gend(10)))
(Intercept)           x 
  -1.291464    2.067586 
> coef(lm(y~x,data=gend(100)))
(Intercept)           x 
   1.396720    1.997408 
> coef(lm(y~x,data=gend(1000)))
(Intercept)           x 
  0.9864317   1.9999570 
> coef(lm(y~x,data=gend(10000)))
(Intercept)           x 
  0.9595726   2.0000065 

Here I generated two trend stationary variables with $ \ alpha_1 = 2 $.