Return to Answer

I edited the question to include the assumption of orthogonality and iid of independent variable

Source Link

edit approved May 1, 2014 at 2:06

4.6k
4
31
65

Although not clearly stated, it appears that the $X$'s are independent of the white noises, and also that they are not autocorrelated. Then $$E[X_t e_t \mid F_{t-1}] = E[X_t(\epsilon_t + \theta_1 \epsilon_{t-1} + \theta_2 \epsilon_{t-2})\mid \{X_{t-1}, e_{t-1}, X_{t-2}, e_{t-2}, ...\}]$$

$$=E[X_t[\epsilon_t + X_t\theta_1 \epsilon_{t-1} + X_t\theta_2 \epsilon_{t-2})\mid \{X_{t-1}, e_{t-1}, X_{t-2}, e_{t-2}, ...\}]$$

$$=E[X_t]\cdot E[\epsilon_t] + \theta_1E[X_t]\cdot E[\epsilon_{t-1}\mid \{e_{t-1}\}] +\theta_2 E[X_t]\cdot E[\epsilon_{t-2}\mid \{e_{t-1}, e_{t-2}\}]$$

The first term is zero but the other two are not: $\epsilon_{t-1}$ is not independent of $e_{t-1}$ and $\epsilon_{t-2}$ is not independent of $(e_{t-1}, e_{t-2})$. But non-independence implies that

$$E[\epsilon_{t-1}\mid \{e_{t-1}\}] \neq E[\epsilon_{t-1}]=0,\;\; E[\epsilon_{t-2}\mid \{e_{t-1}, e_{t-2}\}] \neq E[\epsilon_{t-2}]=0$$

So $$E[X_t e_t \mid F_{t-1}] \neq 0$$ and therefore it is not a martingale-difference sequence. There is no need to actually calculate the expected values.

$$=E[X_t[\epsilon_t + X_t\theta_1 \epsilon_{t-1} + X_t\theta_2 \epsilon_{t-2})\mid \{X_{t-1}, e_{t-1}, X_{t-2}, e_{t-2}, ...\}]$$

$$=E[X_t]\cdot E[\epsilon_t] + \theta_1E[X_t]\cdot E[\epsilon_{t-1}\mid \{e_{t-1}\}] +\theta_2 E[X_t]\cdot E[\epsilon_{t-2}\mid \{e_{t-1}, e_{t-2}\}]$$

The first term is zero but the other two are not: $\epsilon_{t-1}$ is not independent of $e_{t-1}$ and $\epsilon_{t-2}$ is not independent of $(e_{t-1}, e_{t-2})$. But non-independence implies that

$$E[\epsilon_{t-1}\mid \{e_{t-1}\}] \neq E[\epsilon_{t-1}]=0,\;\; E[\epsilon_{t-2}\mid \{e_{t-1}, e_{t-2}\}] \neq E[\epsilon_{t-2}]=0$$

So $$E[X_t e_t \mid F_{t-1}] \neq 0$$ and therefore it is not a martingale-difference sequence. There is no need to actually calculate the expected values.

$$E[X_t e_t \mid F_{t-1}] = E[X_t(\epsilon_t + \theta_1 \epsilon_{t-1} + \theta_2 \epsilon_{t-2})\mid \{X_{t-1}, e_{t-1}, X_{t-2}, e_{t-2}, ...\}]$$

$$=E[X_t[\epsilon_t + X_t\theta_1 \epsilon_{t-1} + X_t\theta_2 \epsilon_{t-2})\mid \{X_{t-1}, e_{t-1}, X_{t-2}, e_{t-2}, ...\}]$$

$$=E[X_t]\cdot E[\epsilon_t] + \theta_1E[X_t]\cdot E[\epsilon_{t-1}\mid \{e_{t-1}\}] +\theta_2 E[X_t]\cdot E[\epsilon_{t-2}\mid \{e_{t-1}, e_{t-2}\}]$$

The first term is zero but the other two are not: $\epsilon_{t-1}$ is not independent of $e_{t-1}$ and $\epsilon_{t-2}$ is not independent of $(e_{t-1}, e_{t-2})$. But non-independence implies that

$$E[\epsilon_{t-1}\mid \{e_{t-1}\}] \neq E[\epsilon_{t-1}]=0,\;\; E[\epsilon_{t-2}\mid \{e_{t-1}, e_{t-2}\}] \neq E[\epsilon_{t-2}]=0$$

So $$E[X_t e_t \mid F_{t-1}] \neq 0$$ and therefore it is not a martingale-difference sequence. There is no need to actually calculate the expected values.

Source Link

created Apr 30, 2014 at 23:45

Alecos Papadopoulos

60.8k
8
154
278

$$=E[X_t[\epsilon_t + X_t\theta_1 \epsilon_{t-1} + X_t\theta_2 \epsilon_{t-2})\mid \{X_{t-1}, e_{t-1}, X_{t-2}, e_{t-2}, ...\}]$$

$$=E[X_t]\cdot E[\epsilon_t] + \theta_1E[X_t]\cdot E[\epsilon_{t-1}\mid \{e_{t-1}\}] +\theta_2 E[X_t]\cdot E[\epsilon_{t-2}\mid \{e_{t-1}, e_{t-2}\}]$$

The first term is zero but the other two are not: $\epsilon_{t-1}$ is not independent of $e_{t-1}$ and $\epsilon_{t-2}$ is not independent of $(e_{t-1}, e_{t-2})$. But non-independence implies that

$$E[\epsilon_{t-1}\mid \{e_{t-1}\}] \neq E[\epsilon_{t-1}]=0,\;\; E[\epsilon_{t-2}\mid \{e_{t-1}, e_{t-2}\}] \neq E[\epsilon_{t-2}]=0$$

So $$E[X_t e_t \mid F_{t-1}] \neq 0$$ and therefore it is not a martingale-difference sequence. There is no need to actually calculate the expected values.