# Proving for an AR(2) process that $E[X_t | F_{t-1}]=E[X_t | F_{t-2}]=E[X_t | F_{t-3}]$

An exercise states:

Using the law of iterated expectations applied to an AR(2) process, verify that $E_{t−k} . . . E_{t−1} (X_t ) = E(X_t |F_{t−k} )$ for $k = 1, 2, 3$ where $E_{t−k} (X_t ) = E(X_t |F_{t−k} ) .$

I dislike the way the question is posed and I think it can be reduced to proving:

$$E[X_t | F_{t-1}]=E[X_t | F_{t-2}]=E[X_t | F_{t-3}]$$

I proceeded in this way:

$$E[X_t|F_{t-1}] =E[ \phi_{1}X_{t-1} + \phi_{2}X_{t-2} + \epsilon_t | F_{t-1}] = \phi_{1}X_{t-1} + \phi_{2}X_{t-2} \tag{1}$$

by linearity of the expected value and because $E[ \epsilon_t] = 0$ (white noise).

Then:

$$E[X_t|F_{t-2}] =E[ \phi_{1}X_{t-1} + \phi_{2}X_{t-2} + \epsilon_t | F_{t-2}] = E[\phi_{1}X_{t-1}|F_{t-2}] + \phi_{2}X_{t-2} = \phi_{1}^2X_{t-2} + \phi_{2}\phi_1X_{t-3} + \phi_{2}X_{t-2}$$

Now it seems that substituting the AR(2) again in $(1)$ we would get the required result (the equality between the two conditional expected values):

$$\phi_{1}X_{t-1} + \phi_{2}X_{t-2} = \phi_{1}^2X_{t-2} + \phi_{2}\phi_1X_{t-3} + \phi_{2}X_{t-2} + \epsilon_{t-1}$$

except for that pesky $\epsilon_{t-1}$ .

Where did I go wrong?

What you think is an equivalent statement, it is not.
Think intuitively: if it was the case that $E[X_t | F_{t-1}]=E[X_t | F_{t-2}]=E[X_t | F_{t-3}]$ (it is not), it would mean that new information (as reflected in the fact that $F_{t-3} \subset F_{t-2} \subset F_{t-1}$), would not change your "average anticipation" of what the value of $X_t$ might be (It would be as though the white noise was zero- does this raise a flag?)

Mathematically, you derived correctly

$$E[X_t\mid F_{t-1}] = \phi_{1}X_{t-1} + \phi_{2}X_{t-2} \tag{1}$$

while we have

$$E[X_t|F_{t-2}] =E[ \phi_{1}X_{t-1} + \phi_{2}X_{t-2} + \epsilon_t \mid F_{t-2}] = \phi_{1}E[ X_{t-1} \mid F_{t-2}]+ \phi_{2}X_{t-2} \tag{2}$$

For $(1)$ to be equal to $(2)$ we must have

$$X_{t-1} = ?\; E[ X_{t-1} \mid F_{t-2}]$$

Can we? (we could, if the white noise was zero...)

This tedious exercise asks you to verify that

$$E\Big(E\Big[E(\phi_{1}X_{t-1} + \phi_{2}X_{t-2}+\epsilon_t\mid F_{t-1})\mid F_{t-2}\Big]\mid F_{t-3}\Big) = E\Big(\phi_{1}X_{t-1} + \phi_{2}X_{t-2}+\epsilon_t\mid F_{t-3}\Big)$$

so that I guess, you "see with your own eyes" that the law of iterated expectations holds (which is essentially part of the definition of the conditional expectation, see this post if you are interested).

Start "getting rid" of expected values by using the functional form of the $X$'s when necessary, in order to express the future that is not covered by the immediately available sigma-algebra, with the present and the past that is.