Simple Conditional Variance Proof Question

Question

So I have the following best linear predictor:

$y_{t+1} = a + b y_t + v_t $ , where $b$ is a measure of persistence and $v_t$ is noise (independent of $y_t$).

Variance is persistent across generations so $\sigma_{yt}^2 = \sigma_{yt+1}^2 = \sigma_y^2$

I'm trying to show that

$\sigma_y^2 = b^2 \sigma_y^2 + \sigma_v^2$

Which should be relatively simple, but I feel like I'm missing something really obvious.

The only thing I can think of is to show the equivalency:

$\sigma_y^2 = \sigma_{yt+1}^2$

Can someone point me in the right direction?

Answer 1

Hint: Try to simplify

$$ V(y_{t+1})=V(a+by_t+v_t) $$ using (i) rules of how the variance operator works, and (ii) the information given.

Also note that what you are trying to show has $\sigma^2_y$ on both sides. Try to solve for it to get an expression for it that does not involve itself.

Answer 2

Your formula is incorrect, you have $V(aX)=a^2\sigma^2$, so $V(b^2\sigma_y^2)=b^4\sigma_y^2$.

Second, the variance of a constant is zero: $V(a)=0$.

Then you get by independence of $y_t,v_t,a$:

$$V(y_{t+1})=V(a+b^2y_t+v_t)=V(a)+V(y_t)+V(v_t)=b^4\sigma_y^2+\sigma_v^2$$

Answer 3

For the variance to be persistent across generations, then it must be true that $\sigma^2(v_{t+1}) = \sigma^2(v_t) = \sigma^2(v)$ as well. Given this, you can start from the definitions of $\sigma^2$: $$ \sigma^2(y_t) \equiv E\left[ (y_t-E(y_t))^2 \right] $$ $$ \sigma^2(v_t) \equiv E\left[ (v_t-E(v_t))^2 \right] $$ Now plug in $y_{t+1} = a + b^2 y_t + v_t$ and expand the square in the expression for $\sigma^2(y_t)$ to get an expression for $\sigma^2(y_{t+1})$.

You will find that the $a$ terms drop out, and that you are left with b^4 times the expression for $\sigma^2(y_t)$ plus the expression for $\sigma^2(v_t)$. The result you want almost follows; there is a $b^4$ instead of $b^2$ in the right result.