# How to find the conditional variance?

Suppose I have the following time series process with $$a_t$$ being an independent white noise with mean $$0$$ and variance $$\sigma^2$$:

$$z_t = \frac{1}{3}a_t + \frac{1}{3}a_{t-1} + \frac{1}{3}a_{t-2}$$

How do I compute the variance conditional on $$a_{t-2} = 5$$

Also, the unconditional variance in this problem is $$\frac{\sigma^2}{3}$$ correct? Thank you!

• Are you looking for the variance (conditional or unconditional) of what? Of $z_t$? Commented Jan 28, 2021 at 23:33
• Yes. The variance of $z_t$ conditional on $a_{t-2} = 5$ Commented Jan 28, 2021 at 23:41

• $$\frac{1}{3}a_t$$ has variance $$\frac19\sigma^2$$
$$a_t,a_{t-1},a_{t-2}$$ are mutually independent so
• $$\mathrm{Var}(z_t)=\frac39\sigma^2=\frac13\sigma^2$$
• $$\mathrm{Var}(z_t\mid a_{t-2}=5)=\frac29\sigma^2$$
• So Basically since $Var(a_{t-2})$ is given we exclude it when calculating the variance of $z_t$ which is then just the sum of the variance of the other two random variables that were not given? Commented Jan 29, 2021 at 18:39
• Thank you very much! How would I calculate the $Cov[z_t,z_{t-1}]$? Commented Jan 30, 2021 at 0:11
• $\mathrm{Cov}[z_t,z_{t-1}]$ $= \mathrm{Cov}\left[\frac{1}{3}a_t + \frac{1}{3}a_{t-1} + \frac{1}{3}a_{t-2},\frac{1}{3}a_{t-1} + \frac{1}{3}a_{t-2}+ \frac{1}{3}a_{t-3}\right]$ $= \mathrm{Var}\left[ \frac{1}{3}a_{t-1} \right] +\mathrm{Var}\left[ \frac{1}{3}a_{t-2} \right]$ $=\frac29\sigma^2$ Commented Jan 30, 2021 at 11:49