So suppose you have two sequences $\{Y_t\}$ and $\{Z_t\}$ and they are both iid and independent from each other. Now suppose I have a time series $\{X_t\}$ such that...

$$\{X_t\} = Y_t(1-Y_{t-1})Z_t$$

What would be the variance of this time series?

If my understanding of this is correct the process would go.

\begin{align} \newcommand{\Var}{{\rm Var}} \Var(Y_t(1-Y_{t-1})Z_t) &= \Var(Z_tY_t-Z_tY_tY_{t-1}) \\ &= \Var(Z_tY_t) + \Var(Z_tY_tY_{t-1}) \end{align}


My question mainly focusing around how to deal with the expanded version $\Var(Z_tY_t-Z_tY_tY_{t-1})$ as well as what to do afterwards. I know that $\Var(X-Y) = \Var(X) + \Var(Y)$ when $X$ and $Y$ are independent however I don't know if that carries when $X$ and/or $Y$ are products of iid random variables. As well how to decompose the variance of three sequences.


You know that for any random variable $X$, we have that $$\operatorname{var}(X) = E[X^2] - \left(E[X]\right)^2,\tag{1}$$ and so $$\operatorname{var}(X_t) = E[X_t^2] - \left(E[X_t]\right)^2.\tag{2}$$ Now, \begin{align}E[X_t] &= E[Y_t(1-Y_{t-1})Z_t]\\ &= E[Y_t]E[1-Y_{t-1}]E[Z_t] &\scriptstyle{\text{independence}}\\ \ &= \mu_Y(1-\mu_Y)\mu_Z\tag{3} \end{align} while \begin{align}E[X_t^2] &= E[Y_t^2(1-Y_{t-1})^2Z_t^2]\\ &= E[Y_t^2]E[(1-Y_{t-1})^2]E[Z_t^2] &\scriptstyle{\text{independence}}\\ &= (\sigma_Y^2+\mu_Y^2)\cdot\left(\operatorname{var}(1-Y_{t-1})+(1-\mu_Y)^2\right)\cdot(\sigma_Z^2+\mu_Z^2) &\scriptstyle{\text{applying } (1)}\\ &= (\sigma_Y^2+\mu_Y^2)\cdot\left(\sigma_Y^2+(1-\mu_Y)^2\right)\cdot(\sigma_Z^2+\mu_Z^2)\tag{4} \end{align} I will leave it to you to substitute $(3)$ and $(4)$ into $(2)$ to find a formula for $\operatorname{var}(X_t)$.

Note: The i.i.d. assumption about the $Y$ time series and the independence of the $Y$ and $Z$ series leads to $Y_t$, $1-Y_{t-1}$ and $Z_t$ being independent, making the above result a special case of the result shown in this answer of mine regarding the variance of the product of several independent random variables.


You want the bilinearity property of covariance (linear in both arguments).

If $X$ and $Y$ are independent, this implies zero covariance/correlation, and your adding rule, $Var(X-Y) = Var(X) + Var(Y)$, is a special case of this property. But in this case it's more helpful to know the general rule.

\begin{align*} \text{Var}(Y_t(1-Y_{t-1})Z_t) &= \text{Cov}[Y_t(1-Y_{t-1})Z_t, Y_t(1-Y_{t-1})Z_t] \\ &= \text{Cov}[Z_tY_t-Z_tY_tY_{t-1},Z_tY_t-Z_tY_tY_{t-1}] \\ &= \text{Cov}[Z_tY_t,Z_tY_t] - 2\text{Cov}[Z_tY_t,Z_tY_tY_{t-1}] + \text{Cov}[Z_tY_tY_{t-1},Z_tY_tY_{t-1}] \\ &= \text{Var}[Z_tY_t] - 2\text{Cov}[Z_tY_t,Z_tY_tY_{t-1}] + \text{Var}[Z_tY_tY_{t-1}]. \end{align*}

To say any more than this would require more information about these rv sequences. Ill let you break down the last line into expectations.

  • $\begingroup$ Your approach makes this problem way harder than it actually is. For a complete solution, see my answer. $\endgroup$ – Dilip Sarwate Jan 28 '17 at 17:01
  • $\begingroup$ I disagree, but that's subjective. For example he might ask, "Why is $1-Y_{t-1}$ independent from $Y_t$?" This answer addresses the main lack of understanding OP has, which is what happens generally when you take the variance of a linear combination of rvs. This was the mistake he made in his work that he showed. Although your answer brings him to a full answer, it wasn't clear to me that he didn't know about how to split up expectations. $\endgroup$ – Taylor Jan 28 '17 at 17:25

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