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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.

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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.

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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.

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  • $\begingroup$ 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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