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Write out the simple equations

$$\begin{align} Y_j &= a_0 Z_j + a_1 Z_{j-1} + a_2 Z_{j-2}\\ Y_{j-1} &= a_0 Z_{j-1} + a_1 Z_{j-2} + a_2 Z_{j-3} \end{align}$$

There are some very simple cases that make $Y_j \perp Y_{j-1}$ due to the independence assumption of the random variables $\{Z_i\}_{i\in\mathbb{Z}}$. An example is $a_0 \in \mathbb{R}\setminus \{0\},\, a_1 = 0,\, a_2 = 0$. Not sure if you were looking for a complete solution but this should help get you started.

Also, an easy check for RV which are not independent is using the contrapositive form of the common theorem $$X\perp Y \implies E[XY] = E[X]E[Y]$$ Note that the converse of this statement is not true.

Proof

Assertion $a_1a_0 + a_2a_1 = 0 \iff Y_j \perp Y_{j-1}$

Define $\mu = E[Z]$

($\implies$) Suppose $a_1a_0 + a_2a_1 = 0$. There are two cases where this is possible. Case 1, suppose $a_1 = 0$. The equations become

$$\begin{align} Y_j &= a_0 Z_j + a_2 Z_{j-2}\\ Y_{j-1} &= a_0 Z_{j-1} + a_2 Z_{j-3} \end{align}$$

Their $\sigma$-algebras are given by $\sigma(Y_j) = \sigma(Z_j)\cup\sigma(Z_{j-2})$ and $\sigma(Y_{j-1}) = \sigma(Z_{j-1})\cup \sigma(Z_{j-3})$. Thus $Y_j \perp Y_{j-1}$. This could be more gruesomely detailed but I take some for granted. See this for more details including definitions etc.

Case 2, suppose $a_2 = 0$ and $a_0 = 0$. The equations become

$$\begin{align} Y_j &= a_1 Z_{j-1} \\ Y_{j-1} &= a_1 Z_{j-2} \end{align}$$

The same $\sigma$-algebra argument applies more easily but a more elegant solution presents itself in the form of the CDF.

$$\begin{align} F_{Y_j, Y_{j-1}}(y_j, y_{j-1}) &= P(Y_j \leq y_j \text{ and } Y_{j-1} \leq y_{j-1}) \\ & = F_{Z_j}(y_j/a_1)F_{Z_{j-1}}(y_{j-1}/a_1)\\ & = F_{Y_j}(y_j)F_{Y_{j-1}}(y_{j-1}) \end{align}$$

($\impliedby$) Suppose $Y_j \perp Y_{j-1}$ by theorem, we know that $E[Y_j Y_{j-1}] = E[Y_j]E[Y_{j-1}]$ calculating these values separately,

$$\begin{align} E[Y_jY_{j-1}] & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2 )\mu^2 \\ & + (a_1a_0 + a_2a_1)E[Z^2] \end{align}$$

$$\begin{align} E[Y_j]E[Y_{j-1}] & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2)\mu^2\\ &+ (a_1a_0 + a_2a_1)\mu^2 \end{align}$$

In the non-degenerate case when the distribution of $Z$ is not a constant, the variance is strictly positive so that $E[Z^2] - \mu^2 > 0$ and so $E[Z^2] > \mu^2$ and more importantly $E[Z^2] \neq \mu^2$ Thus for the equality $E[Y_j Y_{j-1}] = E[Y_j]E[Y_{j-1}]$ to hold, it must be the case that $a_1a_0 + a_2a_1 = 0$.

Write out the simple equations

$$\begin{align} Y_j &= a_0 Z_j + a_1 Z_{j-1} + a_2 Z_{j-2}\\ Y_{j-1} &= a_0 Z_{j-1} + a_1 Z_{j-2} + a_2 Z_{j-3} \end{align}$$

There are some very simple cases that make $Y_j \perp Y_{j-1}$ due to the independence assumption of the random variables $\{Z_i\}_{i\in\mathbb{Z}}$. An example is $a_0 \in \mathbb{R}\setminus \{0\},\, a_1 = 0,\, a_2 = 0$. Not sure if you were looking for a complete solution but this should help get you started.

Also, an easy check for RV which are not independent is using the contrapositive form of the common theorem $$X\perp Y \implies E[XY] = E[X]E[Y]$$ Note that the converse of this statement is not true.

Proof

Assertion $a_1a_0 + a_2a_1 = 0 \iff Y_j \perp Y_{j-1}$

Define $\mu = E[Z]$

($\implies$) Suppose $a_1a_0 + a_2a_1 = 0$. There are two cases where this is possible. Case 1, suppose $a_1 = 0$. The equations become

$$\begin{align} Y_j &= a_0 Z_j + a_2 Z_{j-2}\\ Y_{j-1} &= a_0 Z_{j-1} + a_2 Z_{j-3} \end{align}$$

Their $\sigma$-algebras are given by $\sigma(Y_j) = \sigma(Z_j)\cup\sigma(Z_{j-2})$ and $\sigma(Y_{j-1}) = \sigma(Z_{j-1})\cup \sigma(Z_{j-3})$. Thus $Y_j \perp Y_{j-1}$. This could be more gruesomely detailed but I take some for granted. See this for more details including definitions etc.

Case 2, suppose $a_2 = 0$ and $a_0 = 0$. The equations become

$$\begin{align} Y_j &= a_1 Z_{j-1} \\ Y_{j-1} &= a_1 Z_{j-2} \end{align}$$

The same $\sigma$-algebra argument applies more easily but a more elegant solution presents itself in the form of the CDF.

$$\begin{align} F_{Y_j, Y_{j-1}}(y_j, y_{j-1}) &= P(Y_j \leq y_j \text{ and } Y_{j-1} \leq y_{j-1}) \\ & = F_{Z_j}(y_j/a_1)F_{Z_{j-1}}(y_{j-1}/a_1)\\ & = F_{Y_j}(y_j)F_{Y_{j-1}}(y_{j-1}) \end{align}$$

($\impliedby$) Suppose $Y_j \perp Y_{j-1}$ by theorem, we know that $E[Y_j Y_{j-1}] = E[Y_j]E[Y_{j-1}]$ calculating these values separately,

$$\begin{align} E[Y_jY_{j-1}] & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2 )\mu^2 \\ & + (a_1a_0 + a_2a_1)E[Z^2] \end{align}$$

$$\begin{align} E[Y_j]E[Y_{j-1}] & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2)\mu^2\\ &+ (a_1a_0 + a_2a_1)\mu^2 \end{align}$$

In the non-degenerate case when the distribution of $Z$ is not a constant, the variance is strictly positive so that $E[Z^2] - \mu^2 > 0$ and so $E[Z^2] > \mu^2$ and more importantly $E[Z^2] \neq \mu^2$ Thus for the equality $E[Y_j Y_{j-1}] = E[Y_j]E[Y_{j-1}]$ to hold, it must be the case that $a_1a_0 + a_2a_1 = 0$.

Write out the simple equations

$$\begin{align} Y_j &= a_0 Z_j + a_1 Z_{j-1} + a_2 Z_{j-2}\\ Y_{j-1} &= a_0 Z_{j-1} + a_1 Z_{j-2} + a_2 Z_{j-3} \end{align}$$

There are some very simple cases that make $Y_j \perp Y_{j-1}$ due to the independence assumption of the random variables $\{Z_i\}_{i\in\mathbb{Z}}$. An example is $a_0 \in \mathbb{R}\setminus \{0\},\, a_1 = 0,\, a_2 = 0$. Not sure if you were looking for a complete solution but this should help get you started.

Also, an easy check for RV which are not independent is using the contrapositive form of the common theorem $$X\perp Y \implies E[XY] = E[X]E[Y]$$ Note that the converse of this statement is not true.

Proof

Assertion $a_1a_0 + a_2a_1 = 0 \iff Y_j \perp Y_{j-1}$

Define $\mu = E[Z]$

($\implies$) Suppose not, so that $a_1a_0 + a_2a_1 = 0$ and $Y_j$ is not independent of $Y_{j-1}$.

$$\begin{align} E[Y_jY_{j-1}] = E[ &a_0^2 Z_j Z_{j-1} + a_0a_1 Z_j Z_{j-2} + a_0 a_2 Z_j Z_{j-3}\\ +& a_1 a_0 Z_{j-1}^2 + a_1^2 Z_{j-1}Z_{j-2} + a_1 a_2 Z_{j-1}Z_{j-3}\\ +& a_2 a_0 Z_{j-2}Z_{j-1} + a_2 a_1Z_{j-2}^2 + a_2^2Z_{j-2}Z_{j-3} ] \\ = & (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2 )\mu^2 + 0 \end{align}$$

$$\begin{align} E[Y_j]E[Y_{j-1}] & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2 + 0)\mu^2\\ & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2 )\mu^2 \end{align}$$

This is a contradiction of the (contrapositive of the) theorem above.

($\impliedby$) Suppose $Y_j \perp Y_{j-1}$ by theorem, we know that $E[Y_j Y_{j-1}] = E[Y_j]E[Y_{j-1}]$ calculating these values separately,

$$\begin{align} E[Y_jY_{j-1}] & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2 )\mu^2 \\ & + (a_1a_0 + a_2a_1)E[Z^2] \end{align}$$

$$\begin{align} E[Y_j]E[Y_{j-1}] & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2)\mu^2\\ &+ (a_1a_0 + a_2a_1)\mu^2 \end{align}$$

In the non-degenerate case when the distribution of $Z$ is not a constant, the variance is strictly positive so that $E[Z^2] - \mu^2 > 0$ and so $E[Z^2] > \mu^2$ and more importantly $E[Z^2] \neq \mu^2$ Thus for the equality $E[Y_j Y_{j-1}] = E[Y_j]E[Y_{j-1}]$ to hold, it must be the case that $a_1a_0 + a_2a_1 = 0$.

Write out the simple equations

$$\begin{align} Y_j &= a_0 Z_j + a_1 Z_{j-1} + a_2 Z_{j-2}\\ Y_{j-1} &= a_0 Z_{j-1} + a_1 Z_{j-2} + a_2 Z_{j-3} \end{align}$$

There are some very simple cases that make $Y_j \perp Y_{j-1}$ due to the independence assumption of the random variables $\{Z_i\}_{i\in\mathbb{Z}}$. An example is $a_0 \in \mathbb{R}\setminus \{0\},\, a_1 = 0,\, a_2 = 0$. Not sure if you were looking for a complete solution but this should help get you started.

Also, an easy check for RV which are not independent is using the contrapositive form of the common theorem $$X\perp Y \implies E[XY] = E[X]E[Y]$$ Note that the converse of this statement is not true.

Proof

Assertion $a_1a_0 + a_2a_1 = 0 \iff Y_j \perp Y_{j-1}$

Define $\mu = E[Z]$

($\implies$) Suppose $a_1a_0 + a_2a_1 = 0$. There are two cases where this is possible. Case 1, suppose $a_1 = 0$. The equations become

$$\begin{align} Y_j &= a_0 Z_j + a_2 Z_{j-2}\\ Y_{j-1} &= a_0 Z_{j-1} + a_2 Z_{j-3} \end{align}$$

Their $\sigma$-algebras are given by $\sigma(Y_j) = \sigma(Z_j)\cup\sigma(Z_{j-2})$ and $\sigma(Y_{j-1}) = \sigma(Z_{j-1})\cup \sigma(Z_{j-3})$. Thus $Y_j \perp Y_{j-1}$. This could be more gruesomely detailed but I take some for granted. See this for more details including definitions etc.

Case 2, suppose $a_2 = 0$ and $a_0 = 0$. The equations become

$$\begin{align} Y_j &= a_1 Z_{j-1} \\ Y_{j-1} &= a_1 Z_{j-2} \end{align}$$

The same $\sigma$-algebra argument applies more easily but a more elegant solution presents itself in the form of the CDF.

$$\begin{align} F_{Y_j, Y_{j-1}}(y_j, y_{j-1}) &= P(Y_j \leq y_j \text{ and } Y_{j-1} \leq y_{j-1}) \\ & = F_{Z_j}(y_j/a_1)F_{Z_{j-1}}(y_{j-1}/a_1)\\ & = F_{Y_j}(y_j)F_{Y_{j-1}}(y_{j-1}) \end{align}$$

($\impliedby$) Suppose $Y_j \perp Y_{j-1}$ by theorem, we know that $E[Y_j Y_{j-1}] = E[Y_j]E[Y_{j-1}]$ calculating these values separately,

$$\begin{align} E[Y_jY_{j-1}] & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2 )\mu^2 \\ & + (a_1a_0 + a_2a_1)E[Z^2] \end{align}$$

$$\begin{align} E[Y_j]E[Y_{j-1}] & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2)\mu^2\\ &+ (a_1a_0 + a_2a_1)\mu^2 \end{align}$$

In the non-degenerate case when the distribution of $Z$ is not a constant, the variance is strictly positive so that $E[Z^2] - \mu^2 > 0$ and so $E[Z^2] > \mu^2$ and more importantly $E[Z^2] \neq \mu^2$ Thus for the equality $E[Y_j Y_{j-1}] = E[Y_j]E[Y_{j-1}]$ to hold, it must be the case that $a_1a_0 + a_2a_1 = 0$.

Write out the simple equations

$$\begin{align} Y_j &= a_0 Z_j + a_1 Z_{j-1} + a_2 Z_{j-2}\\ Y_{j-1} &= a_0 Z_{j-1} + a_1 Z_{j-2} + a_2 Z_{j-3} \end{align}$$

There are some very simple cases that make $Y_j \perp Y_{j-1}$ due to the independence assumption of the random variables $\{Z_i\}_{i\in\mathbb{Z}}$. An example is $a_0 \in \mathbb{R}\setminus \{0\},\, a_1 = 0,\, a_2 = 0$. Not sure if you were looking for a complete solution but this should help get you started.

Also, an easy check for RV which are not independent is using the contrapositive form of the common theorem $$X\perp Y \implies E[XY] = E[X]E[Y]$$ Note that the converse of this statement is not true.

Proof

Assertion $a_1 = 0 \iff Y_j \perp Y_{j-1}$

($\implies$) Suppose not, so that $a_1 = 0$ and $Y_j$ is not independent of $Y_{j-1}$.

$$\begin{align} E[Y_j Y_{j-1}] & = E[ a_0^2 Z_j Z_{j-1} + a_0 a_2Z_jZ_{j-3} + a_2 a_0 Z_{j-2} + a_2^2Z_{j-2}Z_{j-3}] \\ & = E[Y_j]E[Y_{j-1}] \end{align}$$

This is a contradiction of the theorem above.

($\impliedby$) Suppose $Y_j \perp Y_{j-1}$ by theorem, we know that $E[Y_j Y_{j-1}] = E[Y_j]E[Y_{j-1}]$ calculating these values separately,

$$\begin{align} E[Y_jY_{j-1}] = E[ &a_0^2 Z_j Z_{j-1} + a_0a_1 Z_j Z_{j-2} + a_0 a_2 Z_j Z_{j-3}\\ & a_1 a_0 Z_{j-1}^2 + a_1^2 Z_{j-1}Z_{j-2} + a_1 a_2 Z_{j-1}Z_{j-3}\\ & a_2 a_0 Z_{j-2}Z_{j-1} + a_2 a_1Z_{j-2}^2 + a_2^2Z_{j-2}Z_{j-3} ] \end{align}$$

Notice that in the terms $a_1 a_0 Z_{j-1}^2$ and $a_2 a_1Z_{j-2}^2$ the RV is squared inside the expectation. Due to Jensen's inequality, the expectation of the squared value is not equal (in general) to the squared value of the expectation. Thus, $a_1 = 0$ for the theorem to hold.

Write out the simple equations

$$\begin{align} Y_j &= a_0 Z_j + a_1 Z_{j-1} + a_2 Z_{j-2}\\ Y_{j-1} &= a_0 Z_{j-1} + a_1 Z_{j-2} + a_2 Z_{j-3} \end{align}$$

There are some very simple cases that make $Y_j \perp Y_{j-1}$ due to the independence assumption of the random variables $\{Z_i\}_{i\in\mathbb{Z}}$. An example is $a_0 \in \mathbb{R}\setminus \{0\},\, a_1 = 0,\, a_2 = 0$. Not sure if you were looking for a complete solution but this should help get you started.

Also, an easy check for RV which are not independent is using the contrapositive form of the common theorem $$X\perp Y \implies E[XY] = E[X]E[Y]$$ Note that the converse of this statement is not true.

Proof

Assertion $a_1a_0 + a_2a_1 = 0 \iff Y_j \perp Y_{j-1}$

Define $\mu = E[Z]$

($\implies$) Suppose not, so that $a_1a_0 + a_2a_1 = 0$ and $Y_j$ is not independent of $Y_{j-1}$.

$$\begin{align} E[Y_jY_{j-1}] = E[ &a_0^2 Z_j Z_{j-1} + a_0a_1 Z_j Z_{j-2} + a_0 a_2 Z_j Z_{j-3}\\ +& a_1 a_0 Z_{j-1}^2 + a_1^2 Z_{j-1}Z_{j-2} + a_1 a_2 Z_{j-1}Z_{j-3}\\ +& a_2 a_0 Z_{j-2}Z_{j-1} + a_2 a_1Z_{j-2}^2 + a_2^2Z_{j-2}Z_{j-3} ] \\ = & (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2 )\mu^2 + 0 \end{align}$$

$$\begin{align} E[Y_j]E[Y_{j-1}] & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2 + 0)\mu^2\\ & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2 )\mu^2 \end{align}$$

This is a contradiction of the (contrapositive of the) theorem above.

($\impliedby$) Suppose $Y_j \perp Y_{j-1}$ by theorem, we know that $E[Y_j Y_{j-1}] = E[Y_j]E[Y_{j-1}]$ calculating these values separately,

$$\begin{align} E[Y_jY_{j-1}] & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2 )\mu^2 \\ & + (a_1a_0 + a_2a_1)E[Z^2] \end{align}$$

$$\begin{align} E[Y_j]E[Y_{j-1}] & = (a_0^2 + a_0a_1 + a_0 a_2 + a_1^2 + a_1 a_2 + a_2 a_0 + a_2^2)\mu^2\\ &+ (a_1a_0 + a_2a_1)\mu^2 \end{align}$$

In the non-degenerate case when the distribution of $Z$ is not a constant, the variance is strictly positive so that $E[Z^2] - \mu^2 > 0$ and so $E[Z^2] > \mu^2$ and more importantly $E[Z^2] \neq \mu^2$ Thus for the equality $E[Y_j Y_{j-1}] = E[Y_j]E[Y_{j-1}]$ to hold, it must be the case that $a_1a_0 + a_2a_1 = 0$.