Q.7.39 - A First Course in Probability

Question

Let $X_1, . . .$be independent with common mean $\mu$and common variance ${\sigma }_2$, and set $Y_n=X_n+X_{n+1}+X_{n+2}$. For $j\ge 0$, find $Cov\left(Y_n,Y_{n+j}\right).$

Step-by-Step Solution

Answer

The value of $Cov\left(Y_n,Y_{n+j}\right)$ is $0.$

Step 1: Given Information

Independent variable $=X_1$

Mean $=\mu$

Common variance $={\sigma }_2$

Set function $Y_n=X_n+X_{n+1}+X_{n+2}$

For $j\ge 0$, find $Cov\left(Y_n,Y_{n+j}\right)$

Step 2: Explanation

From the information, observe that $X_1,\dots ..$ be independent with common mean $\mu$ and common variance ${\sigma }^2$

we have that,

$Cov\left(Y_n,Y_n\right)=Var\left(X_n+X_{n+1}+X_{n+2}\right)$

$=Var\left(X_n\right)+Var\left(X_{n+1}\right)+Var\left(X_{n+2}\right)$

$={\sigma }^2+{\sigma }^2+{\sigma }^2$

$=3{\sigma }^2$

Due to the variance of sum of $n$ independent variables with common distribution

Step 3: Explanation

If $j=1$, we have that

$Cov\left(Y_n,Y_{n+1}\right)=Cov\left(X_n+X_{n+1}+X_{n+2},X_{n+1}+X_{n+2}+X_{n+3}\right)$

$=Var\left(X_{n+1}\right)+Var\left(X_{n+2}\right)$

$={\sigma }^2+{\sigma }^2$

$=2{\sigma }^2$

If $j=2,$we have that,

$Cov\left(Y_n,Y_{n+2}\right)=Cov\left(X_n+X_{n+1}+X_{n+2},X_{n+2}+X_{n+3}+X_{n+4}\right)$

$=Var\left(X_{n+2}\right)$

$={\sigma }^2$

For $j \ge 3,$ we see that, $Cov\left(Y_n,Y_{n+j}\right)=0$

Since the definition of $Y_n$ and basic properties of covariance to obtain the required.

Step 4: Final Answer

Hence, the value of $Cov\left(Y_n,Y_{n+j}\right)$ is $0.$