# Calculate some weird RV's variance

• July 10th 2010, 08:04 AM
JoachimAgrell
Calculate some weird RV's variance
Can you help me with something? I get stuck everytime i try to solve the following problem.

Let Xi, 1<=i<=10 be independent random variables such that EXi = 10 and Var Xi = 6. Let $Y=X_1X_2+X_2X_3+...+X_{10}X_{11}$
Calculate Var Y.

I tried to solve it using the fact that Var Y = Cov(Y,Y). That leads to:

$Var Y = Cov(Y,Y) = Cov(\sum_{i=1}^{10}X_iX_{i+1},\sum_{j=1}^{10}X_jX_ {j+1})=$
$=\sum_{i,j}Cov(X_iX_{i+1},X_jX_{j+1})=$
$=\sum_{i=j}Cov(X_iX_{i+1},X_jX_{j+1})+2\sum_{i

The second term can be decomposed in two types:
1. $Cov(X_iX_{i+1},X_{i+1}X_{i+2})$
2. $Cov(X_iX_{i+1},X_{j}X_{j+1}), j>i+1$
The second type is zero because the variables are independent. The first type can be calculated:

$Cov(X_iX_{i+1},X_{i+1}X_{i+2})=E(X_iX_{i+1}^2X_{i+ 2})-E(X_i)E(X_{i+1})^2E(X_{i+2})$
$=EX_iEX_{i+2}(E(X_{i+1}^2)-E(X_{i+1})^2)=EX_iEX_{i+2}Var(X_{i+1})$

Ok. But here's the problem: look what i get when I calculate Cov(XiXi+1,XiXi+1):

$Cov(X_iX_{i+1},X_iX_{i+1})=\mathbb{E}(X_i^2X_{i+1} ^2)-\mathbb{E}(X_i)^2\mathbb{E}(X_{i+1})^2=$
$E(X_i^2)E(X_{i+1}^2)-E(X_i)^2E(X_{i+1})^2=$
$=E(X_i^2)E(X_{i+1}^2)-E(X_i)^2E(X_{i+1})^2+E(X_i)^2E(X_{i+1}^2)-E(X_i)^2E(X_{i+1}^2)=$
$=E(X_{i})^2Var(X_{i+1})+E(X_{i+1}^2)Var(X_i)$

And I can't get rid of this E(X_{i+1)^2)!

I appreciate any help. Thanks
• July 10th 2010, 02:38 PM
SpringFan25
i haven't not checked your working at all, but:

its always true that:
$V(X) = E(X^2) - (E(X))^2)$

so
$6 = E(X^2) - 100$
$E(X^2) = 106$
• July 10th 2010, 03:45 PM
JoachimAgrell
Jeepers! How could i not realize it was that simple?

Thank you so much!