Evaluating covariance for independent variables.

Question

Let $x_i,x_j$, two independent variables.

Consider this calculation:

$$\text{Cov}(x_i, x_ix_j) = E(x_i^2x_j) - E(x_i)E(x_ix_j) = E(x_i^2x_j) - E(x_i^2)E(x_j) \\= E(x_i^2)E(x_j) - E(x_i^2)E(x_j) = 0$$

Questions:

1. Why is true that $E(x_i)E(x_i) = E(x_i^2)$?
2. Why is true that $E(x_i^2x_j) = E(x_i^2)E(x_j)$. We don't know that $x_i^2$ and $x_j$ are independent.

Answer 1

It is wrong. You have $$ Cov ( X_i, X_iX_j) = E[X_i^2 X_j] - E[X_i]E[X_iX_j] $$ since $X_i$ and $X_j$ are independent we can split up the expected values: $$ Cov ( X_i, X_iX_j) = E[X_i^2]E[ X_j] - E[X_i]^2 E[X_j] $$ Observe the difference with your computations. The second term has the "square" outside. Finally you can write: $$ Cov ( X_i, X_iX_j) = \left(E[X_i^2] - E[X_i]^2\right) E[X_j] = Var[X_i] E[X_j]. $$

Then the covariance is 0 if and only if, either $Var[X_i]=0$ or $E[X_j]=0$.

Conclusion:

1) It is not true.

2) Yes, they are of course still independent. Take a look at this post: Are functions of independent variables also independent?

Answer 2

$E[X^2]\not = E[X]E[X]$ in general. Consider a random variable taking the values $-1$ and $+1$ with equal probability. Then $E[X]=0$ and $E[X^2]=1$.