Is the autocovariance function of a sequence identically zero if the sequence is iid?

Question

My professor gives the following definition for the autocovariance function. $$\rho(i,j) = Cov(X_i , X_j)$$$\\$If I have a sequence that is iid, when i compute $\rho(n,n+1)$ for $n \geq 0$, I found that it is always zero (since they are independent). Is this accurate?

Answer 1

Yes, since $X_i$ and $X_j$ are independent for $i \neq j$, their covariance is $0$, and hence the autocovariance function has value $0$ for all $i \neq j$. Note that identical distribution is not needed for this result to hold; independence suffices. In fact, all that is needed is for $X_i$ and $X_j$ to be uncorrelated for all $i \neq j$, which is true when the random variables are independent, but can also happen without independence.

Answer 2

Yes, except for a minor technical condition pertaining to the existence of moments, eg Cauchy distribution.