# Does the covariance of i.i.d. random vectors/multivariate random variables have any zero terms?

If we have i.i.d. random variables, $$X$$ and $$Y$$, then $$\text{Cov}(X,Y)=0$$.

But let's say we have i.i.d. random vectors $$\boldsymbol{X}$$ and $$\boldsymbol{Y}$$, where $$\boldsymbol{X}=(X_{1},...,X_{p})$$ and $$\boldsymbol{Y}=(Y_{1},...,Y_{q})$$. Do we have any properties analogous to the univariate case: $$\text{Cov}(X,Y)=0$$, that always hold true, as a result of these random vectors being i.i.d.? Something like $$\text{Cov}(X_i,X_j)=0$$ if $$i\neq j$$, or $$\text{Cov}(X_i,Y_i)=0$$, or maybe the independence of $$\boldsymbol{X}$$ and $$\boldsymbol{Y}$$ gives a covariance matrix with zero terms everywhere except the diagonal (these are all just guesses, and not necessarily true). Are there any properties like this for i.i.d. random vectors?

## 1 Answer

If the random vectors $$X, Y$$ are independent, then, as functions of independent random variables are independent (see Functions of Independent Random Variables), $$X_i$$ and $$Y_j$$ are independent, since selecting one given component from a vector is a function. But the covariance of independent random variables are zero, that gives you the result.

In your question you speculate about a condition $$i \not= j$$, but that is unnecessary here. $$X_i$$ and $$Y_i$$ re also independent. The covariance matrix $$\DeclareMathOperator{\C}{\mathbb{C}} \C(x,y)$$ will be the zero matrix, also the diagonal.