Suppose I have $n$ variables $X: X_1, X_2, ..., X_n$ that are independent from each other.

Which means that: if $i≠j$, then $\text{Cov}(X_i, X_j) = 0$

As a consequence, I'm wondering if their Covariance Matrix Sigma should be a diagonal matrix...

Someone to confirm this last point??


PS: Covariance matrix sigma defined in Wikipedia: https://en.wikipedia.org/wiki/Covariance_matrix

  • 9
    $\begingroup$ Yes, the covariance matrix is diagonal. I am not sure where your source of confusion is since the condition $\operatorname{cov}(X_i,X_j)=0$ for all $i \neq j$ is exactly the condition that is needed to claim that the covariance matrix is diagonal. $\endgroup$ May 6 '15 at 12:51
  • 1
    $\begingroup$ It seemed obvious to me but I just wanted to be absolutely sure. $\endgroup$
    – tmangin
    May 6 '15 at 14:15

Independence implies zero correlation (but the converse doesn't hold):

$\:\:E(XY)=\int\int \,x\,y\, f(x,y)\, dy \,dx$

$\qquad\qquad=\int\,\int x\,y \,f(x)\,f(y)\, dy\, dx\quad$ (independence)

$\qquad\qquad=\int\, y\, f(y)\, dy\,\cdot\,\int\, x \,f(x)\, dx$


Hence $\text{Cov}(X,Y)=E(XY)-E(X)E(Y)=E(X)E(Y)-E(X)E(Y)=0$

Consequently each off-diagonal term in the covariance matrix should be 0.

  • $\begingroup$ That's assuming the random variables have pdfs? $\endgroup$
    – BCLC
    Jun 12 '15 at 18:48
  • 2
    $\begingroup$ @BCLC Yes but you can construct a very similar but more general argument of the same form, as long as the expectations all exist. $\endgroup$
    – Glen_b
    Jun 13 '15 at 0:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.