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. $\endgroup$ Commented May 6, 2015 at 12:51
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    $\begingroup$ It seemed obvious to me but I just wanted to be absolutely sure. $\endgroup$
    – tmangin
    Commented May 6, 2015 at 14:15

1 Answer 1


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
    Commented Jun 12, 2015 at 18:48
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    $\begingroup$ @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
    Commented Jun 13, 2015 at 0:33

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