Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know that the correlation function between random variables $X$ and $Y$ is defined as

$$ \rho_{X,Y}=\mathrm{corr}(X,Y)={\mathrm{cov}(X,Y) \over \sigma_X \sigma_Y} ={E[(X-\mu_X)(Y-\mu_Y)] \over \sigma_X\sigma_Y}. $$

What does happen when $\mathbf{X}$ and $\mathbf{Y}$ are random vectors?

$$ \mathbf{X} = \begin{bmatrix} X_1 \\ X_2 \\ \vdots \\ X_n \end{bmatrix}, \quad \quad \mathbf{Y} = \begin{bmatrix} Y_1 \\ Y_2 \\ \vdots \\ Y_n \end{bmatrix} $$

How is the correlation function defined in this case?

$$ \mathrm{corr}(\mathbf{X},\mathbf{Y}) \, \triangleq \, ? $$

share|cite|improve this question
Then the correlation is a matrix where the $(i,j)$th entry is the correlation between $X_i$ and $Y_j$. – Stefan Hansen Mar 31 '13 at 8:40
up vote 1 down vote accepted

You almost answered your own question. Just put vectors with $X$ and $Y$ and transpose one of them to get a matrix: a covariance matrix. So, $$\mathrm{cov}(\mathbf{X},\mathbf{Y}) \, \triangleq \,(\vec{\mathbf{X}}-E[\vec{\mathbf{X}}]).(\vec{\mathbf{Y}}-E[\vec{\mathbf{Y}}])^{T}$$ is a matrix with entries: $\mathrm{cov}(\mathbf{X},\mathbf{Y})_{ij}=\langle(X_i-\mu_{X_i})(Y_j-\mu_{Y_j})\rangle$.

share|cite|improve this answer
As far as I see, "correlation" and "covariance" are different things in single variable case. Are they the same things in multivariate case? Because, my question was about correlation, not covariance. – hkBattousai Apr 1 '13 at 4:47
@hkBattousai the difference is "in the mean". If you substract the mean from the vector (or scalar, that doesn't matter) then you have covariance if you don't then correlation. It is just terminology. $\mathrm{cov}(\mathbf{X},\mathbf{Y}) \, \triangleq \,(\vec{\mathbf{X}}-E[\vec{\mathbf{X}}]).{(\vec{\mathbf{Y}}-E[\vec{\mathbf{Y}}])‌​}^{T}$ and $\mathrm{cor}(\mathbf{X},\mathbf{Y}) \, \triangleq \,\vec{\mathbf{X}}.\vec{\mathbf{Y}}^{T}$ – Caran-d'Ache Apr 1 '13 at 7:00
Isn't covariance defined as $\, \mathrm{cov}(\mathbf{X},\mathbf{Y}) \, \triangleq \, E\big[(\vec{\mathbf{X}}-E[\vec{\mathbf{X}}])(\vec{\mathbf{Y}}-E[\vec{\mathbf{Y}}‌​])^{T}\big] \,$? You wrote $\, \mathrm{cov}(\mathbf{X},\mathbf{Y}) \, \triangleq \, (\vec{\mathbf{X}}-E[\vec{\mathbf{X}}])(\vec{\mathbf{Y}}-E[\vec{\mathbf{Y}}])^{T} \,$ instead. Are you sure that your definition is correct? – hkBattousai Apr 1 '13 at 11:23
Well, yes sure, expectation should be taken. My fault, I was inattentive and one can correct the comment only during the period of 5 minutes. :( – Caran-d'Ache Apr 1 '13 at 11:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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