2
$\begingroup$

This question is about finding a covariance matrix and I wasn't sure about the final step.

Given a standard $d$-dimensional normal RVec $X=(X_1,\ldots,X_d)$ has i.i.d components $X_j\sim N(0,1)$. Here we treat $X$ as a row vector. Now consider $Y:=\mu+XA$, where $\mu\in\mathbb{R}^m$, $A\in\mathbb{R}^{d\times m}$, so that $Y\in\mathbb{R}^m$. Then (we write the covariance matrix of Y as $C_Y^2$), $$C_Y^2=\mathbb{E}(Y-\mathbb{E}Y)^\intercal(Y-\mathbb{E}Y)=\mathbb{E}(XA)^\intercal XA$$ $$=\mathbb{E}A^\top X^\top XA=A^\top(\mathbb{E}X^\top X)A=A^\top IA=A^\top A.$$

I wasn't sure about how to compute $\mathbb{E}X^\top X$. Since $X$ is a row vector then $X^\top X$ is a column vector multiplied by a row vector which resulted in a matrix, then how can we take the expected value of a matrix? Moreover how can we get the identity matrix as the product?

Also how to choose dimension d and m, it is said (without reason) that it is not good if m>d.

May anyone explain it to me? Thanks.

$\endgroup$

1 Answer 1

3
$\begingroup$

You do it term by term, which you can do because Expectation is a linear operator. You get the identity matrix because the $X_i$ terms are independent and $N(0,1)$ so $E[X_i*X_i]=1$ but $E[X_i*X_j]=0.$

$\endgroup$
1
  • $\begingroup$ Yes you are right, thank you. $\endgroup$
    – user71346
    May 26, 2013 at 4:24

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.