Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For a random vector, I know that if the covariance matrix is non invertible, the random vector doesn't have a pdf. However, is there an intuitive explanation why linear dependence between the variables in a random vector infers non existence of a pdf?


share|cite|improve this question

1 Answer 1

up vote 3 down vote accepted

Linear dependence of a random variable $X$ with values in $\mathbb R^n$ means there exists a strict subspace $V$ of $\mathbb R^n$ such that $\mathbb P(X\in V)=1$. Since $V$ has Lebesgue measure zero, the distribution of $X$ has no density.

share|cite|improve this answer
Thanks, I understand that, but is there an explanation with cdfs? I think that it would somehow mean that the $F_X(x_1,x_2,...)$ will not be differentiable. But how is it intuitively shown when the vector is linearly dependent? – yoki Oct 21 '12 at 10:19
To me, intuition passes by the explanation in my answer more than by CDFs. One can try to translate things into CDFs but this is awkward. – Did Oct 21 '12 at 10:25

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.