# No probability density for linearly dependent random vector?

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?

Thanks.

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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.
 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? – ido 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