I have a series of questions, in various degrees of befuddled muddledness (and they are related to my previous questions: this and this)

First question: how do I do a change of variable if the determinant of the jacobian is singular?

The setting for this question is as follows: I have one $n$-dimensional standard gaussian random variable $u \sim N(0,I)$ and a fixed $v \in \mathbb{R}^n$. Then I define the random variable $$z = u - \frac{u^Tv}{v^Tv}v$$ and I'd like to derive a density for $z$. So the Jacobian: $$\frac{dz}{du} = I - \frac{vv^T}{v^Tv}$$ which turns out to be singular and so $|\frac{dz}{du}|=0$. Does this mean trying to do a change of variable is fundamentally wrong here? Or is there a way to do this?

Second question: Aside from the Jacobian, I'm not sure how to change a standard normal distribution on $u$ to a distribution on $z$. So if the density on $u$ is $$\frac{1}{\sqrt{2\pi}}\exp(-u^Tu/2)$$ is there an inverse function $z^{-1}$ such that $z^{-1}(z) = u$? Then (I think) $$\frac{1}{\sqrt{2\pi}}\exp(-u^Tu/2) du = \frac{1}{\sqrt{2\pi}}\exp(-z^{-T}z^{-1}/2) \left|\frac{dz}{du}\right| du$$ So, is there such a $z^{-1}$? And if there is, what is it?

Third question: ultimately, I'm trying to answer this question. Am I going about this the right way by asking the two questions above? (The person who replied to my question there says something about Haar measure which I'd never heard of before so it's not enlightening to me as a proof.)

  • 7
    $\begingroup$ Roughly speaking, a change of variables must preserve dimension in some way, you are projecting onto a smaller subspace. Your change of variables should be something like $u \mapsto (z, \frac{u^Tv}{v^Tv})$. Then your Jacobian should be invertible. $\endgroup$
    – copper.hat
    May 1, 2012 at 5:32
  • $\begingroup$ @copper.hat Thanks! Could you help out with the second question, too? $\endgroup$
    – JasonMond
    May 2, 2012 at 2:23
  • $\begingroup$ The map $z\mapsto u$ is an orthogonal projection $\pi$ onto the $(n-1)$-dimensional subspace $V:=\langle v^\perp\rangle$. There is a general principle that "pushes forward" measures along with a map. The resulting measure on $V$ will again be Gaussian. $\endgroup$ Oct 3, 2012 at 18:49

1 Answer 1


I think for the second question... If $f$ is the density of v.a. $u$ with $$f(u\;\vert\;\mu={\bf 0},\sigma={\bf I})$$ according to the transformation $z$, then $$f(z\;\vert\;\tilde{\mu},\tilde{\sigma})=f\left(u(z)\;\vert\;\mu={\bf 0},\sigma={\bf I}\right)\left|\frac{\partial u}{\partial z}\right|$$ (respect to $dz$!!!)

P.D.: Excuse my English ;)

  • $\begingroup$ Irrelevant. As explained in the comments, the distribution of $z$ has no density on $\mathbb R^n$. $\endgroup$
    – Did
    Apr 16, 2016 at 10:15

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.