Suppose $\alpha=U\beta$ where $U$ is $N\times K$ with $N > K$. What is the probability density function (PDF) of $\beta$, $p(\beta)$, given that we know that it is proportional to $q$, the PDF of $\alpha$ ? In other words, if $$ p(\beta) = A q(U\beta) \quad \int_{\mathbb{R}^K} p(\beta)d\beta =\int_{\mathbb{R}^N} q(\alpha) d\alpha = 1$$ is there a simple way to calculate $A$?

Of course, we can write $$A^{-1} = \int_{\mathbb{R}^K}p(U\beta)d\beta$$ and then go through the math, but I was wondering if there was a shortcut, maybe a sort of jacobian function of $U$, that would be easier to compute than this integral.

If it helps, I need it when $q$ is a product of independent Laplace distributions on each $\alpha_i$, and $U$ is a matrix with a $1$ and a $-1$ for every line. Note that the answer is not the multivariate laplace distribution.

EDIT: to clarify, $\int_{\mathbb{R}^N}q(\alpha)d\alpha=1$, but we are interested in the value of this integral $\int_\Omega q(\alpha)d\alpha$, where $\Omega=\{U\beta, \quad\beta\in\mathbb{R}^K\}$

  • $\begingroup$ I can't make much sense of this, because I cannot see how an $N$-variate distribution could in any meaningful sense be "proportional to" a $K$-variate distribution. Could you please clarify what you mean? What kind of object is $A$, exactly? $\endgroup$ – whuber Feb 1 '17 at 16:00
  • $\begingroup$ @whuber A is a positive real. The function $\beta \to q(U\beta)$ is not normalized with respect to $\beta$ (but $q$ is with respect to $\alpha$), so we seek the normalizing constant. Does it make more sense now? $\endgroup$ – yannick Feb 1 '17 at 16:10
  • $\begingroup$ Not quite, because more needs to be assumed for these manipulations to be valid. For instance, $\alpha$ as defined must be located within some $K$-dimensional subspace of $\mathbb{R}^N$ and, since $N \gt K$, it cannot have a density. Thus the very meaning of $q$ is doubtful. $\endgroup$ – whuber Feb 1 '17 at 19:11
  • $\begingroup$ @whuber edited. Is it clearer now? $\endgroup$ – yannick Feb 2 '17 at 9:30

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