# measure of a degenerate Gaussian distribution

I want to do computations with a degenerate Gaussian measure, but I do not know how to represent it in a close form.

After starting with a Gaussian random variable and restricting it to a condition, I end up with a Gaussian random variable $$X\sim \mathcal N(0, \Sigma)$$, where $$\Sigma$$ is singular (say I end up with one dimension less). I guess, that one can make computations with density \begin{align} \pi(x) &= \frac{1}{\left(2\pi^{c-1} {|\Sigma|}\right)^{\frac{1}{2}}} \exp \left( -\frac 1 2 x^\intercal \Sigma^- x\right), \end{align} where $$|\Sigma|$$, $$\Sigma^-$$ denote the generalized determinant and generalized inverse respectively. What I have understand so far, is that $$\pi$$ is a density function but with respect to a degenerate measure (a sort of restriction of the Lebesgue measure). I tried working with the dicomposition theory, but didn't get any satisfactory solution so far. Since I can calculate the kernel of $$\Sigma$$, after doing a transformation with respect to the eigenvalues/eigenvectors of $$\Sigma$$, I can become another density $$\widetilde \pi$$ which ignores the first component. My guess would be that I can integrate then with respect to the Lebesgue measure in one dimension less. Does it make sense?

The main questions would be: With respect to which measure is this distribution defined and do it exist a close form for this measure?


Consider the spectral decomposition of $$\Si$$: $$$$\Si=QDQ^T,\tag{0}$$$$ where $$Q$$ is an orthogonal matrix $$n\times n$$ matrix and $$D$$ is the diagonal matrix with diagonal entries $$\si_1^2>0,\dots,\si_k^2>0,0,\dots,0$$, so that $$k$$ is the rank of $$\Si$$. Let $$Y:=Q^TX$$, so that $$X=QY$$ and $$Y=(Y_1,\dots,Y_k,0,\dots,0)\sim N(0,D)$$. So, for any Borel-measurable function $$g\colon\mathbb R^n\to\mathbb R$$ such that $$Eg(X)$$ exists, we have \begin{aligned} Eg(X)&=Eg(Q[Y_1,\dots,Y_k,0,\dots,0]^T) \\ &=\int_{\mathbb R^k}g(Q[y_1,\dots,y_k,0,\dots,0]^T) \\ &\times f_{\si_1}(y_1)\cdots f_{\si_k}(y_k)dy_1\cdots dy_k, \end{aligned} \tag{1} where $$f_\si$$ is the pdf of $$N(0,\si^2)$$.

Now you can compute the probabilities for $$X$$ as follows: $$P(X\in B)=E1(X\in B),$$ where $$B$$ is any Borel subset of $$\mathbb R^n$$.

For an illustration, here is a Mathematica notebook with the calculation of the probability $$P(X_1>X_2+1)$$ given $$(X_1,X_2)\sim N\left((0,0),\left( \begin{array}{cc} 1 & 2 \\ 2 & 4 \\ \end{array} \right)\right)$$:

As you insist on writing the expectation $$Eg(X)$$ as an integral with respect to the Lebesgue measure on the support of the distribution of $$X$$, here it is.

Note that the support of the distribution of $$X$$ is the range(=column space) $$\Si\R^n$$ of $$\Si$$, which is the same as the range $$q\R^k$$ of $$q$$, where $$q$$ is the map $$\begin{equation*} \R^k\ni z=(z_1,\dots,z_k)\mapsto qz:=QJz\in q\R^k=\Si\R^n \end{equation*}$$ and, in turn, $$J$$ is the map $$\begin{equation*} \R^k\ni z=(z_1,\dots,z_k)\mapsto Jz:=(z_1,\dots,z_k,0,\dots,0)\in\R^n. \end{equation*}$$ The equality $$\Si\R^n=q\R^k$$ follows because, in view of (0), the range $$\Si\R^n$$ of $$\Si$$ is the same as that of $$QD^{1/2}$$, which is the same as that of $$q$$ (since the range of $$D^{1/2}$$ is the same as that of $$J$$). Here, we identify matrices with the corresponding linear transformations.

In view of (1), \begin{equation*} \begin{aligned} Eg(X)&=\int_{\R^k}g(qz)\nu(dz) =\int_{\Si\R^n}g(x)\mu(dx), \end{aligned} \tag{2} \end{equation*} where $$\nu(dz):=\pi(z)\la_k(dz)$$, $$\pi(z):=f_{\si_1}(z_1)\cdots f_{\si_k}(z_k)$$, $$\la_k(dz)$$ is the Lebesgue measure on $$\R^k$$, and $$\mu(dx):=\nu(q^{-1}(dx))=\pi(q^{-1}(x))\la_k(q^{-1}(dx))$$.

Since the map $$q$$ is an isometry, for the Lebesgue measure $$\la_{\Si\R^n}$$ over $$\Si\R^n$$ we have $$\la_{\Si\R^n}(dx)=\la_k(q^{-1}(dx))$$. So, by (2), \begin{equation*} \begin{aligned} Eg(X)&=\int_{\Si\R^n}g(x)\pi(q^{-1}(x))\la_{\Si\R^n}(dx). \end{aligned} \tag{3} \end{equation*} In particular, letting $$g:=1_B$$ for any Borel set $$B\subseteq\Si\R^n$$, we get \begin{equation*} \begin{aligned} P(X\in B)&=\int_B \pi(q^{-1}(x))\la_{\Si\R^n}(dx). \end{aligned} \tag{3a} \end{equation*}

I think formulas (3) and, in particular, (3a) are completely useless, though -- because you will actually compute $$Eg(X)$$ and $$P(X\in B)$$ (as in the above Mathematica notebook) by formula (1).

• Thanks for your answer! This is what I guessed, when I meant using $\widetilde \pi$, since $$\widetilde \pi(y) = f_{\sigma 1} (y_1) \cdots f_{\sigma k}(y_k).$$ Thus, $\widetilde \pi$ can be seen as the density function of $Q^\intercal X \sim \mathcal N (0,D)$ "in a sense" (avoiding the zero components). But, is there a way to express the density of $X$ (i.e., $\pi$ in my question) w.r.t. a (degenerate) measure?, i.e., $$P(X\in B) = \int \pi(x) d?,$$ The reason for this need, is that I want to apply later transformations, and applying already one, makes everything more confusing. Mar 15, 2021 at 17:27
• @SkullSoul : I have to teach now. I'll respond to this later. Mar 15, 2021 at 17:39
• @SkullSoul : I have added the measure stuff. Mar 15, 2021 at 20:27
• @losifPinelis: Thanks a lot for the effort!! I understand what you mean and that your explanation help for numerical computations. But my question was slightly different as your answer. I formulate again with the same terminology: Given is a positiv-SEMI definite matrix $\Sigma$, $\mu = 0$ and $$\pi(x) = \frac{1}{2 \pi^{c-1} |\Sigma| } e^{-\frac{1}{2} x^T \Sigma^- x},$$ $|\Sigma|, \Sigma^-$ are generalized determinant and inverse. Is there a way to write the degenerate measure $\zeta$ in a close form, i.e. if $$Pr(X \in B) = \int_B \pi(x) d\zeta,$$ then $$\zeta = ?$$ Mar 18, 2021 at 20:34
• @SkullSoul : Of course, the probabilities $P(X\in B)$ can be expressed as integrals with respect to the Lebesgue measure on the support of the distribution of $X$. I have now added the corresponding formula (3a) -- which is a particular case of the previously given formula (3). Formula (3a) has the correct expression, $\pi(q^{-1}(\cdot))$, for the density of the distribution of $X$ with respect to the Lebesgue measure on the support of the distribution of $X$. Mar 18, 2021 at 21:06