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This wikipedia article https://en.wikipedia.org/wiki/Probability_density_function has the formula ${\displaystyle f_{Y}(y)=\int _{\mathbb {R} ^{n}}f_{X}(\mathbf {x} )\delta {\big (}y-V(\mathbf {x} ){\big )}\,d\mathbf {x}}$ for density for random variable $Y=V(X)$, for a differentiable function $V$.

Since $\delta$, which is supposed to be the delta function, it should be a distribution acting on $f_X$.

What confuses me is it is a composition with a function. Since the integral should be only formal, and thus $\mathbf{x}$ is only formal, I am not sure how this distribution is defined. I guess distributional derivative of some sort of Heaviside function?

What kind of distribution is this? There are many questions on compositions of delta function on this website, but they seem to consider formal calculations.

I think my problem is I don't know how the functional $\delta(y-V(\cdot))$ is defined. The very definition seems to take the variable $\mathbf{x}$

kisten
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  • I wouldn't worry about it. The point is that they are showing you how to get the density for $V: R^n \to R$ – Andrew Mar 15 '23 at 13:07
  • Have a look here. Maybe it helps ? – user111 Mar 15 '23 at 14:43
  • @user111 Thinking about the support of distribution kind of helps? But I think the problem is I don't know how the functional $\delta(y-V(\cdot))$ is defined. The very definition seems to take the variable $\mathbf{x}$. – kisten Mar 16 '23 at 11:10
  • As far as I can tell, $\delta(y-V(\cdot))$ should be defined such that $\int f(x)\delta(y-V(x))dx=f(s)$ where $s$ satisfies $V(s)=y$. It's only well defined I suppose if $V(\cdot)-y$ has only one zero. Or it has no zeros in which case the integral is zero. – K.defaoite Mar 16 '23 at 11:21

1 Answers1

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They mean $$\int_{\Bbb{R}^n} f(x)\delta(y-V(x))dx= \lim_{n\to \infty} \int_{\Bbb{R}^n} f(x)2n\, 1_{|y-V(x)|< 1/n}dx$$

$\delta(t)$ is the distribution which is the limit in the sense of distributions of the sequence of functions $2n \, 1_{|t|<1/n}$.

Writing distributions as limits of sequences of distributions makes composition with functions, change of variables, differentiation (and so on) meaningful and consistent.

reuns
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  • The choice of the kernel $n1_{|y-V(x)|<1/n}$ seems to be somewhat arbitrary...? Is it a good kernel and the definition of "$\delta_{y,V}$" is independent of the choice of the approximation kernel as long as it is a good kernel or something...? – kisten Mar 19 '23 at 18:12
  • @kisten It approximates $\delta$ in the strong dual of $C^0(\Bbb{R})$. If the result depends on the chosen approximation then you should better wonder about $V$, not the chosen approximation. – reuns Mar 19 '23 at 18:24
  • But $V$ is given. The functional under study should depend on $V$. – kisten Mar 20 '23 at 12:04