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I would appreciate if you could help me to find the following integral, thank you.

$$f(u)= \int_{-\infty }^{\infty} \int_{-\infty }^{\infty} \frac{e^{ -(x-a)^2/2b^2} }{{b\sqrt {2\pi}}} \frac{e^{ -(y-c)^2/2d^2} }{{d\sqrt {2\pi}}} \delta (xy-u) dx dy$$

where $\delta()$ is delta function and a, b, c , d are real numbers.

According to MathWorld when a and c are zero the answer would be $$\frac {K_0 (\frac{|u| }{bd})}{ \pi bd}$$

where $K_0()$ is modified Bessel function. Now what if a and c are not zero?

I simplified it as follows (not sure if it is correct)

$$f(u)=\frac1{2\pi bd} \int_{-\infty }^{\infty} \frac 1{|y|} e^{ -(\frac uy-a)^2/2b^2} e^{ -(y-c)^2/2d^2} dy$$

May
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  • So this integrates along the hyperbola $y=u/x$ for non-zero $u$. Are you sure an analytic solution exists? For $u=0$ you are integrating either $x=0$ or $y=0$. You can separate out that case and the integral should split, change variables $z = (x-a)/b$ to get $e^{-x^2/2}/\sqrt{2\pi}$, which should give you 1. – gt6989b Sep 16 '13 at 21:03
  • There is a small difference between your expression and the one of the Normal Product Distribution you cite, i.e. in the denominators of the exponents you should take $2b^2$ and $2d^2$ instead of $2b$ and $2d$. Is it just a typo or you want to compute exactly what you wrote? – Giovanni De Gaetano Sep 16 '13 at 21:07
  • @gt6989b Since there is a solution for the case that a and c are zero, it should have a solution. – May Sep 16 '13 at 21:08
  • @GiovanniDeGaetano Thanks a lot! I edited the question. – May Sep 16 '13 at 21:10
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    just to make it more concrete, how to get rid of $\delta$: substitute $y=u/x$ in the exponentials,then multiply them by $dx/|x|$, and integrate over $x\in\mathbb R$ (not that I know how to do it, bu the integral looks Besselish). Perhaps it's not a smart move to get rid of $\delta$. – user8268 Sep 16 '13 at 21:16
  • Do you have a proof for the $a=c=0$ case? I think that the knowledge of the right integral representation of $K_\nu$ could unlock the problem... – Giovanni De Gaetano Sep 16 '13 at 21:18
  • @GiovanniDeGaetano unfortunately I don't. You are right that can help! – May Sep 16 '13 at 21:19
  • @user8268 using your simplification method, it would be: $$f(u)=\frac1{2bd\pi} \int_{-\infty }^{\infty} e^{ -(\frac uy-a)^2/2b^2} e^{ -(y-c)^2/2d^2} dy$$ – May Sep 16 '13 at 21:29
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    @May There is $1/|y|$ missing (provided I got it right) – user8268 Sep 16 '13 at 21:32
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    The original paper by C. Craig (1936), web.ipac.caltech.edu/staff/fmasci/home/statistics_refs/GaussianProduct1934.pdf contains the distribution for non-zero and different means. Just be careful to follow his notation. – Alecos Papadopoulos Sep 16 '13 at 23:00
  • @AlecosPapadopoulos Thank you. – May Sep 16 '13 at 23:04
  • $\Large\delta\left(xy\ -\ u\right) = {\delta\left(y\ -\ u/x\right) \over \left\vert x \right\vert}$ . – Felix Marin Sep 18 '13 at 20:02

1 Answers1

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In this answer, it is shown that when composing the dirac delta with $g(x)$, we get $$ \int_{\mathbb{R}^n} f(x)\,\delta(g(x))\,\mathrm{d}x=\int_{\mathcal{S}}\frac{f(x)}{|\nabla g(x)|}\,\mathrm{d}\sigma(x) $$ where $\mathcal{S}$ is the surface on which $g(x)=0$ and $\mathrm{d}\sigma(x)$ is standard surface measure on $\mathcal{S}$.

In the given integral, $\mathrm{d}\sigma(x)$ is arclength and $$ |\nabla g(x)|=|(y,x)|=\sqrt{x^2+y^2} $$ If we parameterize the curve $xy=u$ by $y=u/x$, we get that $$ \begin{align} \mathrm{d}\sigma(x) &=\sqrt{1+y'^2}\,\mathrm{d}x\\ &=\sqrt{1+\frac{u^2}{x^4}}\,\mathrm{d}x\\ \end{align} $$ Therefore, the integral becomes $$ \begin{align} &\int_{-\infty}^\infty\frac{e^{ -(x-a)^2/2b^2} }{{b\sqrt {2\pi}}} \frac{e^{ -(u/x-c)^2/2d^2} }{{d\sqrt {2\pi}}}\sqrt{1+u^2/x^4}\frac{\mathrm{d}x}{\sqrt{x^2+u^2/x^2}}\\ &=\int_{-\infty}^\infty\frac{e^{ -(x-a)^2/2b^2} }{{b\sqrt {2\pi}}} \frac{e^{ -(u/x-c)^2/2d^2} }{{d\sqrt {2\pi}}}\frac{\mathrm{d}x}{|x|} \end{align} $$ which is what you got.

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