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I have to calculate the following integral:

$$\frac{c^{-2n}}{\sqrt{2n}}\int_0^\infty\cdots\int_0^\infty \left( \sqrt{x_1 ^2 +\cdots+x_n ^2}\right) (x_1\ldots x_n) \exp{ \left( -\frac{x_1^2+\cdots+x_n^2}{2c^2} \right) } \, dx_1 \cdots dx_n$$

by transforming to spherical polar coordinates. Looking here I found the transformations, but I'm not sure what good this does with this ugly factor of $x_1 \ldots x_n$ sitting there. Is this doable at all or did I screw up somewhere earlier deriving the formula?

This arises when I want to test if the estimator $\hat{c} = \frac{X_\mathrm{RMS}}{\sqrt{2}}$ for the Rayleigh distribution is unbiased, so I'm calculating $\frac{1}{\sqrt{2n}} E\left[\sqrt{X_1^2+\cdots+X_n^2}\right]$ hoping to get $c$.

Michael Hardy
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Spine Feast
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1 Answers1

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Note that $$ \int_0^\infty xe^{-x^2/2}\,\mathrm{d}x=1\tag{1} $$ Thus, multiplying $n$ copies of $(1)$ together yields $$ \int_{\mathbb{R_+^n}}x_1x_2\dots x_n\,e^{|x|^2/2}\,\mathrm{d}x=1\tag{2} $$ Consider the following integral over a quadrant of the surface of the unit sphere: $$ \int_{S_+^{n-1}}x_1x_2\dots x_n\,\mathrm{d}\sigma(x)=\Lambda_{n-1}\tag{3} $$ Integrating $(3)$ times $e^{-r^2/2}$ over quadrants of spheres of radius $r$ and thickness $\mathrm{d}r$ yields $$ \begin{align} \int_{\mathbb{R_+^n}}x_1x_2\dots x_n\,e^{|x|^2/2}\,\mathrm{d}x &=\int_0^\infty\Lambda_{n-1}r^{2n-1}e^{-r^2/2}\,\mathrm{d}r\\ &=2^{n-1}\Lambda_{n-1}\Gamma(n)\tag{4} \end{align} $$ Equating $(2)$ and $(4)$ yields $$ \Lambda_{n-1}=\frac1{2^{n-1}\Gamma(n)}\tag{5} $$ The integral in question adds one more power of $r$ to the integrand in $(4)$ $$ \begin{align} \frac{c^{2n}}{\sqrt{2n}}\int_{\mathbb{R_+^n}}|x|\,x_1x_2\dots x_n\,e^{|x|^2/(2c^2)}\,\mathrm{d}x &=\frac{c}{\sqrt{2n}}\int_{\mathbb{R_+^n}}|x|\,x_1x_2\dots x_n\,e^{|x|^2/2}\,\mathrm{d}x\\ &=\frac{c}{\sqrt{2n}}\int_0^\infty\Lambda_{n-1}r^{2n}e^{-r^2/2}\,\mathrm{d}r\\ &=\frac{c}{\sqrt{2n}}\frac{2^{n-1/2}\Gamma(n+1/2)}{2^{n-1}\Gamma(n)}\\ &=\frac{c}{4^n}\binom{2n}{n}\sqrt{\pi n}\tag{6} \end{align} $$ Asymptotically, $\displaystyle\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}$. Therefore, $\displaystyle\frac{c}{4^n}\binom{2n}{n}\sqrt{\pi n}\sim c$.

robjohn
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  • @robjohn: This is probably the simplest way to go. Very nice. (+1) – user26872 Jun 14 '13 at 21:40
  • The modulus in (2) and after is shorthand for $\sqrt{x_1 ^2 + \cdots x_n ^2}$ right? And what precisely is meant by $d \sigma (x)$ in (3) ? This is all kind of new territory for me :) – Spine Feast Jun 14 '13 at 22:00
  • @DepeHb: yes, $|x|=\sqrt{x_1^2+\dots+x_n^2}$ . $\mathrm{d}\sigma(x)$ is surface area on the sphere. – robjohn Jun 14 '13 at 22:05
  • What is the exact definiton of the set $S^{n-1}_{+}$ ? – Spine Feast Jun 15 '13 at 16:13
  • $\mathbb{R_+^n}$ is the quadrant of $\mathbb{R^n}$ with non-negative coordinates. $S_+^{n-1}$ is the quadrant of $S^{n-1}$ (the unit sphere) with non-negative coordinates: $\left{(x_1,x_2,\dots,x_n):x_j\ge0\text{ and }\sum\limits_{j=1}^nx_j^2=1\right}$. – robjohn Jun 15 '13 at 17:03
  • Ah, I was thrown off by the word quadrant. In reality, it's kind of an "$2^n$-drant" right? – Spine Feast Jun 15 '13 at 17:10
  • @DepeHb: yes. octant is similarly hampered. I think that quadrant is commonly used in $\mathbb{R}^n$. – robjohn Jun 15 '13 at 18:31
  • Another thing that's not very clear to me is this: how do you use the result for the unit sphere ($\Lambda_{n-1}$) in integration over spheres of arbitrary radius. I would imagine that if $\Lambda_{n-1} (1) := \Lambda_{n-1}$ and $\Lambda_{n-1} (r)$ would be that integral over sphere of radius $r$, there is some sort of factorisation in the form $\Lambda_{n-1}(r) = (something) \Lambda_{n-1} (1)$, but I don't see how to get it - if I have $\Lambda_{n-1} (r) = \int_{S_{+}^{n-1} (r)} x_1 \cdots x_n d\sigma(r)$ where $d\sigma(r)$ is the surface element of a sphere of radius $r$, then – Spine Feast Jun 15 '13 at 19:30
  • how do I relate $\sigma(r)$ to $\sigma(1)$. It seems that it should be $d\sigma(r)=r^n \sigma(1)$ for an $n$-dimensional sphere, but how to prove this? – Spine Feast Jun 15 '13 at 19:38
  • Actually, it seems to be $d\sigma(r) = r^{n-1} d\sigma(1)$ for a sphere in $n$ dimensions, which is a $(n-1)$-dimensional object. – Spine Feast Jun 15 '13 at 19:48
  • @DepeHb: yes, the surface area of the projection of a small patch on $S_+^{n-1}$ onto the sphere of radius $r$ is $r^{n-1},\mathrm{d}\sigma$ where $\mathrm{d}\sigma$ is the surface area on the unit sphere. – robjohn Jun 15 '13 at 22:16
  • Could you point me to a proof of this fact? – Spine Feast Jun 15 '13 at 23:09
  • When you scale an $n$ dimensional object by a factor of $m$ its $n$-dimensional measure scales by $m^n$. This is because the Jacobian is the determinant of an $n\times n$ identity matrix times $m$. $S^{n-1}$ is $n-1$ dimensional. See Substitution for multiple variables. – robjohn Jun 15 '13 at 23:27