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The integrand is nothing but a Gaussian/Normal distribution of $b$, where $b \in \mathbb{R}$ and $a,x \in \mathbb{R}^p$. Is there a way to compute the integral without expanding the square, which has $(1+p)^2$ terms?

Also,how about $\int_{[-1,1]\times \cdots \times [-1,1]} x\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(a^Tx-b)^2}dx$? The result of this integral should be a vector in $\mathbb{R}^p$.

I guess the solution is to first write the integrand as a multivariate Gaussian in $\mathbb{R}^p$. I tried but got stuck.

David Tan
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  • Based on Robert's reply, I realized the integrals should be defined on, say, $[-1,1] \times, \cdots, \times [-1,1]$. – David Tan May 13 '14 at 07:27

1 Answers1

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The first integral diverges, because the integrand is constant in directions orthogonal to $a$. The second is even worse: it grows in magnitude in directions orthogonal to $a$.

EDIT: If you integrate over $[-1,1]^p$, you get a really complicated answer involving exponentials and the error function. Here's the case $p=2$, according to Maple:

$$1/2\,{\frac {\sqrt {2}\sqrt {\pi } \left( -a_{{1}}-a_{{2}}+b \right) {{\rm erf}\left(-1/2\,\sqrt {2}a_{{1}}-1/2\,\sqrt {2}a_{{2}}+1/2\,\sqrt {2}b\right)} }{a_{{2}}a_{{1}}}}-1/2\,{\frac {\sqrt {2}\sqrt {\pi } \left( -a_{{1}}+ a_{{2}}+b \right) {{\rm erf}\left(-1/2\,\sqrt {2}a_{{1}}+1/2\,\sqrt {2}a_{{2}}+1/2\,\sqrt {2}b\right)} }{a_{{2}}a_{{1}}}}-1/2\,{\frac {\sqrt {2}\sqrt {\pi } \left( a_{{1}}-a _{{2}}+b \right) {{\rm erf}\left(1/2\,\sqrt {2}a_{{1}}-1/2\,\sqrt {2}a_{{2}}+1/2\,\sqrt {2}b\right)} }{a_{{2}}a_{{1}}}}+1/2\,{\frac {\sqrt {2}\sqrt {\pi } \left( a_{{1}}+a _{{2}}+b \right) {{\rm erf}\left(1/2\,\sqrt {2}a_{{1}}+1/2\,\sqrt {2}a_{{2}}+1/2\,\sqrt {2}b\right)} }{a_{{2}}a_{{1}}}}-{\frac { \left( -{{\rm e}^{-1/2\,{a_{{1}}}^{2}-1/2 \,{a_{{2}}}^{2}-1/2\,{b}^{2}}}-{{\rm e}^{-1/2\,{a_{{1}}}^{2}+2\,a_{{1} }b-1/2\,{a_{{2}}}^{2}+2\,a_{{2}}b-1/2\,{b}^{2}}}+{{\rm e}^{-1/2\,{a_{{ 1}}}^{2}+2\,a_{{2}}a_{{1}}+2\,a_{{1}}b-1/2\,{a_{{2}}}^{2}-1/2\,{b}^{2} }}+{{\rm e}^{-1/2\,{a_{{1}}}^{2}+2\,a_{{2}}a_{{1}}-1/2\,{a_{{2}}}^{2}+ 2\,a_{{2}}b-1/2\,{b}^{2}}} \right) {{\rm e}^{-a_{{2}}a_{{1}}-a_{{1}}b- a_{{2}}b}}}{a_{{2}}a_{{1}}}} $$

Robert Israel
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