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I have an integral over a subspace of $\mathbb{R}^n \times \mathbb{R}^n$ with an integrand of the form $$\exp\left(-\frac{1}{2}\left[||u^2|| + \langle u, v \rangle + ||v||^2\right]\right)$$ The subspace is exactly the space for which $u_{i} = u_{n-i}$ (assume $n$ is even). In other words, $v$ is a true $n$-dimensional vector, whereas $u$ is two copies of an $n/2$ dimensional vector.

In order to evaluate this integral, I have been thinking about it as a density over ALL of $\mathbb{R}^n \times \mathbb{R}^n$ of two correlated Gaussians with covariance given by the block matrix $$\begin{bmatrix} A & B \\ B^t & C \end{bmatrix} $$ where $A,C$ are $n \times n$ identity matrices and $B$ is the block matrix $$\begin{bmatrix} I_{n/2} & I_{n/2} \\ 0 & 0 \end{bmatrix} $$ However, given that this is degenerate, I am having trouble finishing the computation. Would greatly appreciate any tips!

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  • $\begingroup$ Just want to make sure you don't actually mean inner product of u and v to have a coefficient of 2--that's how these integrals usually look because inner product of u+v with itself yields a coefficient of 2 on the middle term. This would make Iosef's answer even simpler. $\endgroup$ – Sheridan Grant Sep 15 at 22:34
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$\newcommand{\R}{\mathbb{R}}$ Let $U:=\{u\in\R^n\colon u_i=u_{n-i}\ \forall i\}$ be your $n/2$-dimensional subspace. I am assuming that your integral is with respect to the product of the Lebesgue measures on $U$ and $\R^n$, and I will denote those measures by $du$ and $dv$, respectively. So, if $\cdot$ denotes the dot product, your integral is \begin{align} I&:=\int_U du\,\int_{\R^n}dv\,\exp\big(-(\|u\|^2+u\cdot v+\|v\|^2)/2\big) \\ &=\int_U du\,\int_{\R^n}dv\,\exp\big(-(3\|u\|^2/4+\|v+u/2\|^2)/2\big) \\ &=\int_U du\,\int_{\R^n}dw\,\exp\big(-(3\|u\|^2/4+\|w\|^2)/2\big) \\ &=(2\pi)^{n/2}\int_U du\,\exp\big(-3\|u\|^2/8\big) \\ &=(2\pi)^{n/2}\int_{\R^{n/2}} dt\,\exp\big(-3\|t\|^2/8\big) \\ &=(2\pi)^{n/2}(2\pi)^{n/4}(4/3)^{n/4}. \end{align} The penultimate equality here holds because both the Euclidean norm and the Lebesgue measure are rotation invariant, whereas the dimension of $U$ is $n/2$; in fact, this is how the Lebesgue measure on $U$ can/should be defined: by the condition that \begin{equation} \int_U du\,f(u)=\int_{\R^{n/2}} dt\,f(Tt) \end{equation} for all nonnegative Borel-measurable functions $f\colon U\to\R$, where $T\colon\R^{n/2}\to U$ is a linear isometry.

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    $\begingroup$ Ah, I see. It is as simple as just factoring and completing the square? Thanks for your quick + helpful answers Iosif! $\endgroup$ – DJA Sep 15 at 20:29
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    $\begingroup$ @DJA : From my viewpoint, the main thing here is to properly/conveniently define the Lebesgue measure on a linear subspace. $\endgroup$ – Iosif Pinelis Sep 15 at 21:45

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