We evaluate an integral having to do with vector averages over all orientations in an n-dimensional space.

Problem definition

Let \(\hat{v}\) be a unit vector in \(n\)-dimensions and consider the orientation average of

\begin{eqnarray} \tag{1} \label{1} J \equiv \langle \hat{v} \cdot \vec{a}_1 \hat{v} \cdot \vec{a}_2 \ldots \hat{v} \cdot \vec{a}_k \rangle \end{eqnarray}

where \(\vec{a}_1, \ldots, \vec{a}_k\) are some given fixed vectors. For example, if all \(\vec{a}_i\) are equal to \(\hat{x}\), we want the orientation average of \(v_x^k\).


We’ll evaluate our integral using parameter differentiation of the multivariate Gaussian integral. Let

\begin{eqnarray} \nonumber I &=& \frac{1}{(2 \pi)^{n/2}} \int e^{- \frac{\vert \vec{v} \vert^2}{2} + \sum_{i=1}^k \alpha_i \vec{v} \cdot \vec{a}_i} d^nv \\ \tag{2} \label{2} &=& \exp \left [- \frac{1}{2} \vert \sum_{i=1}^k \alpha_i \vec{a}_i \vert^2 \right] \end{eqnarray}

The expression in the second line follows from completing the square in the exponent in the first — for review, see our post on the normal distribution, here. Now, we consider a particular derivative of \(I\) with respect to the \(\alpha\) parameters. From the first line of (\ref{2}), we have

\begin{eqnarray} \tag{3} \label{3} \partial_{\alpha_1}\ldots \partial_{\alpha_k}I \vert_{\vec{\alpha}=0} &=& \frac{1}{(2 \pi)^{n/2}} \int e^{- \frac{\vert \vec{v} \vert^2}{2}} \prod_{i=1}^k \vec{v} \cdot \vec{a}_i d^n v \\ &\equiv & \frac{1}{(2 \pi)^{n/2}} \int_0^{\infty} e^{- \frac{\vert \vec{v} \vert^2}{2}} v^{n + k -1} dv \int \prod_{i=1}^k \hat{v} \cdot \vec{a}_i d \Omega_v \\ &=& \frac{2^{k/2 - 1}}{\pi^{n/2}} \Gamma(\frac{n+k}{2}) \times \int \prod_{i=1}^k \hat{v} \cdot \vec{a}_i d \Omega_v \end{eqnarray}

The second factor above is almost our desired orientation average \(J\) — the only thing it’s missing is the normalization, which we can get by evaluating this integral without any \(\vec{a}\)s.

Next, we evaluate the parameter derivative considered above in a second way, using the second line of (\ref{2}). This gives,

\begin{eqnarray} \tag{4} \label{4} \partial_{\alpha_1}\ldots \partial_{\alpha_k}I \vert_{\vec{\alpha}=0} &=& \partial_{\alpha_1}\ldots \partial_{\alpha_k} \exp \left [- \frac{1}{2} \vert \sum_{i=1}^k \alpha_i \vec{a}_i \vert^2 \right] \vert_{\vec{\alpha}=0} \\ &=& \sum_{\text{pairings}} (\vec{a}_{i_1} \cdot \vec{a}_{i_2}) (\vec{a}_{i_3} \cdot \vec{a}_{i_4})\ldots (\vec{a}_{i_{k-1}} \cdot \vec{a}_{i_k}) \end{eqnarray}

The sum here is over all possible, unique pairings of the indices. You can see this is correct by carrying out the differentiation one parameter at a time.

To complete the calculation, we equate (\ref{3}) and (\ref{4}). This gives

\begin{eqnarray} \tag{5}\label{5} \int \prod_{i=1}^k \hat{v} \cdot \vec{a}_i d \Omega_v = \frac{\pi^{n/2}} {2^{k/2 - 1}\Gamma(\frac{n+k}{2})}\sum_{\text{pairings}} (\vec{a}_{i_1} \cdot \vec{a}_{i_2}) (\vec{a}_{i_3} \cdot \vec{a}_{i_4})\ldots (\vec{a}_{i_{k-1}} \cdot \vec{a}_{i_k}) \end{eqnarray}

Again, to get the desired average, we need to divide the above by the normalization factor. This is given by the value of the integral (\ref{5}) when \(k = 0\). This gives,

\begin{eqnarray}\tag{6}\label{6} J = \frac{1}{2^{k/2}}\frac{\Gamma(n/2)}{\Gamma(\frac{n+k}{2})} \sum_{\text{pairings}} (\vec{a}_{i_1} \cdot \vec{a}_{i_2}) (\vec{a}_{i_3} \cdot \vec{a}_{i_4})\ldots (\vec{a}_{i_{k-1}} \cdot \vec{a}_{i_k}) \end{eqnarray}


Consider the case where \(k=2\) and \(\vec{a}_1 = \vec{a}_2 = \hat{x}\). In this case, we note that the average of \(\hat{v}_x^2\) is equal to the average along any other orientation. This means we have

\begin{eqnarray}\nonumber \tag{7} \label{7} \langle \hat{v}_x^2 \rangle &=& \frac{1}{n} \sum_{i=1}^n \langle \hat{v}_x^2 + \hat{v}_y^2 + \ldots \rangle \\ &=& \frac{1}{n} \end{eqnarray}

We get this same result from our more general formula: Plugging in \(k=2\) and \(\vec{a}_1 = \vec{a}_2 = \hat{x}\) into (\ref{6}), we obtain

\begin{eqnarray}\nonumber \tag{8} \label{8} \langle \hat{v}_x^2 \rangle &=& \frac{1}{2}\frac{\Gamma(n/2)}{\Gamma(\frac{n}{2} + 1)} \\ &=& \frac{1}{n} \end{eqnarray}

The two results agree.

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Jonathan Landy Avatar Jonathan Landy Jonathan grew up in the midwest and then went to school at Caltech and UCLA. Following this, he did two postdocs, one at UCSB and one at UC Berkeley.  His academic research focused primarily on applications of statistical mechanics, but his professional passion has always been in the mastering, development, and practical application of slick math methods/tools. He worked as a data-scientist at Square for four years and is now working on a quantitative investing startup.

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