Can anybody please suggest how I can compute the moment generating function of the inner product of two Gaussian random vectors, each distributed as $\mathcal N(0,\sigma^2)$, independent of each other? Is there some standard result available for this? Any pointer is highly appreciated.


First let's address the case $\Sigma = \sigma\mathbb{I}$. At the end is the (easy) generalization to arbitrary $\Sigma$.

Begin by observing the inner product is the sum of iid variables, each of them the product of two independent Normal$(0,\sigma)$ variates, thereby reducing the question to finding the mgf of the latter, because the mgf of a sum is the product of the mgfs.

The mgf can be found by integration, but there's an easier way. When $X$ and $Y$ are standard normal,

$$XY = ((X+Y)/2)^2 - ((X-Y)/2)^2$$

is a difference of two independent scaled Chi-squared variates. (The scale factor is $1/2$ because the variances of $(X\pm Y)/2$ equal $1/2$.) Because the mgf of a chi-squared variate is $1/\sqrt{1 - 2\omega}$, the mgf of $((X+Y)/2)^2$ is $1/\sqrt{1-\omega}$ and the mgf of $-((X-Y)/2)^2$ is $1/\sqrt{1+\omega}$. Multiplying, we find that the desired mgf equals $1/\sqrt{1-\omega^2}$.

(For later reference, notice that when $X$ and $Y$ are rescaled by $\sigma$, their product scales by $\sigma^2$, whence $\omega$ should scale by $\sigma^2$, too.)

This should look familiar: up to some constant factors and a sign, it looks like the probability density for a Student t distribution with $0$ degrees of freedom. (Indeed, if we had been working with characteristic functions instead of mgfs, we would obtain $1/\sqrt{1 + \omega^2}$, which is even closer to a Student t PDF.) Never mind that there is no such thing as a Student t with $0$ dfs--all that matters is that the mgf be analytic in a neighborhood of $0$ and this clearly is (by the Binomial Theorem).

It follows immediately that the distribution of the inner product of these iid Gaussian $n$-vectors has mgf equal to the $n$-fold product of this mgf,

$$(1 - \omega^2 \sigma^4)^{-n/2}, \quad n=1, 2, \ldots.$$

By looking up the characteristic function of the Student t distributions, we deduce (with a tiny bit of algebra or an integration to find the normalizing constant) that the PDF itself is given by

$$f_{n,\sigma}(x) = \frac{2^{\frac{1-n}{2}} |x|^{\frac{n-1}{2}} K_{\frac{n-1}{2}}\left(\frac{|x|}{\sigma ^2}\right)}{\sqrt{\pi } \sigma ^4 \Gamma \left(\frac{n}{2}\right)}$$

($K$ is a Bessel function).

For instance, here is a plot of that PDF superimposed on the histogram of a random sample of $10^5$ such inner products where $\sigma=1/2$ and $n=3$:


It's harder to confirm the accuracy of the mgf from a simulation, but note (from the Binomial Theorem) that

$$(1 + t^2 \sigma^4)^{-3/2} = 1-\frac{3 \sigma ^4 t^2}{2}+\frac{15 \sigma ^8 t^4}{8}-\frac{35 \sigma ^{12} t^6}{16}+\frac{315 \sigma ^{16} t^8}{128}+\ldots,$$

from which we may read off the moments (divided by factorials). Due to the symmetry about $0$, only the even moments matter. For $\sigma=1/2$ we obtain the following values, to be compared to the raw moments of this simulation:

 k    mgf           simulation/k!
 2    0.09375       0.09424920
 4    0.00732422    0.00740436
 6    0.00053406    0.00054128
 8    0.00003755    0.00003674
10    2.58 e-6      2.17 e-6

As to be expected, the high moments of the simulation will begin departing from the moments given by the mgf; but at least up through the tenth moment, there is excellent agreement.

Incidentally, when $n=2$ the distribution is bi-exponential.

To handle the general case, begin by noting that the inner product is a coordinate-independent object. We may therefore take the principal directions (eigenvectors) of $\Sigma$ as coordinates. In these coordinates the inner product is the sum of independent products of independent Normal variates, each component distributed with a variance equal to its associated eigenvalue. Thus, letting the nonzero eigenvalues be $\sigma_1^2, \sigma_2^2, \ldots, \sigma_d^2$ (with $0 \le d \le n$), the mgf must equal

$$\left(\prod_{i=1}^d (1 - \omega^2\sigma_i^4)\right)^{-1/2}.$$

To confirm that I made no error in this reasoning, I worked out an example where $\Sigma$ is the matrix

$$\left( \begin{array}{ccc} 1 & \frac{1}{2} & -\frac{1}{8} \\ \frac{1}{2} & 1 & -\frac{1}{4} \\ -\frac{1}{8} & -\frac{1}{4} & \frac{1}{2} \end{array} \right)$$

and computed that its eigenvalues are

$$\left(\sigma_1^2, \sigma_2^2, \sigma_3^2\right) = \left(\frac{1}{16} \left(17+\sqrt{65}\right),\frac{1}{16} \left(17-\sqrt{65}\right),\frac{3}{8}\right)\approx \left(1.56639,0.558609,0.375\right).$$

It was possible to compute the PDF by numerically evaluating the Fourier Transform of the characteristic function (as derived from the mgf formula given here): a plot of this PDF is shown in the following figure as a red line. At the same time, I generated $10^6$ iid variates $X_i$ from the Normal$(0,\Sigma)$ distribution and another $10^6$ iid variates $Y_i$ in the same way, and computed the $10^6$ dot products $X_i\cdot Y_i$. The plot shows the histogram of these dot products (omitting some of the most extreme values--the range was from $-12$ to $15$):

Histogram and PDF

As before, the agreement is excellent. Furthermore, the moments match well through the eighth and reasonably well even at the tenth:

 k    mgf           simulation/k!
 2     1.45313       1.45208
 4     2.59009       2.59605
 6     5.20824       5.29333
 8    11.0994       11.3115
10    24.4166       22.9982


(Added 9 August 2013.)

$f_{n,\sigma}$ is an instance of the variance-gamma distribution, which originally was defined as " the normal variance-mean mixture where the mixing density is the gamma distribution." It has a standard location ($0$), asymmetry parameter of $0$ (it is symmetric), scale parameter $\sigma^2$, and shape parameter $n/2$ (according to the Wikipedia parameterization).

  • 1
    $\begingroup$ Hello whuber, thanks a lot for the detailed explanation. I have one doubt, though. When $\Sigma$ is general, the terms in the sum expansion of the inner product are not iid anymore; hence the mgf of the sum is no more the product of the mgfs. Then, how do we generalize the above analysis to a more general Sigma? $\endgroup$
    – abhibhat
    Mar 9 '13 at 9:57
  • $\begingroup$ I added a new section to provide some of the (easy) details of this generalization, to make it clear that nothing new is involved here. You can also use basic properties of mgfs to write down the mgf in the case where the data have nonzero means, too, thereby resolving the problem in full generality. $\endgroup$
    – whuber
    Mar 9 '13 at 21:41

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