Given a function $f(\mu)$ (satisfying certain properties), it is possible to find a function $g(x)$ such that $f(\mu)=\int_{-\infty}^{\infty} g(x)\phi(x-\mu) dx $, where $\phi$ is the standard normal density. In fact, $g=\widehat{\widehat{f}/\widehat{\phi}}$, where hat denotes the Fourier transform (might be missing some constants there, depending on convention).

I'm trying to figure out if something similar can be derived for the standard deviation instead of the mean, i.e. given some $f(\sigma)$, find $g$ s.t. $f(\sigma)=\int_{-\infty}^{\infty} g(x)\frac{1}{\sigma}\phi\left(\frac{x}{\sigma}\right)dx=\mathbb{E}g(\sigma X)$ where $X\sim\mathcal{N}(0,1)$.

If we pretend that $X$ is positive for a second, then writing $g(\sigma X)=g(\exp(\log(\sigma)+\log(X)))$ we have, defining $g_0(z)=g(\exp(z))$, that $\mathbb{E}g(\sigma X)=\mathbb{E}g_0(\log(\sigma)+\log(X))$ which (after finding the distribution of $\log(X)$) would look like a convolution, and the same approach as was used above for $\mu$ might work.

Now, $X$ isn't positive, but it might be reasonable to say that $g$ should be symmetric anyway -- no reason for it not to be. So we can look for $g$ such that $f(\sigma)=\mathbb{E}g(\sigma |X|)$, and then the above approach will require computing the Fourier transform of the density of the log of the "folded" normal.

But before I try diving into that -- does this sound familiar to anyone? Am I missing something that would make this either much easier or impossible? (Should I be asking this over at mathoverflow instead?)

Bonus question -- multivariate version of the same problem, i.e. with $f(\Sigma)$. This is really the version I need to solve, and since the absolute value approach doesn't seem like it would work here, I'm trying to avoid it...


1 Answer 1


The short answer is yes and what you want to do is work on the topological group $\mathbb{R}^*$ equipped with the Haar measure $\mu(A) = \int_A \frac{1}{|x|}dx$.

It essentially computes what you computed with log because $e^x : \mathbb{R} \rightarrow \mathbb{R}^*_{\geq 0}$ is a map of topological groups. It takes the interval $[a,b]$ to $[e^a,e^b]$. The size of $[e^a, e^b]$ in $\mathbb{R}^*$ with the Haar measure above is $b=log(e^b)-log(e^a)=a$ which is the measure of $[a,b]$ in $\mathbb{R}$.

Take $\Psi (y) = \frac{1}{|y|}\phi(\frac{1}{y})$. To compute $f(\sigma)$ as a convolution consider

$$\begin{eqnarray*} f(\sigma) &=& \frac{1}{\sigma}\int_{-\infty}^{\infty} g(x)\phi(\frac{x}{\sigma})dx\\ &=& \frac{1}{\sigma}\int_{-\infty}^{\infty}g(x)\Psi(\frac{\sigma}{x})\frac{\sigma}{|x|}dx\\ &=& \int_{\mathbb{R}^*}g(x)\Psi(\sigma x^{-1})d\mu\\ &=& (g * \Psi)(\sigma). \end{eqnarray*}$$

Now you can try the same strategy for computing $g$ by using Fourier theory for $\mathbb{R}^*$.

  • $\begingroup$ Very helpful, thanks! Sadly this probably won't work in $>1$ dimensions, since it amounts to trying to write $f(\Sigma)$ as a convolution with functions of rank 1 matrices; i.e., if $f(\Sigma)=\int_{\mathbb{R}^d} g(x)\exp\left(-\frac{1}{2} x^T \Sigma^{-1} x\right) dx$, assuming $g$ is symmetric, for some $g_0$ we can rewrite $f(\Sigma)=\int_{\mathbb{R}^d} g_0(xx^T)\exp\left(-\frac{1}{2} \mathrm{trace}(\Sigma^{-1} xx^T)\right) dx=\int_{\Omega} g_0(X)h(\Sigma^{-1}X) dX$ where $\Omega$ is the set of symmetric rank 1 matrices, and $h(Z)=\exp(-\mathrm{trace}(Z)/2)$. That seems impossible... $\endgroup$
    – martin
    Feb 12, 2014 at 11:02

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