# Radon transform of the function $h(x_1,\ldots,x_n) = x_1 g(x_1,\ldots,x_n)$, where $g$ is the density of multivariate Gaussian $N(\mu,\Sigma)$

Given an absolutely integrable function $$f:\mathbb R^n \to \mathbb R$$, let $$R[f]$$ be its Radon transform defined for every $$(w,b) \in (\mathbb R^n \setminus \{0\}) \times \mathbb R$$ by

$$R[f](w,b) := \int_H f(x)ds(x) = \int_{\mathbb R^n}f(x)\delta(b-x^\top w)\,dx,$$ where $$\delta$$ is the Dirac distribution and $$ds$$ is area element on the hyperplane $$H:=\{x \in \mathbb R^n \mid x^\top w = b\}.$$ In particular, if $$P$$ is a probability distribution on $$\mathbb R^n$$ with density $$f$$ and $$z \sim P$$, then we can inteprete $$R[f](w,b)$$ as the density of the random variable $$z^\top w$$ evaluated at the point $$b$$.

Now, let $$g:\mathbb R^n \to \mathbb R$$ be the density of the multivariate Gaussian distribution $$N(\mu,\Sigma)$$ with mean $$\mu \in \mathbb R^n$$ and covariance matrix $$\Sigma \in \mathbb R^{n \times n}$$. Since $$z^\top w \sim N(\mu^\top w,w^\top \Sigma w)$$ for $$z \sim N(\mu,\Sigma)$$, it is clear by virtue of the previous remark that, $$R[g](w,b) = \varphi\left(\frac{b-\mu^\top w}{\|w\|_\Sigma}\right), \tag{1}$$ where $$\varphi$$ is the density of the standard Gaussian distribution $$N(0,1)$$, and $$\|w\|_\Sigma := (w^\top \Sigma w)^{1/2}$$.

Consider the function $$h:\mathbb R^n \to \mathbb R$$ defined by $$h(x)=x_1 g(x)$$ for every $$x=(x_1,\ldots,x_n) \in \mathbb R^n$$.

Question. In the spirit of (1), what is an analytic formula for $$R[h](w,b)$$ ?

In the special case where $$d=1$$ so that $$\Sigma=\sigma>0$$ is just a scalar, a simple computation gives

$$R[h](w,b) = \int_{-\infty}^{+\infty} x\varphi(x)\delta(b-wx)\,dx = \frac{b}{w}\varphi(\frac{b-w\mu}{2\sigma w}) = \frac{b}{w}R[g](w,b).$$

## Update

$$R[h](w,b) = (\alpha(b-\mu_2)+\mu_1)R[g](w,b),$$

where $$\mu_1 := e_1^\top \mu$$, $$e_1=(1,0,\dots,0) \in \mathbb R^n$$, $$\mu_2 := w^\top \mu$$, $$\alpha := \dfrac{e_1^\top \Sigma w}{w^\top \Sigma w}$$.

Of course, this can be generalized by replacing $$e_1$$ with any unit-vector $$v$$.

$$\newcommand{\si}{\sigma}\newcommand{\Si}{\Sigma}\newcommand{\ep}{\varepsilon}\newcommand{\vpi}{\varphi}\newcommand{\R}{\mathbb R}$$In this "Gaussian" setting especially, it is convenient to approximate the delta function by the normal distribution $$N(0,\ep^2)$$ with $$\ep\downarrow0$$, so that
$$\begin{equation*} R[f](w,b)=\lim_{\ep\downarrow0}R_\ep[f](w,b), \tag{1}\label{1} \end{equation*}$$ where $$\begin{equation*} R_\ep[f](w,b):=\int_{\R^n}dx\,\vpi_\ep(w^\top x-b)f(x), \end{equation*}$$ $$\begin{equation*} \vpi_\ep(t):=\frac1\ep \vpi\Big(\frac t\ep\Big), \end{equation*}$$ and $$\vpi$$ is the standard normal density.
Now, for $$h(x)\equiv x_1 g(x)$$ and $$g$$ the density of $$N(\mu,\Si)$$, we can write $$\begin{equation*} R_\ep[h](w,b) =\int_{\R^n}dx\,g(x) x_1 \vpi_\ep(w^\top x-b) =E e_1^\top X\, \vpi_\ep(w^\top X-b), \end{equation*}$$ where $$e_1:=[1,0,\dots,0]^\top\in\R^{n\times1}=\R^n$$ and $$X\sim N(\mu,\Si)$$ .
Note that the joint distribution of $$e_1^\top X$$ and $$w^\top X$$ is bivariate normal with respective means $$\begin{equation*} \mu_1:=e_1^\top\mu\quad\text{and}\quad\mu_2:=w^\top\mu, \tag{2}\label{2} \end{equation*}$$ respective standard deviations $$\begin{equation*} \si_1:=\sqrt{e_1^\top\Si e_1} \quad\text{and}\quad \si_2:=\sqrt{w^\top\Si w}, \tag{3}\label{3} \end{equation*}$$ and correlation $$\begin{equation*} \rho:=\frac{e_1^\top\Si w}{\si_1\si_2}. \tag{4}\label{4} \end{equation*}$$
So, straightforward calculations yield $$\begin{equation*} R_\ep[h](w,b) = \frac{ \rho \si _1 \si _2 (b-\mu _2)+\mu _1 (\si _2^2+\ep ^2)}{\sqrt{2 \pi } (\si _2^2+\ep ^2){}^{3/2}}\, \exp\Big\{-\frac{(b-\mu _2){}^2}{2 (\si _2^2+\ep ^2)}\Big\}. \end{equation*}$$ Finally, by \eqref{1}, $$\begin{equation*} R[h](w,b) = \frac{\rho\si_1(b-\mu_2)+\mu_1\si_2}{\sqrt{2\pi}\,\si _2^2}\, \exp\Big\{-\frac{(b-\mu_2)^2}{2 \si _2^2}\Big\}, \end{equation*}$$ with $$\mu_1,\mu_2,\si_1,\si_2,\rho$$ given by \eqref{2}--\eqref{4}.