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 $N(\mu,\Sigma)$

Question

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

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

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Update

The accepted answer shows that

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

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 $v$.

Answer 1

$\newcommand{\si}{\sigma}\newcommand{\Si}{\Sigma}\newcommand{\ep}{\varepsilon}\newcommand{\vpi}{\varphi}\newcommand{\R}{\mathbb R}$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}.