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Let $P$ be a "nice" distribution on $\mathbb R^m$ (e.g., multivariate Gaussian, etc.), with density $p$. Let $H := \{x \in \mathbb R^m \mid x^\top w = b\}$ be a hyperplane in $\mathbb R^m$ with unit-normal $w \in \mathbb R^m$. Let $R$ be the Radon transform of $p$ w.r.t $H$ by $$ R := \int_H p(x)\,ds(x), $$ where $ds(x)$ is the surface-area element on $H$. Finally, let $X_1,\ldots,X_n$ be an iid sample from $P$.

Question. Is there a simple statistical estimator $\widehat R_n := s(X_1,\ldots,X_n)$ which converges to $R$ in the limit $n \to \infty$ ?

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$\newcommand\th\theta\newcommand\R{\mathbb R}$Suppose that we have a parametric setting, that is, the unknown distribution $P$ belongs to a known parametric family $(P_\th)$ of distributions parameterized by a sufficiently low-dimensional parameter $\th$; this may be the case if $P$ is Gaussian. Then you can get (say) a maximum likelihood estimate (MLE) $\hat\th_n$ of $\th$ and estimate $R$ by $$\hat R_n:=\int_H p_{\hat\th_n}\,ds,$$ where $p_{\hat\th_n}$ is the density of $P_{\hat\th_n}$.

The MLE is usually consistent, so that you will have $\hat\th_n\to\th$ in probability (as $n\to\infty$). If now $p_\th$ is continuous on $\th$ uniformly on compact subsets of $\R^m$, you will get $p_{\hat\th_n}\to p_\th$ uniformly on compact subsets of $\R^m$. If, moreover, you can control the tails of the densities $p_\th$ so as to have their local uniform integrability (as you would have in the Gaussian case), then you will end up with the desired conclusion $\hat R_n\to R$ in probability.


If the setting is nonparametric but the dimension $m$ is small, then you can use nonparametric (say kernel) estimators of the unknown density $p$, instead of the parametric estimators. However, then, naturally, the convergence $\hat R_n\to R$ will be substantially more problematic.

If the setting is completely nonparametric and the dimension $m$ is not small (say $m\ge10$), then the situation will of course be even worse. Here you will need the sample size $n$ to be at least as big as something like $100^m\ge10^{20}$.

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