I noticed [this](https://mathoverflow.net/questions/380070/is-a-mixture-of-real-analytic-functions-again-analytic) post. But still I'd like to follow up with a specific case I have in mind. Say $p(x| \theta)$ is the density of a Gaussian distribution on $\mathbb{R}^n$ with mean $\theta$ and known covariance $\Sigma$. Let
$$f(x) = \int\_{\Theta} p(x| \theta) \Lambda(d\theta),$$
where $\Theta$ is a proper subset of $\mathbb{R}^n$, and $\Lambda$ is an arbitrary probability measure over $(\Theta, \mathcal{B}(\Theta))$, not necessarily continuous w.r.t. the Lebesgue measure. Let $g(x)$ be the density of another Gaussian on $\mathbb{R}^n$ with mean $\mu$ and known covariance $\Sigma + \Psi$. Is it possible for $g(x)$ and $f(x)$ to agree on a set of positive Lebesgue measure?



Mixture of gaussian density agree with another gaussian on positive measure
$\newcommand\R{\mathbb R}\renewcommand\th{\theta}\newcommand{\Th}{\Theta}\newcommand{\Si}{\Sigma}
\newcommand\La\Lambda\newcommand{\C}{\mathbb C}$The answer is: This will be so if (and only if) $\Lambda$ itself is a (possibly degenerate) Gaussian distribution.


Let us prove the "only if" part. Here it does not really matter whether it is a priori assumed that $\Th$ is a proper subset of $\R^n$. Also, by substitutions $x\leftrightarrow\Si^{1/2}x$ and $\th\leftrightarrow\Si^{1/2}\th$, without loss of generality $\Si=I\_n$, the identity matrix. So,  

\begin{equation\*}
 f(x)=(2\pi)^{-n/2}\int\_{\R^n}\exp\Big(-\frac12\sum\_{j=1}^n(x\_j-\th\_j)^2\Big)\,\La(d\th)
\end{equation\*}
for $x=(x\_1,\dots,x\_n)\in\R^n$. Since
\begin{equation\*}
 \sum\_{j=1}^n(x\_j+iy\_j-\th\_j)^2=\sum\_{j=1}^n(x\_j-\th\_j)^2-\sum\_{j=1}^n y\_j^2
 +2i\sum\_{j=1}^n(x\_j-\th\_j)y\_j, \tag{2}\label{2}
\end{equation\*}
the function $f$ can be extended to a holomorphic function on $\C^n$. Similarly, the function $g$ can be extended to a holomorphic function on $\C^n$.


So, $f$ and $g$ are real-analytic on each (straight) line in $\R^n$.


Suppose now $f=g$ on some subset $A$ of $\R^n$ of Lebesgue measure $|A|>0$. Without loss of generality, the set $A$ is bounded.


Take any $x\in\R^n$. Then
\begin{equation\*}
 0<|A|=|A-x|=\int\_{S^{n-1}}du\,\int\_0^\infty r^{n-1}\,dr\,1(ru\in A-x),
\end{equation\*}
where $\int\_{S^{n-1}}du$ is the integral with respect to the surface measure on the unit sphere $S^{n-1}$ in $\R^n$. So, by the Tonelli theorem, there is some $u\_x\in S^{n-1}$ such that $\int\_0^\infty r^{n-1}\,dr\,1(ru\_x\in A-x)>0$ and hence
\begin{equation\*}
 |A\cap(x+\R\_+ u\_x)|=\int\_0^\infty dr\,1(x+ru\_x\in A)=\int\_0^\infty dr\,1(ru\_x\in A-x)>0. 
\end{equation\*}
So, the set $A\cap(x+\R\_+ u\_x)$ is an infinite and bounded subset of the line $x+\R u\_x$, and $f=g$ on this infinite bounded subset of a line. Recalling that $f$ and $g$ are real-analytic on each line in $\R^n$, we conclude that $f=g$ on the line $x+\R u\_x$, and hence $f(x)=g(x)$.


So, $f=g$ (on the entire space $\R^n$), which can be rewritten as the identity
\begin{equation\*}
 e^{-\|x\|^2/2}\int\_{\R^n}\exp\Big(\sum\_{j=1}^n \th\_j x\_j\Big)\,L(d\th)
 =C\exp\big(-x^\top Bx+x^\top\mu\big) 
\end{equation\*}
for all $x\in\R^n$,
where $\|\cdot\|$ is the Euclidean norm,
$$L(d\th):=e^{-\|\th\|^2/2}\La(d\th),$$
$B$ is some positive-definite $n\times n$ real matrix, $\mu$ is some vector in $\R^n$, and $C:=L(\R^n)=(2\pi)^{n/2}f(0)$. So, for (joint) moment generating function (mgf) $M\_{L/C}$ of the probability measure $L/C$, some symmetric matrix $R$, and all $x\in\R^n$ we have
\begin{equation\*}
 M\_{L/C}(x)=\exp\big(-x^\top Rx+x^\top\mu\big)
\end{equation\*}
Any mgf is log convex. So, $R$ must be positive semidefinite. Thus, $M\_{L/C}$ is the mgf of a (possibly degenerate) Gaussian distribution. So, $L/C$ is a (possibly degenerate) Gaussian distribution. So, $\La$ is a (possibly degenerate) Gaussian distribution.


Vice versa, if $\La$ is a (possibly degenerate) Gaussian distribution, then $f$ is clearly a Gaussian distribution. $\quad\Box$


