Probability of overlap 2 gaussians with gaussian means I have the following problem:
Let $X$ and $Y$ be two independent gaussian distributions, with same mean $\mu$ and standard deviations $b$ and $c$ ($X \sim \mathcal{N}(\mu,b)$ and $ Y \sim \mathcal{N}(\mu,c)$).
Let $U$ and $V$ be two "gaussian" distributions with resp. means $X$ and $Y$ and same standard deviation $\sigma$ ($U \sim \mathcal{N}(X,\sigma)$ and $V \sim \mathcal{N}(Y,\sigma)$).
I want to compute the probability of overlap between $U$ and $V$, which I define as :
$P(U>X+\frac{|X−Y|}{2})+P(V<Y− \frac{|X−Y|}{2})$
More precisely, I would like to express this quantity as a function of $\mu$, $b$, $c$ and $\sigma$, or at least prove that the probability of overlap grows with $b+c$ and is inversely proportional to $\sigma$).
Maybe this could be useful (but I didn't figure out how) : $X - Y$ is gaussian with mean $0$ and s.d. $b+c$ (as a sum of 2 independent gaussians), therefore $E(|X−Y|)= \sqrt{2(b+c)/\pi}$ (general result for a centered gaussian).
The context : in my experiment, $X$ and $Y$ are sampled once, then $U$ and $V$ are sampled hundreds of times with the obtained means, and I want to know the probability to sample for $U$ a value that is "more probable" for $V$ and vice versa, ie to be in the yellow or red region :

(this drawing might be biased, since according to an answer here, the resulting distribution of $U$ and $V$ might in fact be gaussians with same mean $\mu$ and different standard deviations... But I am quite confused because I do not know how to define the probability of overlap in my experiment anymore...)

If the event of interest is that $U$ and $V$ are "more probable" for the alternative Normal, conditional on $X=x$ and $Y=y$, it could be written as
$$\varphi((U-y)/\sigma)>\varphi((U-x)/\sigma)\quad\text{and}\quad\varphi((V-y)/\sigma)<\varphi((V-x)/\sigma)$$
where $\varphi(\cdot)$ is the standard Normal density. Expanding the density leads to
$$(U-y)^2<(U-x)^2\quad\text{and}\quad(V-x)^2<(V-y)^2$$
and, if $y>x$,
$$2U>x+y\quad\text{and}\quad 2V<x+y\,,$$
while, if $y<x$,
$$2U<x+y\quad\text{and}\quad 2V>x+y\,,$$
Conditionally on $X=x$ and $Y=y$, since $U\sim\mathcal N(x,\sigma^2)$ and $V\sim\mathcal N(y,\sigma^2)$, this event has probability
$$\left.\begin{aligned}\Phi((x-y)/2\sigma)^2 &\text{ if } ~y>x\\\\\Phi((y-x)/2\sigma)^2 &\text{ if } ~y<x\end{aligned}\right\} = \Phi(-|x-y|/2\sigma)^2$$
The final probability thus writes as
$$\mathbb E[\Phi(-|x-y|/2\sigma)^2]=\int \Phi(-|z|/2\sigma)^2\,e^{-z^2/2(b^2+c^2)}\,\text dz/\sqrt{2\pi(b^2+c^2)}$$
In the event $4\sigma^2=b^2+c^2$, this integral enjoys a closed-form expression since
$$e^{-z^2/2(b^2+c^2)}\propto\frac{\text d}{\text dz}\Phi(-z/2\sigma)$$
