# 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

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 $$y>x$$, $$2U>x+y\quad\text{and}\quad 2V while, if $$y, $$2Ux+y\,,$$ 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 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 $$e^{-z^2/2(b^2+c^2)}\propto\frac{\text d}{\text dz}\Phi(-z/2\sigma)$$