Mean of split-normal distribution I came across with the split-normal distribution, with PDF
$$
\begin{array}{l}
f\left(x ; \mu, \sigma_{1}, \sigma_{2}\right)=A \exp \left(-\frac{(x-\mu)^{2}}{2 \sigma_{1}^{2}}\right) \quad \text { if } x<\mu \\
f\left(x ; \mu, \sigma_{1}, \sigma_{2}\right)=A \exp \left(-\frac{(x-\mu)^{2}}{2 \sigma_{2}^{2}}\right) \quad \text { otherwise }
\end{array}
$$
where
$$
A=\sqrt{2 / \pi}\left(\sigma_{1}+\sigma_{2}\right)^{-1}
$$
What is the expected value?
 A: The result is hidden behind the notation and becomes plainer when we generalize the situation.  Although you can just mechanically apply Calculus to the problem, it's not needed: basic geometric operations suffice and reveal the basic underlying ideas.

A "half-normal" distribution is the tail to the right or left of the Normal mean/median/mode $\mu.$  Its pdf therefore is given by a formula like
$$f_+(x;\mu,\sigma) = 2\, \frac{1}{\sqrt{2\pi\sigma^2}}\,\exp\left(-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2\right)\,\mathcal{I}(x \ge \mu)\tag{1}$$
for the right tail and a similar expression (which I will call $f_{-}$), with "$x\ge\mu$" replaced by "$x\le\mu,$" for the left tail.  (The function $\mathcal I$ is the indicator, equal to $1$ wherever its argument is true and otherwise equal to $0.$)
This is just the right tail of the standard Normal distribution, stretched horizontally by a factor of $\sigma,$ then translated horizontally by an amount $\mu.$
The "$2$" in the front doubles everything to compensate for taking just one-half of the distribution.  The factor of $1/\sqrt{2\pi\sigma^2}$ is the usual constant needed to normalize the Normal density to integrate to unity.
Let's consider a mixture
$$f(x;\mu,\sigma,\tau,p) = p f_+(x;\mu,\sigma) + (1-p) f_{-}(x;\mu,\tau).\tag{2}$$
This takes a proportion $p$ of the right tail of a Normal distribution and the complementary proportion $1-p$ of the left tail of a possibly different Normal distribution: but they are both centered right at $\mu.$  We wish to find the mean of this distribution.
After shifting and rescaling one of the tails, it is evident this calculation will come down to finding the value of
$$E_+ = \int_0^\infty x\,\left(\frac{1}{\sqrt{2\pi}} e^{-x^2/2}\right)\,\mathrm{d}x = \frac{-1}{\sqrt{2\pi}} e^{-x^2/2}\left|^{\infty}_0\right. = \frac{1}{\sqrt{2\pi}}.$$
With this in mind, we may immediately compute

$$\eqalign{
E(\mu,\sigma,\tau,p) &= \int_{-\infty}^{\infty} x\, f(x;\mu,\sigma,\tau,p)\,\mathrm{d}x\\
& = \mu + 2(p\,\sigma E_{+} - (1-p)\,\tau E_{+}) \\
&= \mu + 2(p(\sigma+\tau)-\tau)E_{+}.\tag{3}
}$$

The right hand side is the weighted mixture of the stretched, translated tails of the two half-normal distributions.  The translation obviously adds the term $\mu$ while the weights $p$ and $1-p$ multiply their respective terms, each of which must be multiplied by the amount of stretching ($\sigma$ or $\tau$) involved (with the value for the left tail negated).
We can now answer the question.  If the coefficients of $pf_+$ and $(1-p)f_{-1}$ are both to equal a common value $A,$ then by $(1)$ $p$ is proportional to $\sigma$ and $1-p$ is proportional to $\tau,$ showing
$$p = \frac{\sigma}{\sigma+\tau},\quad 1-p = \frac{\tau}{\sigma+\tau}.$$
Plugging these into $(1)$ shows the common coefficient is
$$A = 2\,\frac{1}{\sqrt{2\pi\sigma^2}}\, p\ =\ 2\,\frac{1}{\sqrt{2\pi\tau^2}}\, (1-p)\ =\ \frac{2}{\sqrt{\pi}} \frac{1}{\sigma+\tau},$$
as claimed, and this value of $p$ simplifies the general result $(3)$ to
$$E\left(\mu,\sigma,\tau, \frac{\sigma}{\sigma+\tau}\right) = \mu + 2\left(\frac{\sigma}{\sigma+\tau}\left(\sigma+\tau\right) -\tau\right)E_{+} = \mu + (\sigma-\tau)\,\sqrt{\frac{2}{\pi}}.\tag{4}$$

You can check that we needed only the following facts about the standard Normal distribution:

*

*It is symmetric around $0.$


*It has a finite expectation (which therefore, by $(1)$, equals 0).
This determines the location-scale family of Normal distributions with mean $\mu$ and scale factor $\sigma.$  Thus,

The same result $(4)$ holds when any mean-zero symmetric distribution is used in place of the standard Normal distribution.

The only change is that the factor of $\sqrt{2/\pi}$ must be replaced by twice the normalizing constant of the distribution.  (Extra care is needed when this distribution is not continuous at zero; that is, when it has positive probability of equaling zero.)
