# 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?

• Hint: because $\mu$ is both the mean and median of the two parent Normals, the answer is staring right at you.
– whuber
Apr 6 '20 at 18:36
• I see it in Wikipedia (en.wikipedia.org/wiki/Split_normal_distribution), but I don't see how to get there. Apr 6 '20 at 19:02
• Yes, I see the point now: the use of a common factor "$A$" in both terms disguises the imbalance. (+1).
– whuber
Apr 6 '20 at 20:54

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:

1. It is symmetric around $$0.$$

2. 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 symmetric distribution of zero expectation is used in place of the standard Normal.

The only change is that the factor of $$\sqrt{2/\pi}$$ must be replaced by twice the normalizing constant of the distribution.