# Expectation of "mixed-variables" obtained from a N(0,1) variable

I have a variable $z$ which is normally distributed with zero mean and unit variance. I should derive $$E[z^+]$$ $$E[z^-]$$ $$E[z^+z^+]$$ $$E[z^-z^-]$$ $$E[z^+z^-]$$ where $z^+ =max(z, 0)$ and $z^- =min(z, 0)$. For the first two I know that $$E[z^+] = -E[z^-] = \frac{1}{\sqrt{2\pi}}$$ (though I don't know how to derive the solution), whereas I don't have the solution for the other three expectations.

Anyone who could help me deriving these expectations? Many thanks!

• Determine the densities for $z^+$ and $z^-$. Then write down the integrals for the expected values you wish to compute. Then evaluate the integrals. Of course, this presumes you know some basics in probability and calculus. May 9, 2016 at 15:30
• The other three are easily and very simply related to $E[z^2]$. Exploiting the symmetry of the distribution about $0$ and drawing pictures will give an instant geometric solution. For an elegant algebraic demonstration, use $z^2 = (z^{+}+z^{-})^2$, expand, and take expectations.
– whuber
May 9, 2016 at 15:30
• There's more than one way to skin a cat. The more ways you learn, the better off you'll be. "Circus tricks" are nice when they apply, but they are no substitute for learning and being able to apply fundamentals having more general applicability. May 9, 2016 at 15:37
• @Mark Yes, seeing multiple solutions to a problem is instructive. But it's unclear what you are reacting to. Do you refer to your proposal as "circus tricks" because neither $z^{+}$ nor $z^{-}$ has a density? (In a generalized sense, well beyond "basics in calculus," they do--is that the "trick"?) I would view techniques such as looking at plots of PDFs or reasoning from symmetry to be far more fundamental (and insightful) than rote application of change-of-variables formulas.
– whuber
May 9, 2016 at 16:36
• Thank you, both inputs were helpful. @Mark, I don't know why I didn't think immediately of integrating the densities, I thought there was some hidden trick that would have allowed to get immediately to the solution and discarded the idea of solving it "how you are supposed to do it". whuber, That suggestion of expanding and taking expectations was eye-opening, thanks! May 9, 2016 at 21:35

The final product is trivial (when $z^+>0$, what's $z^-$? from there it should be clear). Note that the other two products must have the same variance as each other.
Then use whuber's suggestion* of writing $z=z^++z^-$ and take the expectation of the square.