 # Expected value of the maximum of two standard half-normal distributions

Let $$X_1, X_2$$ be i.i.d. random variables that follow the standard normal distribution $$\mathcal{N}(0,1^2)$$. In the previous post, I have found the expected value of $$\min(|X_1|, |X_2|)$$. Now it’s time to find the value of $$Z = \max(|X_1|, |X_2|)$$.

Let’s denote the absolute values of $$X_1, X_2$$ as $$Y_1, Y_2$$:

$Y_1=|X_1|, \quad Y_2=|X_2|.$

Thus, $$Y_1$$, $$Y_2$$ follow the standard half-normal distribution. The CDF of this distribution is well known:

$F_Y(y) = \operatorname{erf}\bigg( \frac{y}{\sqrt{2}} \bigg), \quad \textrm{for}\,\,\, y\geq 0,$

where $$\operatorname{erf}$$ is the error function. Let $$\Phi$$ be the CDF of the standard normal distribution:

$F_X(x) = \Phi(x) = \frac{1}{2} \Bigg( 1 + \operatorname{erf} \bigg( \frac{x}{\sqrt{2}} \bigg) \Bigg).$

Let also $$\phi$$ be the PDF of the standard normal distribution:

$\phi(x) = \Phi'(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$

It’s easy to see that

$F_Y(y) = 2\Phi(y)-1.$

Suppose $$U, V$$ are independent random variables with CDFs $$F_U$$ and $$F_V$$, $$W = \max(U,V)$$. Let’s express the CDF $$F_W$$ of $$W$$ via $$F_U$$ and $$F_W$$:

$\begin{split} F_W(w) & = \mathbb{P}(W \leq w) = \mathbb{P}(\max(U, V) \leq w) = \mathbb{P}(U \leq w, V \leq w) = \\ & = \mathbb{P}(U \leq w)\cdot \mathbb{P}(V \leq w) = F_U(w)\cdot F_V(w). \end{split}$

Now let’s apply this rule to $$F_Z$$:

$F_Z(z) = F_Y^2(z) = (2\Phi(z)-1)^2 = 4\Phi^2(z) - 4\Phi(z) + 1.$

Now we can calculate the PDF of $$Z$$:

$f_Z(z) = F_Z'(z) = (4\Phi^2(z) - 4\Phi(z) + 1)' = 8\phi(z)\Phi(z) - 4\phi(z) = 4\phi(z) (2\Phi(z) - 1).$

Since $$Y = |X|\geq 0$$, the expected value of $$Z$$ is a definite integral from $$0$$ to $$\infty$$ of $$xf_Z(x)dx$$:

$\mathbb{E}(Z) = \int_0^\infty xf_Z(x)dx = \int_0^\infty 4x \phi(x) (2\Phi(x) - 1) dx$

First of all, let’s calculate the indefinite integral of this expression:

$\int 4x \phi(x) (2\Phi(x) - 1) dx = 2\sqrt{\frac{2}{\pi}} \Big( \frac{\operatorname{erf}(x)}{\sqrt{2}} - e^{-x^2/2}(\operatorname{erf}(x/\sqrt{2})) \Big).$

(The result is easy to derive following the trick that we used in the previous post.)

Now we are ready to finish the calculation of $$\mathbb{E}(Z)$$:

$\begin{split} \mathbb{E}(Z) & = \int_0^\infty 4x \phi(x) (2\Phi(x) - 1)dx =\\ & = 2\sqrt{\frac{2}{\pi}} \Big( \frac{\operatorname{erf}(x)}{\sqrt{2}} - e^{-x^2/2}(\operatorname{erf}(x/\sqrt{2})) \Big) \Bigg|_0^\infty =\\ = \frac{2}{\sqrt{\pi}} \approx 1.12837916709551. \end{split}$

Hooray, we have solved the initial problem:

$\mathbb{E}(\max(X_1, X_2)) = \frac{2}{\sqrt{\pi}} \approx 1.12837916709551.$

We can check the correctness of this result using the following R script that uses a Monte-Carlo simulation to evaluate the first three digits after the decimal point of the result:

set.seed(42)
round(mean(pmin(abs(rnorm(100000000)), abs(rnorm(100000000)))), 3)
# 1.128