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: