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:

round(mean(pmin(abs(rnorm(100000000)), abs(rnorm(100000000)))), 3)
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