Expected value of the minimum of two standard half-normal distributions



Let \(X_1, X_2\) be i.i.d. random variables that follow the standard normal distribution \(\mathcal{N}(0,1^2)\). One day I wondered, what is the expected value of \(Z = \min(|X_1|, |X_2|)\)? It turned out to be a fun exercise. Let’s solve it together!

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 \(F_V\), \(W = \min(U,V)\). Let’s express the CDF \(F_W\) of \(W\) via \(F_U\) and \(F_W\):

\[\begin{split} F_W(w) & = \mathbb{P}(W \leq w) = 1 - \mathbb{P}(W > w) = 1 - \mathbb{P}(\min(U, V) > w) = 1 - \mathbb{P}(U > w, V > w) = \\ & = 1 - \mathbb{P}(U > w)\cdot \mathbb{P}(V > w) = 1 - (1-F_U(w))(1-F_V(w)) = \\ & = F_U(w) + F_V(w) - F_U(w)F_V(w). \end{split} \]

Now let’s apply this rule to \(F_Z\):

\[F_Z(z) = 2F_Y(z) - F_Y^2(z) = 2(2\Phi(z) - 1) - (2\Phi(z) - 1)^2 = -4\Phi^2(z)+8\Phi(z)-3. \]

Now we can calculate the PDF of \(Z\):

\[f_Z(z) = F_Z'(z) = (-4\Phi^2(z)+8\Phi(z)-3)' = -8\phi(z)\Phi(z)+8\phi(z) = 8\phi(z) (1-\Phi(z)). \]

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

\[\mathbb{E}(Z) = \int_0^\infty xf_Z(x)dx = \int_0^\infty 8x \phi(z) (1-\Phi(z))dx \]

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

\[\begin{split} \int 8x \phi(z) (1-\Phi(z))dx & = \int 8x \frac{1}{\sqrt{2\pi}}e^{-x^2/2} (1 - \frac{1}{2}(1+\operatorname{erf}(x/\sqrt{2}))dx =\\ & = \int -2\sqrt{\frac{2}{\pi}} x e^{-x^2/2} (\operatorname{erf}(x/\sqrt{2})-1)dx =\\ & = \int -2\sqrt{\frac{2}{\pi}} \bigg( 2\frac{e^{-x^2}}{\sqrt{2\pi}} + xe^{-x^2/2}(\operatorname{erf}(x/\sqrt{2})-1) - 2\frac{e^{-x^2}}{\sqrt{2\pi}} \bigg)dx =\\ & = \int -2\sqrt{\frac{2}{\pi}} \bigg( 2\frac{e^{-x^2}}{\sqrt{2\pi}} - \Big( -xe^{-x^2/2}(\operatorname{erf}(x/\sqrt{2})-1) + e^{-x^2/2}\frac{1}{\sqrt{2}} \frac{2}{\sqrt{\pi}} e^{-x^2/2} \Big) \bigg)dx =\\ & = \int -2\sqrt{\frac{2}{\pi}} \bigg( \Big(\frac{\operatorname{erf}(x)}{\sqrt{2}} \Big)' - \Big( e^{-x^2/2} (\operatorname{erf}(x/\sqrt{2})-1) \Big)' \bigg)dx =\\ & = \int \bigg( -2\sqrt{\frac{2}{\pi}} \Big( \frac{\operatorname{erf}(x)}{\sqrt{2}} - e^{-x^2/2}(\operatorname{erf}(x/\sqrt{2})-1) \Big) \bigg)'dx =\\ & = -2\sqrt{\frac{2}{\pi}} \Big( \frac{\operatorname{erf}(x)}{\sqrt{2}} - e^{-x^2/2}(\operatorname{erf}(x/\sqrt{2})-1) \Big). \end{split} \]

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

\[\begin{split} \mathbb{E}(Z) & = \int_0^\infty 8x \phi(z) (1-\Phi(z))dx =\\ & = -2\sqrt{\frac{2}{\pi}} \Big( \frac{\operatorname{erf}(x)}{\sqrt{2}} - e^{-x^2/2}(\operatorname{erf}(x/\sqrt{2})-1) \Big) \Bigg|_0^\infty =\\ & = -2\sqrt{\frac{2}{\pi}} \bigg( \frac{1}{\sqrt{2}} - 0 \bigg) - \Bigg( -2\sqrt{\frac{2}{\pi}} \bigg( 0 -e^0 (-1) \bigg) \Bigg) =\\ & = -\frac{2}{\sqrt{\pi}} + \frac{2}{\sqrt{\pi}} \sqrt{2} = \frac{2(\sqrt{2}-1)}{\sqrt{\pi}} \approx 0.467389954510218. \end{split} \]

Hooray, we have solved the initial problem:

\[\mathbb{E}(\min(X_1, X_2)) = \frac{2(\sqrt{2}-1)}{\sqrt{\pi}} \approx 0.467389954510218. \]

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)
# 0.467