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Let $\Phi: \mathbb{R} \to \mathbb{R}$ be the cumulative distribution function of a standard normal variable $N(0,1)$ and $\phi: \mathbb{R} \to \mathbb{R}$ be its density function. I am interested in evaluating the following integral

$$ \int_{0}^{\infty} \Phi(x) \phi(x) \, dx. $$

Note that if the integral is changed to $\int_{\mathbb{R}} \Phi(x) \phi(x) \,dx $, then by using the facts that $\Phi- \frac{1}{2}$ is an odd function and that $\phi$ is an even function,

\begin{eqnarray} \int_{\mathbb{R}} \Phi(x) \phi(x) \,dx & = & \int_{\mathbb{R}} \Big( \Phi(x)- \frac{1}{2} \Big) \phi(x) \, dx + \frac{1}{2} \\ & = & \frac{1}{2}. \end{eqnarray}

However, this trick does not work in this case when the limits of integration are $0$ to $\infty$. Integration by substitution (using $u=-x$) does not work either. Any ideas?

Intuitively, the answer should be $\frac{1}{4}$, by some kind of symmetry. But I can’t show it.

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Notice that $\phi(x) = \Phi^{\prime}(x)$. Thus if we make the substitution $u = \Phi(x)$, we get that $\mathrm{d}u = \phi(x)\mathrm{d}x$. Since $\Phi(0) = 0.5$ and $\Phi(\infty) = 1$, we get \begin{align*} \int_{0}^{\infty}\Phi(x)\phi(x)\mathrm{d}x = \int_{1/2}^{1}u\mathrm{d}u = \frac{1}{2} - \frac{1}{8} = \frac{3}{8} \end{align*}

At your case, we have $\Phi(-\infty) = 0$ and $\Phi(\infty) = 1$. Consequently, it results that \begin{align*} \int_{\textbf{R}}\Phi(x)\phi(x)\mathrm{d}x = \int_{0}^{1}u\mathrm{d}u = \frac{1}{2} \end{align*}

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