How to calculate $\int_{-\infty}^{\infty} \Phi(y+c)^{k-1}d(\Phi(y))$?

Question

How to calculate $\int_{-\infty}^{\infty} \Phi(y+c)^{k-1}d(\Phi(y))$? When I check the list of integrals of Gaussian functions, I only find $k-1=2$.

Answer 1

Let $X_0, X_1, \cdots, X_{k-1}$ denote $k$ independent unit-variance normal random variables, and suppose that $E[X_0]=0$, and $E[X_i]=-c$, $1 \leq i \leq k-1$. Let $C$ denote the event that $X_0 > \max_i X_i$. Then, $$P(C\mid X_0 = y) = P(\max_i X_i \leq y) = \prod_{i=1}^{k-1}P(X_i \leq y) = [\Phi(y+c)]^{k-1}.$$ Consequently, $$P(C) = \int_{-\infty}^\infty [\Phi(y+c)]^{k-1}\,\mathrm d[\Phi(y)] = \int_{-\infty}^\infty [\Phi(y+c)]^{k-1}\phi(y)\,\mathrm dy$$ where $\phi(y)$ is the standard normal density function. As whuber says in a comment, there is no closed-form expression for the value of this integral but it is nicely behaved and amenable to numerical evaluation. Evaluating integrals of this type is of great interest to communications engineers -- more truthfully, accurate evaluation of $1-P(C)$ is of great interest to communications engineers-- because $1-P(C)$ corresponds to the probability of error in orthogonal signaling. See, for example, this answer of mine for some details.

Answer 2

This integral is equal to $\mathbb{E}(\Phi^{k-1}(Y+c))$ for $Y$ follows the distribution $\mathcal{N}(0,1)$.

By applying this answer with $a = 1, n = k-1$, we can have a closed-form expression for the integral via a multivariate normal probability.

$$\color{red}{\text{Integral} =\mathbb{E}(\Phi^{k-1}(Y+b)) = \Phi_{k-1} \left(\mathbf{l}, \mathbf{u};\mathbf{0}_{k-1};\Sigma \right)}$$ with

• $\Phi_{k-1}() $ the multivariate normal probability
• $\mathbf{l} \in \mathbb{R}^{k-1}$ vector with all elements are equal to $-\infty$
• $\mathbf{u} \in \mathbb{R}^{k-1}$ vector with all elements are equal to $b$
• $\mathbf{0}_{k-1} \in \mathbb{R}^n$ vector of $0$
• $\Sigma \in \mathbb{R}^{(k-1) \times (k-1)}$, its elements can be defined in $(4)$ of the answer