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

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 $$k-1=2$$.

• Use numerical methods. The integrand is very nicely behaved and it's simple to predict where most of its mass lies.
– whuber
Sep 15, 2021 at 13:32

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.

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
• +1 But although expressing this in terms of $\Phi_k$ can be useful, I doubt that anyone would consider $\Phi_k$ a "basic function", hence your expression is really not "closed-form" Sep 21 at 20:28
• @JarleTufto I know there is at least one published paper, written by a professor, state that formulas composing of many multivariate normal probabilities $\Phi_n$ are ‘closed-form’.
– NN2
Sep 22 at 0:50