A convolution integral involving Gaussian density

Question

Can the following integral be simplified to a closed-form expression: $\int_a^r \phi(x; \mu_1) \Phi(r-x; \mu_2) dx$, where $\phi$ and $\Phi$ are the density and distribution function of Gaussian with unit variance. I would like to be able to compute this without resorting to numerical qudrature.

Answer 1

Let $X_i \sim \mathcal{N}(\mu_i, 1), i = 1, 2$, and they are independent. Then

$$ \begin{align} &~~~~~ \int_a^r \phi(x;\mu_1) \Phi(r-x; \mu_2) dx \\ &= \int_a^r f_{X_1}(x) \Pr\{X_2 \leq r-x\} dx \\ &= \int_a^r \Pr\{X_2 \leq r-x|X_1 = x\} f_{X_1}(x) dx \\ &= \Pr\{X_2 \leq r - X_1, a < X_1 < r\} \\ &= \Pr\{X_1 + X_2 \leq r, X_1 < r\} - \Pr\{X_1 + X_2 \leq r, X_1 < a\} \end{align} $$

So depends on what "closed-form" solution means.

Here note that $$ (X_1 + X_2, X_1) \sim \mathcal{N} \left(\begin{bmatrix} \mu_1 + \mu_2 \\ \mu_1 \end{bmatrix}, \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\right)$$

So the integral itself can be expressed as the difference of two bivariate normal CDF in general. If bivariate normal CDF is considered not a closed-form solution, then this will imply that this integral has no closed-form solution in general.