Given two independent normal random variables $X \sim \mathcal{N_{\mu_1,\sigma_1^2}}$ and $Y \sim \mathcal{N_{\mu_2,\sigma_2^2}}$, I want to compute this quantity:
$\hat{J} = \sum_{v \in \mathbb{Z}} \left( \int_{v-0.5}^{v+0.5} f(x | \mu_1,\sigma_1^2) \; dx \cdot \int_{v-0.5}^{v+0.5} f(y | \mu_2,\sigma_2^2) \; dy\right)$
$f(x | \mu_1,\sigma_1^2)$ denotes the normal PDF. Note that, here, functions are integrated and the results are multiplied.
The following quantity is much easier to compute:
$\tilde{J} = \sum_{v \in \mathbb{Z}} \left( \int_{v-0.5}^{v+0.5} f(x | \mu_1,\sigma_1^2) \; \cdot f(x | \mu_2,\sigma_2^2) \; dx\right)$
$ = \int_{- \infty}^{+\infty} f(x | \mu_1,\sigma_1^2) \; \cdot f(x | \mu_2,\sigma_2^2) \; dx$ $ = f(\mu_1 | \mu_2,\sigma_1^2+\sigma_2^2)$
Note that here the functions are multiplied first and then integrated. The last transformation is not trivial: It can be shown, that the product of two normal densities is a normal density (integrates to one in the previous equation) and a scale factor that is normal as well, but depends only on the distribution parameters (given here).
Obviously, in general, $\tilde{J} \neq \hat{J}$. However, when $\sigma_1$ and $\sigma_2$ are restricted to values such that function values are below 1, $\tilde{J}$ appears to approximate $\hat{J}$ quite well.
Comparison of integrate-then-multiply to multiply-then-integrate
Intuitively, this makes sense to me, as changing the PDFs to densitites that are constant on $(v-0.5,v+0.5], v \in \mathbb{Z}$ (continuous representation of a discrete distribution ), makes the approximation hold with equality .
Is there a way to derive an error bound for $\tilde{J} \approx \hat{J}$? For example, an upper bound on the absolute error depending on the distribution parameters?