> If the output is discretized after the Gaussian step, does that mean the privacy proof is literally unchanged from the continuous Gaussian mechanism, or does the discretization add a new error term somewhere? Not literally unchanged, no. If they do it cleanly, the discretization is folded into the mechanism so you prove DP for the rounded output directly, often by showing the discrete release is just a post-processing or a close coupling of the continuous one, but I’d be surprised if there were zero extra bookkeeping. The annoying part is exactly where the rounding noise lives, because once you quantize, you need to show it doesn’t create a spike that the Gaussian tail bound never saw. So maybe the headline is “same guarantees,” but the proof still has to pay for the discretization somehow, even if the final bound looks identical.