3 ms·
Well, I said "at most", so it holds for all sigma_i. As you know, when the sigma_i are very different, then the accuracy is approximately given by the best acc
by jessriedel 10y ago
Well, I said "at most", so it holds for all sigma_i. As you know, when the sigma_i are very different, then the accuracy is approximately given by the best accuracy (smallest sigma_i), so there's very little improvement from averaging.
- wolfgke 10y ago> Well, I said "at most", OK, you are right. This can indeed be shown.