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Dec 7, 2020 at 19:37 vote accept Arnab
Dec 7, 2020 at 19:37 comment added Arnab Ah! ... Yes, of course. That should have been obvious. :(
Dec 6, 2020 at 18:12 comment added Yuval Peres No, since in your formulation $\theta$ is a function of $X$, the average $$\frac{1}{n}\sum_i f_{\theta(X) }(X_i)$$ is also a function of $X$.
Dec 4, 2020 at 22:10 comment added Arnab Thanks much. But viewing $f_\theta(x)$ as $f(x, \theta)$ and then to be able to apply Mcdiarmid, don't I need $\theta$ and $x$ to be independent?
Dec 4, 2020 at 17:56 history answered Yuval Peres CC BY-SA 4.0