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Jun 7, 2015 at 18:04 history edited Zen CC BY-SA 3.0
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Jun 7, 2015 at 13:42 history edited P.Windridge CC BY-SA 3.0
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Jun 7, 2015 at 13:41 comment added P.Windridge (I accept your point though, and have edited accordingly!)
Jun 7, 2015 at 13:36 comment added P.Windridge Well, if $X$ and $Y$ are independent random variables then $g(X)$ and $h(Y)$ are also independent random variables :)
Jun 7, 2015 at 13:19 comment added Dilip Sarwate +1 Actually, the last equality follows from $\mathbb{E}[g(X)h(Y)] = \mathbb{E}[g(X)]\mathbb{E}[h(Y)]$ for $X, Y$ independent. $\mathbb{E}[XY] = \mathbb{E}[X]\mathbb{E}[Y]$ also holds for uncorrelated random variables $X$ and $Y$, but the Chernoff bound does not follow from this.
Jun 7, 2015 at 12:39 history answered P.Windridge CC BY-SA 3.0