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Jan 17, 2019 at 16:30 answer added Robert Israel timeline score: 1
Jan 17, 2019 at 16:15 comment added Robert Israel In that case $\mathbb E[W] = \sum_{i=0}^\infty S(i)^2$ while $\mathbb E[X] = \sum_{i=1}^\infty S(i)$.
Jan 17, 2019 at 16:04 comment added Robert Israel Do you mean $S(i) = \mathbb P(X \ge i)$ rather than $I$?
Jan 17, 2019 at 14:42 history edited Vilhelm Agdur CC BY-SA 4.0
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Jan 17, 2019 at 14:18 history asked Vilhelm Agdur CC BY-SA 4.0