Feanor

Krakow, Poland

Age: 24

Mar
31
comment Expected number of simple, unordered cycles in a random graph
@jojopil: Thanks for pointing this out! Corrected.
Mar
31
revised Expected number of simple, unordered cycles in a random graph
added 291 characters in body
Mar
15
comment The logic behind "better safe than sorry"
@MrLister: Sorry for the typo. I'd edit that if I could.
Mar
14
comment The logic behind "better safe than sorry"
@Mr.ShinyandNew安宇: You're right, I should have said it was redundant, not illogical.
Mar
12
comment The logic behind "better safe than sorry"
So, "better safe than sorry" could be expanded to "it's better to put in some effort into being safe then to suffer the possible consequences of not doing so, because in most cases/in case at hand the expected value of the loss is greater the cost of preventing it?"
Mar
12
comment The logic behind "better safe than sorry"
@MrLister: I made up the neutral state just for the sake of exposition. What I was trying to argue is that obviosly good is better than bad, so it makes little sense to point this out. I wouldn't say "better rich than poor" or "better healthy than sick" because these contribute nearly nothing. One the other hand, I might say "better poor than sick" or "better bothered by safety measures now than suffering misfortune in the future", because these make non-obvious comparisons.
Mar
12
awarded Notable Question
Mar
12
awarded Yearling
Mar
12
awarded Yearling
Mar
11
awarded Nice Question
Mar
11
comment The logic behind "better safe than sorry"
@Mr.ShinyandNew安宇: I state that it is illogical to use this expression, since it doesn't bring much into a conversation. If it wasn't obvious that being safe is better than being sorry, no one would bother with safety measures of any sort. The expression is useful only if you consider it in the context of the effort it takes to stay safe, and the magnitude of possible loss and its probability. Apparently, this context is implicitly there, which I didn't realise.
Mar
11
answered When $E(X|Y)E(Y) = E(XY)$?
Mar
11
comment When $E(X|Y)E(Y) = E(XY)$?
@Did: Of course not so, thanks for noticing. I seem to have written some nonsense above.
Mar
11
awarded Scholar
Mar
11
accepted The logic behind "better safe than sorry"
Mar
11
awarded Popular Question
Mar
10
asked The logic behind "better safe than sorry"
Mar
9
comment When $E(X|Y)E(Y) = E(XY)$?
(1) is true, for a proof you might want to use $E(XY|Y) = E(X|Y)Y$ (and maybe you need some integrability assumptions); (2) is false, take $X=-Y$; (3) is false because (2) was false
Mar
9
comment When $E(X|Y)E(Y) = E(XY)$?
Yes. $E(E(X|Y)) = E(X)$.
Mar
9
comment When $E(X|Y)E(Y) = E(XY)$?
Still, you would need $E(X|Y)$ to be a.s. equal to a constant, which you can't get unless $X$ and $Y$ are independent.
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