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Jan
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awarded  Notable Question
Oct
29
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Oct
29
comment How to do a statistically sound ranking using varying numbers of likes/dislikes?
True, hence I was rather aiming for the Bayesian approach of the accepted answer.
Oct
29
comment How to do a statistically sound ranking using varying numbers of likes/dislikes?
Actually, I found my question answered (for the most part) at [this CrossValidated question][1]. [1]: stats.stackexchange.com/questions/15979/…
Oct
26
comment How to do a statistically sound ranking using varying numbers of likes/dislikes?
Also, thinking about it, wouldn't most (simple) models of a user's voting behavior pretty much boil down to a simple "likes to dislikes ratio" test? E.g., model $\Pr[like\ x_i]=\delta_i$, $\Pr[dislike\ x_i]=\epsilon_i$, and $\Pr[ignore\ x_i]=1-\delta_i-\epsilon_i$ with like hypothesis $H_0: \delta_i>\epsilon_i$ and dislike hypothesis $H_1: \epsilon_i>\delta_i$?
Oct
26
comment How to do a statistically sound ranking using varying numbers of likes/dislikes?
Thanks for your answer. If I understand correctly you propose a likelihood test to decide whether a particular item is generally liked ($H_0$) or disliked ($H_1$). I agree that this should be possible. However, I do not see how this helps to rank all items accordingly. Do you suggest ranking by likelihood ratios? If so, can you point me to some literature that discusses why such an approach avoids the fallacies given in the opening post?
Oct
25
comment How to do a statistically sound ranking using varying numbers of likes/dislikes?
A possible answer would be highly appreciated. ;)
Oct
25
comment How to do a statistically sound ranking using varying numbers of likes/dislikes?
@Seyhmus Güngören: Do you mind to elaborate?
Oct
25
asked How to do a statistically sound ranking using varying numbers of likes/dislikes?
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Aug
27
comment Positive definite Hessians from strictly convex functions
A proof of Conjecture 2 in dimension d=1: Assume such an open interval exists, that is, there is a (closed, non-empty) interval $[a,b]$ such that $f''(t)=0$ for all $t \in [a,b]$. By Taylor expansion, we find $f(b) = f(a) + f'(a)\cdot(b-a)$. Let $c=(a+b)/2$. Again using Taylor expansion, we find $f(c) = f(a) + f'(a)\cdot(c-a) = f(a) + f'(a)\cdot(b-a)/2$. Hence, $(f(a)+f(b))/2 = f(c)$, which contradicts the strict convexity of $f$.
Aug
27
comment Positive definite Hessians from strictly convex functions
Thanks, this is a very convincing counter example. (I wasn't even aware that there are nowhere dense sets of positive Lebesgue measure.) I took the liberty of updating my question.
Aug
27
revised Positive definite Hessians from strictly convex functions
added 23 characters in body; edited tags
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