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 Mar 23 comment How is the mode of an exponential distribution zero? @rolve obviously nothing is ever really exponentially distributed... Mar 23 comment How is the mode of an exponential distribution zero? @JuhoKokkala no, that =0 is obviously a typo.... Feb 25 comment Is it true that for any square matrix $A$, if $A^4=\text{Id}$ then $A^2=-\text{Id}$ or $A^2=\text{Id}$? @runaround this is the best trolling comment I have seen in ages. Jan 29 comment Expected value of Stock Price, Poisson Process @AndréTerra that goes without saying... Jan 4 comment What is $\sum_ {i=1}^{10}2$? Oh god, all these people jumping on this trivial question. Well done the commentor. Dec 20 comment 3 contestants choosing a smallest number to win a car Fixing an N, you can do better, as shown in the n=3 case above, what you offered is not optimal when assuming this upper bound, but I see your point. Nov 24 comment Why is integrability needed in fundamental principle 'you can't beat the system'? For conditonal expectation to exist, do you not need the r.v. to be $L^1$? In this case, you need $C_n(X_n-X_{n-1})$ to be integrable. The easiest condition is to assume that both $C_n$ and $X_n-X_{n-1}$ are both $L^2$ and use Cauchy-Schwarz. Maybe you can try to construct counter examples where both are integrable but the products fail to be? Note Saz's answer assumed X.Y is integrable. If you can show the product is integrable than $L^1$ is okay but in many cases, it is easier to show $L^2$. Nov 20 comment Find a graph where $|E| = 2|V|-3$ and $\deg (v) = 3$ for every $v$ That drastically changes the triviality of the problem... Nov 20 comment Find a graph where $|E| = 2|V|-3$ and $\deg (v) = 3$ for every $v$ It did not occur to you to start with the graph with just 2 vertices and 1 edge? I fail to comprehend how you could have spent hours. What did you try? Nov 20 comment Find a graph where $|E| = 2|V|-3$ and $\deg (v) = 3$ for every $v$ you certainly didn't try very hard. Nov 18 comment Does $0$ correlation imply independence for marginally normal distributions? @Henry How did you derive that? Nov 12 comment increasing process well, this follows from the Tanaka's formula (trivially), but I am guessing that is not the answer you are looking for. So maybe you want to look into a proof for Tanaka's formula. Oct 24 comment 3 contestants choosing a smallest number to win a car So this just seems like a 3 player prisoner dilemma problem. The only stable point is everyone chooses 1 and nobody win. If two player sample over 1 to N uniformly, i am going to pick 1 every time... Oct 24 comment 3 contestants choosing a smallest number to win a car If the players play this, one of the player should always choose 1 and have greater than 0.5 percent chance of winning Oct 24 comment 3 contestants choosing a smallest number to win a car I got the same numbers, so does this mean this game has no nash equilibrium? Oct 24 comment 3 contestants choosing a smallest number to win a car I am slightly confused. In prisoner's dilemma, both criminal should turn their friend in when they are not colluding. Your last sentence suggestions collusion? While this is in a way, optimal, it is certainly unstable. Oct 24 comment 3 contestants choosing a smallest number to win a car the optimal strategy cannot be over a large number surely? If 2 players do this, i can just choose 1 and 2 with probability 1/2 to beat this strategy. At the top, you said your assumption is the players are not cooperating, but settling for a distribution which make the prob of winning as close to 1/3 as possible is collusion. Oct 24 comment 3 contestants choosing a smallest number to win a car I don't quite know how to write down the max min (?) problem. Oct 24 comment 3 contestants choosing a smallest number to win a car @stochasticboy321 I am vaguely aware of the concept of nash equilibrium concept but I am not sure how we prove it in this case. in a perfectly rational world, the players would probably collude... Oct 24 comment 3 contestants choosing a smallest number to win a car @stochasticboy321 the person who chooses 2. Basically I think we want a strategy which maximises the individual chance of winning but penalises people from deviating. If the choice was (1,1,1), then nobdy wins.