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 Dec 4 reviewed Reject Improving Von Neumann's Unfair Coin Solution Dec 4 reviewed Reject Existence of iid random variables Dec 4 reviewed Approve $\sigma$-algebra $\mathcal{G} \subset \mathcal{F}$ is countably generated iff there is a random variable $X$ such that $\mathcal{G} = \sigma(X)$. Dec 4 reviewed Reject Probability of of an event happening at least once in a sequence of independent events? Dec 4 reviewed Approve Probability of achieving a given density of IID random variables Dec 4 reviewed Reject Can you make money on coin tosses when the odds are against you? Dec 2 awarded Enlightened Dec 2 awarded Nice Answer Nov 29 awarded Nice Answer Nov 27 reviewed Looks OK Product of complex measures Nov 25 reviewed Approve How to eliminate $\theta$? Nov 25 reviewed Leave Closed How to solve for a distribution function over which another function is integrated? Nov 25 reviewed No Action Needed simplyfying boolean algebra 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 24 reviewed No Action Needed Proof that a certain language is Turing Decidable Nov 23 awarded Pundit 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 reviewed Approve Find a graph where $|E| = 2|V|-3$ and $\deg (v) = 3$ for every $v$ 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.