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 11m comment For a martingale $X$ does uniform integrability imply integrability of $\sup |X_{n}|$? Not sure S is finite for the example at the end. 2h comment Choosing numbers at random - expected value calculation For example the probability to draw an odd number at the second draw is 24/48 if the first number drawn is odd, which happens with probability 25/49 and is 25/48 if the first number drawn is even, which happens with probability 24/49, thus the overall probability of an odd number at the second draw is (24/48)x(25/49)+(25/48)x(24/49)=25/49. Likewise for every draw. 2h comment Choosing numbers at random - expected value calculation The number of numbers one draws from is decreasing, yes, hence the probability to draw an odd number at a given time T becomes a mixture of proportions odd/(odd+even) present at time T, these proportions are random since they depend on the previous draws but... their average is exactly rigorously 25/49. 5h comment Choosing numbers at random - expected value calculation Yeah, I just did. 5h comment Choosing numbers at random - expected value calculation By symmetry, the probability that any number chosen is odd is 25/49 hence E(X)=6x25/49=150/49=3.0612... 6h comment Find the generating function for given series The generating function $g$ for the (finite) sequence $(1,3,5,7,9)$ is the function defined by$$g(s)=1+3s+5s^2+7s^3+9s^4.$$ 6h comment Central Limit Theorem for Triangular Array of Dependent Bernoulli Variables. Assuming each sequence $(B_{j,n})_j$ is independent and replacing the conditional expectations in the LHS of (1) by their value, one sees that it converges almost surely to $\sigma^2=\frac{p^3+(1-p)^3}{p(1-p)}$ hence indeed $(S_n)$ satisfies a central limit theorem, but with a variance $\sigma^2>1$. 6h comment Why does the Dominated Convergence Theorem fail for evaluating $\lim_{n \to \infty} \ n^2 \int_{0}^{1} xe^{-n^2x^2}dx$? Funny, now that you do not assert what $g$ is (first you blindly copied $g$ from the other answer but now that somebody told you this $g$ is incorrect, you suggest none), the answer is, strictly speaking, empty. 15h comment Help integrating the transition probability of the Brownian Motion density function. "the cumulative probability for some range" Ambiguous expression, please explain what you mean. "$Pr(T \leq 10, \ -\infty \leq X \leq 5)$" ?? $T$ and $X$ are not even defined, please explain what you mean. The integral you are computing is $E(L)$, where $$L=\int_0^{10}\mathbf 1_{X_t<5}dt$$ is the time spent by the process $(X_t)$ below $5$ between times $0$ and $10$. 15h comment Expectation at 2nd draw from urn. It seems one is after the number of red balls drawn from the urn in two rounds, not the number of red balls in the urn after two draws. 15h revised Expectation at 2nd draw from urn. added 15 characters in body 15h comment Do Kolmogorov's axioms really need only disjointness rather than pairwise disjointness? ... 'pairwise disjoint' obviously all mean the same thing (but I seem to understand you now backtracked from the assertion that some sources were using any of them for anything else than the fact that $A_i\cap A_j=\varnothing$ for every $i\ne j$, hence it might be time to leave this fascinating subject). 15h comment Do Kolmogorov's axioms really need only disjointness rather than pairwise disjointness? ... if really they deducted points for using "disjoint" instead of "pairwise disjoint" to mean that $A_i\cap A_j=\varnothing$ for every $i\ne j$ and not for another reason you would have missed, and if they did not specify in their lectures that they wished to use "pairwise disjoint" (not in accordance with the literature) to mean "disjoint", then they are liable to some explanations for this departure from maths standards. Note the two conditionals. // Re your answer, it might be the most confused part of the whole page since 'disjoint', 'mutually disjoint' (your invention) and ... 15h comment Do Kolmogorov's axioms really need only disjointness rather than pairwise disjointness? The trouble is that, as often, you make circles, rehashing ad infinitum the same trivia already explained clearly to you. Sooo... once again: 1. When one mentions disjoint sets $(A_i)$, one refers to the situation where $A_i\cap A_j=\varnothing$ for every $i\ne j$. 2. To state that $\bigcap\limits_i A_i=\varnothing$, one says that the collection $(A_i)$ has an empty intersection. Thus, "pairwise disjoint" is unnecessary and might be mostly used in hastily written internet sources. All this is crystal clear since at least @Christian's answer, but you do not listen. // Re your professor, ... 16h comment Variance of Normal Distribution @mathlover I do not understand, nobody chooses variance (or standard deviation) as a parameter of binomial distributions, say, so what are you asking? 17h comment Do Kolmogorov's axioms really need only disjointness rather than pairwise disjointness? Not even sure whom you are talking to--but do you have any canonical source for "pairwise disjoint"? 23h comment Why does the Dominated Convergence Theorem fail for evaluating $\lim_{n \to \infty} \ n^2 \int_{0}^{1} xe^{-n^2x^2}dx$? The supremum is not always equal to 1/(xe), only bounded by 1/(xe), and this ruins the argument about Lebesgue dominated convergence theorem. 1d revised Set all measurable real functions on $[0,1]$ with metric $\int_{0}^1 \min \{1,|f(t)−g(t)|\}dt$ is Fréchet without nonzero continuous linear functional deleted 36 characters in body 1d comment How do you do obtain the sum $244 + 132$ in base $5$? @CiaPan "Fallacious" is too strong. I was not aware of the approach using such intermediary steps when I discovered this answer but, after having thought about it, I fail to see what could go wrong using them. Maybe you could be more specific about it? 1d comment How do you do obtain the sum $244 + 132$ in base $5$? At least the result is correct (the only one on the page, so far).