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 Aug 29 awarded Teacher Aug 20 comment Definition of an absolutely continuous random variable @ zoli - Thanks. Aug 19 comment Proof - Limits of CDF @ André Nicolas - Thanks a lot. Aug 18 comment Proof - Limits of CDF @ André Nicolas - Thanks a lot. Btw, is $y\uparrow\infty$ is same as $y\to\infty$ for any $y\in\mathbb{R}$? (I believe yes) Aug 18 comment Definition of an absolutely continuous random variable @ zoli - I understand the condition for qualifying $Q_X$ as absolute continuous w.r.t. to $\mu$ and how (I've some familiarity with Radon-Nikodym theorem) it implies existence of a density function. What I don't get is why an r.v. $X$ is qualified absolute continuous based on whether its distribution function i.e. $Q_X$ is absolute continuous? Pls help! Aug 18 comment Proof - Limits of CDF @ André Nicolas - Great stuff. Can't explain how much I appreciate it as I'm self studying these topics. Hope you could shed some light on the 2nd part of my post about the validity of my proof. Aug 17 comment Proof of the properties of limits of CDFs @ Stefan Hansen - Due to lack of space in comment area, I've put my concern as a separate post in this site. It would be great if you could respond to my post, especially concern (1) in that post is what I mentioned here. Aug 17 asked Proof - Limits of CDF Aug 16 comment Proof of the properties of limits of CDFs @ Stefan Hansen, oeda, - Considering $\mu$ as probability measure $P$, I understand how $P\left(\bigcup_{n=1}^\infty (-\infty, n]\right) = \lim_{n\to\infty} P((-\infty, n])$, but not sure if $\lim_{n\to\infty} P((-\infty, n]) = F(n)$ where $F$ is cdf. The confusion is because, the domain of cdf is $\mathbb{R}$ (real number) but for the countably infinite addition $n$ is positive integers. How can they equate!! Can you pls help! Aug 11 awarded Benefactor Aug 11 awarded Scholar Aug 11 accepted Probability in heterogeneous sample space Aug 8 comment Probability in heterogeneous sample space @ Tryss - I agree to both the points you've mentioned. In fact, the subset relation between '6' and 'H' was the catch that I terribly missed and the whole thing got me haunted. Thanks a lot. Not sure if you care, but this deserves the bounty. Aug 8 comment Probability in heterogeneous sample space @ Bruce Trumbo – We all should understand that unless originator of a post is satisfied, no answer however amazingly/accurately written serve its very core purpose and we should be mature enough to accept a down vote after all it’s the perception of the voter. In all fairness, Tryss’ answer without his last comment was incomplete, at least to me. I admit the down-voting may be a bit early (just like I admit his last comment deserves a bounty now whether anyone cared or not). Hope this clears the air and keeps us in good spirit to help each other. Aug 7 awarded Critic Aug 7 awarded Promoter Aug 7 comment 1 sample space with 2 probability example? I don't think "it is possible to define more probability measure on one probability space" by Berci is very precise statement. Probability space is a tuple $(\Omega, \mathcal{F}, P)$. If you change $P$ keeping $\Omega$ same, it creates different probability spaces but with same sample space. In all fairness, I guess the intention of the question was to know if we can have different $P$ with same $\Omega$ and as Berci pointed out correctly that it is very much possible. You can have several coins each with different probability of head all producing same sample space. Aug 6 revised Probability in heterogeneous sample space edited tags Aug 6 comment Probability in heterogeneous sample space @ Tryss : If$P(6) = P(6$ and $H)$, then $H$ has to be sample space $\Omega$ and thus $P(H)=P(\Omega)=1$ which contradicts your evaluation of $P(H)=1/2$. You see the logic is falling apart- or am I missing something!! Aug 6 comment Probability in heterogeneous sample space @ true blue anil - Both questions are asked by me and remain unanswered. Further, I neither want to count nor the space is equi-probable - rather I would like a mathematical derivation of it. Well, now it looks like not so trivial question after all!! Thanks for your help.