Dahn Jahn
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 Oct 30 comment product of random variables with different distributions What he's using there is $P(A|B)P(B) = P(A\cap B) = P(A) P(B)$ where the first equality is from the definition of conditional probability and the second from indepenence of $A$ and $B$ Oct 30 answered product of random variables with different distributions Oct 30 revised Prove $\sum^n_{i=1} (2i-1)=n^2$ by induction Added MathJax Oct 30 suggested approved edit on Prove $\sum^n_{i=1} (2i-1)=n^2$ by induction Oct 30 comment product of random variables with different distributions I am sure someone else will supply the math, but intuitively, how would you expect the distribution to behave? Such thinking may help greatly with formulating a more rigorous answer. Oct 30 comment Show that collection of finite dimensional cylinder sets is an algebra but not $\sigma$-algebra You say "it might be easier" - is it even possible to prove this through unions? Oct 29 asked When does continuity with probability one imply mean-square continuity Oct 27 accepted Why do we care about bijections in contability? Oct 27 comment Why do we care about bijections in contability? I suppose my confusion led me to pose the wrong question - perhaps what I really should've asked is why does the definition of countably infinite even consider a bijection, intuitively there seems to be no need for it, we only need to know whether $A$ is "smaller" (or better isn't "bigger") than $\mathbb N$. If you could comment on this a bit, that would be great, either way, I will accept this answer, as it answers my question. Oct 27 asked Why do we care about bijections in contability? Oct 25 comment Proving Independence Intuitively Thanks, I agree with the equality, but I am still not there, how does this prove that the distributions are independent? Oct 25 comment Proving Independence Intuitively I guess what I'd like to see is some mathematical notation, as the problem is precisely the perceived tenuous link between english words (no matter how convinced I am of their validity) and mathematical notation that I know how to work with and can identify as a sufficient proof Oct 25 comment Proving Independence Intuitively But why is it sufficient? Good points about $a.s$, thanks, it's good to never forget that, my brain just automatically disregards anything of measure zero. Oct 25 comment Strong and weak extrema I upvoted the other answer and accepted my own. Not sure if my answer is the one that made me understand, because (surprisingly) I wrote it after understanding the problem, but I feel like this explanation (albeit very similar) would help me more. Oct 25 accepted Strong and weak extrema Oct 25 asked Proving Independence Intuitively Oct 25 comment the probability that there's an actual tornado if the alarm goes off (discrete math) Yes. If the test correctly diagnoses the disease in 99% of times it is still entirely possible that you have less than 1% chance of having the disease, depending on how rare it actually it (which makes the false positives much more common). Also it makes you aware of how different $P(A|B)$ and $P(B|A)$ can be, which is something non-mathematicians often confuse. Oct 25 comment the probability that there's an actual tornado if the alarm goes off (discrete math) You're welcome. Bayes theorem is arguably one of the most amazing things you will probably encounter in an introductory probability lessons not based on measure theory (which I am guessing is your case), so I highly recommend actually understanding it. If you ever get a positive result for a scary disease, you'll thank your understanding of math and calm down a bit :) Oct 25 revised the probability that there's an actual tornado if the alarm goes off (discrete math) More precise link Oct 25 answered the probability that there's an actual tornado if the alarm goes off (discrete math)