Activity for DNB
| Type | On... | Excerpt | Status | Date |
|---|---|---|---|---|
| Edit | Post #282643 |
Post edited: |
— | over 3 years ago |
| Edit | Post #282966 |
Post edited: |
— | over 3 years ago |
| Comment | Post #284218 |
How does your answer address my question? If it doesn't, please delete your answer? You expounded the algebra that I grokked, but not the proof strategy that I was asking about. (more) |
— | over 3 years ago |
| Edit | Post #282645 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283252 |
Post edited: |
— | over 3 years ago |
| Comment | Post #283899 |
"bad English" — This is ungracious. English is not my first language. You can't deny that fluency in English is hard. Or else whole world is fluent! "basic failures of reading comprehension" — Honestly, I have no idea what you mean. Where's evidence? Which posts are you referring? "... really, it te... (more) |
— | over 3 years ago |
| Comment | Post #283899 |
I think you're referring me? No offense, but your post appears inequitable. "[tortured synonym for “know”]" — My teacher used these words. And I don't mean "know". Prognosticating a step isn't the same as KNOWING a step. I know now that GME (GameStop stock price) rocketed to $488, but I didn't progn... (more) |
— | over 3 years ago |
| Edit | Post #283887 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283887 | Initial revision | — | over 3 years ago |
| Question | — |
Does "both girls, ≥ 1 winter girl" = "both girls, ≥ 1 winter child" let you generalize the problem statement that postulated both children as girls? What does "the fact that "both girls, at least one winter girl" is the same event as both girls, at least one winter child"" imply about the problem statement below that posited merely girls? I feel that this fact lets us breadthen the problem statement. Can it? Adam Bailey's answer details the ... (more) |
— | over 3 years ago |
| Edit | Post #283886 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283886 | Initial revision | — | over 3 years ago |
| Question | — |
How would you intuit, and soothsay to rewrite, "both girls, ≥ 1 winter girl" as "both girls, ≥ 1 winter child"? 1. Please see the red underline. How can you intuit that "both girls, at least one winter girl" as "both girls, at least one winter child"? I ask this for my 15 y.o. 2. This step feels fey, sibylline! How would you prognosticate to use this fact? We would've never augured to construe "both girls, ... (more) |
— | over 3 years ago |
| Edit | Post #283388 |
Post edited: |
— | over 3 years ago |
| Comment | Post #283387 |
I'd rather not "rephrase the issue you're having", because I want to stick to the text in the book. Have you thought of complaining to the author at blitzstein@stat.harvard.edu? (more) |
— | over 3 years ago |
| Edit | Post #283255 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283255 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283255 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283255 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283118 |
Post edited: |
— | over 3 years ago |
| Edit | Post #282942 |
Post edited: |
— | over 3 years ago |
| Edit | Post #282942 |
Post edited: |
— | over 3 years ago |
| Edit | Post #282942 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283388 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283260 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283260 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283260 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283388 |
Post edited: |
— | over 3 years ago |
| Comment | Post #283399 |
"That also means such questions probably aren't appropriate for this site; consider asking a teacher instead." This feels unmannerly? Some of us aren't wealthy enough to afford private tutors. (more) |
— | over 3 years ago |
| Comment | Post #283387 |
@Moshi I don't think "By definition" answers my question? Why not define these two probabilities the other way around? (more) |
— | over 3 years ago |
| Edit | Post #283388 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283388 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283388 | Initial revision | — | over 3 years ago |
| Question | — |
How do I prove Simpson's Paradox, scilicet $P(A|B) > P(A|B^C)$? $\forall a,b,c,d > 0, aP(A|B^C)$! I can multiply the blue and orange inequalities together, but their product is in the opposite direction too! 2. I can't simply multiply the purple inequality by the green inequality, because the green inequality's in the SAME direction as $P(A|B)>P(A|B^C)$. ... (more) |
— | over 3 years ago |
| Edit | Post #283387 | Initial revision | — | over 3 years ago |
| Question | — |
Why "only 1/10,000 men with wives they abuse subsequently murder them" ≠ P(A|G,M) & "50% of husbands who murder their wives abused them” ≠ P(G|A)? All emboldings are mine. See my red side line — the solution identifies "only 1 in 10,000 men with wives they abuse subsequently murder their wives" (in the problem statement) with $\color{red}P(G|A)$. See my green underline — the solution identifies "50% of husbands who murder their wives prev... (more) |
— | over 3 years ago |
| Edit | Post #283371 | Initial revision | — | over 3 years ago |
| Question | — |
How can I exploit symmetry to intuit P(A wins) + P(B wins) = 1, without performing algebra? I'm not asking about algebra that I can execute. How can I intuit P(A wins) + P(B wins) = 1 most quickly, without algebra? Any 15 year old can calculate that for the $p = 1/2$ case, $\dfrac{i}{N} + {\color{limegreen}\dfrac{N - i}{N}} = 1$. $p \neq 1/2$ case, $\dfrac{1 - (q/p)^i}{1 - (q/p)^N} ... (more) |
— | over 3 years ago |
| Comment | Post #283299 |
Sorry. I think I was editing the old question, but I copypasted my draft for my new question in the wrong box. What would you like me to do? Create a new question for the the Alice question? (more) |
— | over 3 years ago |
| Comment | Post #283317 |
"For this equation to hold (in general) the exponent must be k = 0 and not k + 1 = 1". What do you mean These are equivalent. $k = 0 \iff k + 1 = 1$. (more) |
— | over 3 years ago |
| Comment | Post #282945 |
I don't think you answered by second question above on the "simple geometric interpretation"? (more) |
— | over 3 years ago |
| Edit | Post #283297 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283297 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283297 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283297 | Initial revision | — | over 3 years ago |
| Question | — |
How to vaticinate calculating $w_{k + 1} = w_1(1 + \dfrac{1- p}{p} + \dots + [\dfrac{1- p}{p}]^k)$ separately for $p = 1 - p$ and $p \neq 1 - p$? Please see the orange, green, red underlines. 1. Why does $\color{limegreen}{w{k + 1} = w1(1 + \dfrac{1- p}{p} + \dots + p\dfrac{1- p}{p}]^k)}$ beget the two cases of $\color{red}p = 1 - p$ and $\color{red}p \neq 1 - p$? 2. Even after reading the solution, I wouldn't have prognosticated to calcul... (more) |
— | over 3 years ago |
| Edit | Post #283295 |
Post edited: |
— | over 3 years ago |
| Edit | Post #283295 | Initial revision | — | over 3 years ago |
| Question | — |
In $w_{k + 1} - w_k = (\frac{1 - p}{p})^{exponent}(w_1 - w_0)$, why isn't exponent $k + 1$? Please see the $r^k$ underlined in red, which is $(\frac{1 - p}{p})^k$ as defined by the green underlines. 1. How do you deduce that the exponent must be $k$? Why isn't the exponent $k + 1$? 2. Is this question related to the Fence Post Error? Have I committed it? Image alt text Tsitsikli... (more) |
— | over 3 years ago |
| Edit | Post #283260 |
Post edited: |
— | over 3 years ago |