Post History
#7: Post edited
by
DNB
·
2021-09-18T08:18:31Z (over 3 years ago)
If A & B are joint, why can Arby recoup some of his loss only when $P_{Arby}(A \cup B) < P_{Arby}(A) + P_{Arby}(B)$? But not $P_{Arby}(A \cup B) > P_{Arby}(A) + P_{Arby}(B)$?
- If A & B are joint, can Arby recoup some of his loss only when $P_{Arby}(A \cup B) < P_{Arby}(A) + P_{Arby}(B)$?
1. Please see the sentence alongside the red line below. As the authors didn't write this sentence for the first case ( $P_{Arby}(A \cup B) < P_{Arby}(A) + P_{Arby}(B)$), I'm assuming that if A & B _ARE_ joint, Arby _CAN_ recoup some of his loss. Correct?- 2. If I'm correct above, then why do these 2 cases differ? Scilicet, please see the question in this post's title. What's the intuition?
- Blitzstein, *Introduction to Probability* (2019 2 ed), p 40, Exercise 48. Selected Solutions PDF, p 7.
- 1. Please see the sentence alongside the red line below, but the authors didn't write this sentence for the first case ( $P_{Arby}(A \cup B) < P_{Arby}(A) + P_{Arby}(B)$). Thus if A & B _ARE _joint, can Arby recoup some of his loss in this first class?
- 2. If I'm correct above, then why do these 2 cases differ? Scilicet, please see the question in this post's title. What's the intuition?
- 3. Why does Arby deliberately transact to his loss and detriment? Is Arby stupid? Oughtn't Arby transact the other way to earn money? Please see the sentences alongside my blue line below. Instead, why doesn't Arby SELL an A-certificate and a B-certificate, and buy an A ∪ B certificate?
- 4. Same question as 3. Instead, why doesn't Arby buy an A-certificate and a B-certificate, and sell an A ∪ B certificate?
- 5. In general, $P(A ∪ B) = P(A) + P(B) - P(A ∩ B)$. Thus I see nothing wrong with $\color{limegreen}{P_{Arby} (A ∪ B) \neq P_{Arby}(A) + P_{Arby}(B)}$. Then how does "This problem illustrates the fact that the axioms of probability are not arbitrary, but rather are essential for coherent thought"?
- 
- Blitzstein, *Introduction to Probability* (2019 2 ed), p 40, Exercise 48. Selected Solutions PDF, p 7.
#6: Post edited
by
DNB
·
2021-07-13T06:44:26Z (over 3 years ago)
- 1. Please see the sentence alongside the red line below. As the authors didn't write this sentence for the first case ( $P_{Arby}(A \cup B) < P_{Arby}(A) + P_{Arby}(B)$), I'm assuming that if A & B _ARE_ joint, Arby _CAN_ recoup some of his loss. Correct?
- 2. If I'm correct above, then why do these 2 cases differ? Scilicet, please see the question in this post's title. What's the intuition?
- 
Blitzstein. *Introduction to Probability* (2019 2 ed), p 40, Question 48. Selected Solutions PDF, p 7.
- 1. Please see the sentence alongside the red line below. As the authors didn't write this sentence for the first case ( $P_{Arby}(A \cup B) < P_{Arby}(A) + P_{Arby}(B)$), I'm assuming that if A & B _ARE_ joint, Arby _CAN_ recoup some of his loss. Correct?
- 2. If I'm correct above, then why do these 2 cases differ? Scilicet, please see the question in this post's title. What's the intuition?
- 
- Blitzstein, *Introduction to Probability* (2019 2 ed), p 40, Exercise 48. Selected Solutions PDF, p 7.
#5: Post edited
by
DNB
·
2021-07-13T06:42:22Z (over 3 years ago)
1. Please see the sentence alongside the red line below. As the authors didn't write this for the first case, where $P_{Arby}(A \cup B) < P_{Arby}(A) + P_{Arby}(B)$, I'm assuming that if A & B _ARE_ joint, Arby _CAN_ recoup some of the loss. Am I correct?2. If I'm correct above, then why does these 2 cases differ? Scilicet, please see the question in this post's title. What's the intuition?- 
- Blitzstein. *Introduction to Probability* (2019 2 ed), p 40, Question 48. Selected Solutions PDF, p 7.
- 1. Please see the sentence alongside the red line below. As the authors didn't write this sentence for the first case ( $P_{Arby}(A \cup B) < P_{Arby}(A) + P_{Arby}(B)$), I'm assuming that if A & B _ARE_ joint, Arby _CAN_ recoup some of his loss. Correct?
- 2. If I'm correct above, then why do these 2 cases differ? Scilicet, please see the question in this post's title. What's the intuition?
- 
- Blitzstein. *Introduction to Probability* (2019 2 ed), p 40, Question 48. Selected Solutions PDF, p 7.
#4: Post undeleted
by
DNB
·
2021-07-13T06:41:38Z (over 3 years ago)
#3: Post edited
by
DNB
·
2021-07-13T06:41:33Z (over 3 years ago)
Please expound Jack D'Aurizio's story proof for $\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}$?
- If A & B are joint, why can Arby recoup some of his loss only when $P_{Arby}(A \cup B) < P_{Arby}(A) + P_{Arby}(B)$? But not $P_{Arby}(A \cup B) > P_{Arby}(A) + P_{Arby}(B)$?
I don't understand [Jack D'Aurizio's brusque story proof](https://math.stackexchange.com/a/2327533) of $\color{red}{\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}}$. Can someone please detail it way more?>The LHS accounts for the subsets of $E=\{1,2,\ldots,2n+1\}$ whose cardinality is at most $n$.<br>>For any subset $B\subseteq E$, either $B$ or (exclusive) $E\setminus B$ meets such constraint, hence the LHS equals *half* the number of subsets of $E$, i.e. $\frac{1}{2}\cdot 2^{2n+1} = 2^{2n}$.1. What "constraint" does Jack mean?2. Why does "either $B$ or (exclusive) $E\setminus B$ meets such constraint"?3. Why does the LHS equal "*half* the number of subsets of $E$"?I'm yearning for solely a story proof, without algebra or induction or integrals.
- 1. Please see the sentence alongside the red line below. As the authors didn't write this for the first case, where $P_{Arby}(A \cup B) < P_{Arby}(A) + P_{Arby}(B)$, I'm assuming that if A & B _ARE_ joint, Arby _CAN_ recoup some of the loss. Am I correct?
- 2. If I'm correct above, then why does these 2 cases differ? Scilicet, please see the question in this post's title. What's the intuition?
- 
- Blitzstein. *Introduction to Probability* (2019 2 ed), p 40, Question 48. Selected Solutions PDF, p 7.
#2: Post deleted
by
DNB
·
2021-07-11T20:35:47Z (over 3 years ago)
#1: Initial revision
by
DNB
·
2021-07-11T06:07:51Z (over 3 years ago)
Please expound Jack D'Aurizio's story proof for $\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}$?
I don't understand [Jack D'Aurizio's brusque story proof](https://math.stackexchange.com/a/2327533) of $\color{red}{\sum_{k=0}^{n} \binom{2n+1}{k}=2^{2n}}$. Can someone please detail it way more?
>The LHS accounts for the subsets of $E=\{1,2,\ldots,2n+1\}$ whose cardinality is at most $n$.<br>
>For any subset $B\subseteq E$, either $B$ or (exclusive) $E\setminus B$ meets such constraint, hence the LHS equals *half* the number of subsets of $E$, i.e. $\frac{1}{2}\cdot 2^{2n+1} = 2^{2n}$.
1. What "constraint" does Jack mean?
2. Why does "either $B$ or (exclusive) $E\setminus B$ meets such constraint"?
3. Why does the LHS equal "*half* the number of subsets of $E$"?
I'm yearning for solely a story proof, without algebra or induction or integrals.