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Sep
30
awarded  Explainer
Dec
26
comment Maximum measurable collection and Caratheodory extension
I don't understand the definition of PF-measure-definable (by the way, is this an original concept? Google finds nothing): do you mean "...and there exists a PF-measure definied over $\mathcal{L}$", or over $\mathcal{A}$? I'm also not sure what to make of your requirement that $\mu(S)$ be strictly greater than 0, which pretty much kills completeness of a measure (though this may be your point).
Dec
26
comment the proof of Arrow's Theorem
The merging is done in a way that preserves the order between $a$ and $b$, and between $c$ and $d$ (check!), and in a way that $c\succ_i a$ and $b \succ_i d$ for all $i$, so that we can use unanimity.
Dec
26
comment the proof of Arrow's Theorem
Step 5, you'll have to read and reread carefully until you understand it, no way around it. You start with an arbitrary profile and preferences and manipulate it repeatedly (using IIA) until you can use step 4. I certainly can't explain it better than Reny!
Dec
25
comment the proof of Arrow's Theorem
What exactly isn't clear about the merging process? Note that we are not asking that $a \succ_i b$ for all $i$, only that this occurs iff $c \succ_i' d$.
Dec
25
comment the proof of Arrow's Theorem
I don't know what you mean by transitivity here. (Note the difference between properties of the social preference relation $F(\succ_1,\ldots,\succ_N)$ and the function $F$. The former can be transitive, and always is; I don't know what transitivity means for the latter.)
Dec
25
comment the proof of Arrow's Theorem
I suppose you mean step 5, not figure 5 (which doesn't exist). In any case no, step 5 has full generality: $a$ and $b$ are fixed only in $n$'s preferences.
Dec
25
comment How are these maximization problems same?
@Sush The "why" is correct. It implies the inequality because we have shown that any allocation that satisfies the constraint in the unnumbered equation also satisfies the constraint in (iii), and we are maximizing the same function. Therefore the maximum value in (iii) cannot be smaller than the maximum value in the unnumbered equation. We show the reverse inequality by showing that the solution (a solution) for (iii) also satisfies the unnumbered equation, for a particular choice of the budget allocations.
Dec
25
comment How are these maximization problems same?
I'm afraid you are confused as to what $r$ is. $r$ is just the generic name for a group of goods, much like $i$ is the generic good. In my example, the first group, $r = \{1,2\}$ is the first group, and $r = \{3\}$ is the second group. So we don't take "values of $r$", especially because the partition into $m$ groups is given!
Dec
25
answered the proof of Arrow's Theorem
Dec
23
awarded  Yearling
Dec
20
answered How are these maximization problems same?
Sep
24
awarded  Commentator
Sep
24
comment What is the relevance of the supremum in this question?
@dukenukem Yes, since you are considering continuous functions on a closed and bounded interval, taking the max is the same as taking the supremum (this is the extreme value theorem).
Sep
24
comment Determining if this set defines a metric over and open interval?
Yes, for both questions there is a problem of integrability (that is, $d(x,y)$ could be infinite). However, unboundedness of $f$ or $g$ is not a problem per se: there are unbounded but integrable functions (on both $(0,1)$ and $\mathbb{R}$).
Apr
7
awarded  Organizer
Apr
5
comment Deducing probability from limited statistical data (exercise)
"Obviously" only given monotonicity: in the Brave New World $a$ would be higher than both $p$ and $q$ ("Youth almost unimpaired till sixty, and then, crack! the end."). $\frac{p+q}{2}$ is a good guess but probably not the best one: it interpolates linearly, while something like an exponential distribution seems more appropriate. In what type of math course were you given this exercise?
Apr
5
comment Deducing probability from limited statistical data (exercise)
Assuming the probability of living 15 more years is monotonic in age, we can bound the required probability between $p$ and $q$. I don't see how you can be any more precise than that.
Apr
4
comment Show $f \in L^p$ implies there is a set $E$ and a function $g$ for which $f=f\chi_E+g$ with $m(E)<\infty$ and $|g|\le 1$
Yes, precisely.
Apr
4
comment Show $f \in L^p$ implies there is a set $E$ and a function $g$ for which $f=f\chi_E+g$ with $m(E)<\infty$ and $|g|\le 1$
You are practically done! Just choose the right $g$... hint: it "complements" $f \chi_E$.