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seen Dec 14 at 23:05

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Dec
14
awarded  Caucus
Sep
24
awarded  Autobiographer
Jul
2
awarded  Curious
Jun
13
comment Game theory problem… I think…
I couldn't find a free PDF of Stahl, but maybe you can have a look at Janssen et al. (2005), which directly builds on Stahl. See kelley.iu.edu/mwildenb/costlysearch.pdf
Jun
13
comment Game theory problem… I think…
And in particular, if all consumers have to pay a positive "fine" (search cost), the famous "Diamond paradox" (see Diamond 1971) arises: in equilibrium, both firms will charge the monopoly price for the good(s) they offer.
Jun
13
comment Game theory problem… I think…
You might want to consider reading Stahl (1989), one of the seminal articles in the literature on (sequential) consumer search in economics. I think his setup directly corresponds to the one you have in mind. See jstor.org/stable/1827927?__redirected
Feb
11
answered How to prove the open interval $(1,5)$ is a convex set?
Feb
10
comment How to solve $|x^2-1|-2\ge 2x$
Do you mean the two cases? This follows directly from the right hand side - it will be negative for $x < -1$ and positive for $x \geq -1$. Hence, for $x < -1$, the inequality is automatically fulfilled.
Feb
10
answered How to solve $|x^2-1|-2\ge 2x$
Dec
27
awarded  Yearling
Dec
27
answered If Bob and Alice never met in class, at least one of them missed at least half of the classes
Nov
21
comment Need someone to quickly confirm whether I have this expected value correct
Looks correct to me. $E(Y^2)$ and $E(XY)$ are not needed for that answer, btw.
Nov
17
answered Can't isolate $x$ for this equation
Nov
13
comment Proving constant function given the second derivative.
$f(x) = x^2$ fulfills both criteria, but is not constant.
Oct
28
answered Difficult Derivative?
Oct
28
awarded  Tumbleweed
Oct
26
comment Proving that roots of a quadratic lie between two values
That's overall a nice answer, but to be 100% clean, I think it should read $x \in (x_1, x_2)$ (open instead of closed interval).
Oct
17
awarded  Critic
Oct
16
comment Minimum of set $\{\frac{m}{n} + \frac{4n}{m}\}$
Hint: Can $(2n-m)^2$ be negative?
Oct
9
revised Induction proof problem
added 511 characters in body