Martin
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 Dec14 awarded Caucus Sep24 awarded Autobiographer Jul2 awarded Curious Jun13 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 Jun13 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. Jun13 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 Feb11 answered How to prove the open interval $(1,5)$ is a convex set? Feb10 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. Feb10 answered How to solve $|x^2-1|-2\ge 2x$ Dec27 awarded Yearling Dec27 answered If Bob and Alice never met in class, at least one of them missed at least half of the classes Nov21 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. Nov17 answered Can't isolate $x$ for this equation Nov13 comment Proving constant function given the second derivative. $f(x) = x^2$ fulfills both criteria, but is not constant. Oct28 answered Difficult Derivative? Oct28 awarded Tumbleweed Oct26 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). Oct17 awarded Critic Oct16 comment Minimum of set $\{\frac{m}{n} + \frac{4n}{m}\}$ Hint: Can $(2n-m)^2$ be negative? Oct9 revised Induction proof that $n! > n^3$ for $n \ge 6$, and $\frac{(2n)!}{n! 2^n}$ is an integer for $n \ge 1$ added 511 characters in body