## Interview Question

Strategist Interview

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# Q3: You are playing this game with flipping a coin. You can play it as many times as you want until there is a head. At the nth time, when you get a head, they will pay the 2^n dollars. How much are you willing to play this game? What's the fair value?

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Ok, not sure why the last sentence gets messed up, trying once more here. The condition is not very enlightening, but it is clear if the two functions are plotted in 3D.: _ Exp[a]+Exp[b] only if a > b-Log[-1+Exp[b]]

Anonymous on

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raf's answer to the probability question is wrong. P(X+Y+Z 1) = 5/6

gandu on

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The answer to Q1 above here is wrong: there is no unique answer. To see this just consider the easy examples: _ a=b=0 implies Exp[a] + Exp[b] > Exp[a+b] _ a=b=1 implies Exp[a] + Exp[b] Exp[a]+Exp[b] only if a > b-Log[-1+Exp[b]] . Not very enlightening, but it is clear if the two functions are plotted in 3D.

Anonymous on

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Rewriting here as somehow the comment got messed up. The answer to Q1 above here is wrong: there is no unique answer. To see this just consider the easy examples: _ a=b=0 implies Exp[a] + Exp[b] > Exp[a+b] _ a=b=1 implies Exp[a] + Exp[b] b-Log[-1+Exp[b]] . Not very enlightening, but it is clear if the two functions are plotted in 3D.

Anonymous on

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Q1. exponential is strictly convex on R so as for every convex function : Phi, Phi(a+b) > Phi(a)+Phi(b) Q2. Could anyone tell me if this is correct : ?= P[x+y+z>1] = E[ P(x>1-(y+z)) / y,z] = E[ U(1-(y+z)) if (y+z)<1; 1 otherwise ] where U is the repartition function for a Uniform([0,1]) ?= E[ min((y+z),1) ] = E[ y+z / z<1-y] This last quantity can be computed through integrals using uniform densities : ?= Integral(y=0 to 1) [ y + Integral(z=0 to 1-y) [ z dz ] dy ] = Integral(y=0 to 1) [ y + (1-y)^2/2 dy ] ?= [ y^2/2 - (1-y)^3/6 ] taken between y=1 and y=0 ?= 1/2 + 1/3 = 2/3

raf on