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Entrevista de Strategist

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# Given three random variables (x,y,z) that are i.i.d, what is the probability of x+y+z < 1

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Anónimo en

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The answer is 1/6. Explanation: draw a cube of side length 1 with a base that defined by the points (0,0), (0,1), (1,0), (1,1) and 1 unit high. The volume underneath the plane x+y+z=1 is 1/4 of a pyramid height=1 and base 2, thus the pyramid has volume 1/3 * 1 * 2. Taking 1/4 of this volume is 1/6.

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?= 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 = 5/6

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