Let U and V be independent and identically distributed uniformly on the interval [0, 1]. Show that for 0<x<1; P(x<V<U^2)=1/3 - x + (2/3)*x^3 and hence write down the density of V conditional on the event that V<U^2.
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Originally Posted by djones90 Let U and V be independent and identically distributed uniformly on the interval [0, 1]. Show that for 0<x<1; P(x<V<U^2)=1/3 - x + (2/3)*x^3 and hence write down the density of V conditional on the event that V<U^2. 1. Note that for , and zero otherwise. 2. Draw the square defined by and and shade the region corresponding to . 3. Then . I get , NOT ....
my mistake. Thanks!
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