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Given f(x|y)=1/y, 0<y<1, 0<x<y and f(y)=1, 0<y<1. Calculate...
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You need to remember the formula of variance such that:
`Var(x|y) = E(x^2|y) - (E(x|y))^2`
You need to evaluate `E(x^2|y) ` such that: `int_0^1 y*f(x|y) dy` `E(x^2|y):int_0^1 y*(1/y) dy = int_0^1 dy= 1`
You need to evaluate `(E(x|y))^2` such that: `int_0^1 y^2*f(x|y) dy`
`` `int_0^1 y^2*f(x|y) dy= int_0^1 y^2*(1/y) dy =gt int_0^1 y*f(x|y) dy = 1/2`
`Var(x|y) = 1 - 1/2 =gt Var(x|y) = 1/2 = 0.5`
Hence, evaluating the expectation and variance under the given conditions yields E(`x|y` ) = 0.5 and Var(`x|y` ) = 0.5.
Posted by sciencesolve on February 6, 2012 at 7:08 PM (Answer #1)
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