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What is the expected value of Y given X? E(Y) = E(E(Y) | θ)) = E(P(Y) | θ)) since Y only takes the values 0 or 1 = E(θ) by the Bernoulli distribution = ∫01θp(θ)dθ = ∫012θ2dθ by part (a) = 2/3 Note: this integral is analogous to summation in the discrete case, with probabilities equal to the areas of thin rectangles under the density function's graph: P(θ) = p(θ)dθ.

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