Suppose that Θ takes values in [0,1] and its PDF is of the form
fΘ(θ)=aθ(1−θ)2, for θ∈[0,1],
where a is a normalizing constant.
Use the formula
∫01 θα(1−θ)βdθ = α!β! / (α+β+1)!
to find the following:
a) a= ???? b) E[Θ2]= ?????

Answers

Answer 1

a) To find the value of a, we need to integrate the PDF from 0 to 1 and set it equal to 1 since a is a normalizing constant:


∫01 aθ(1−θ)2dθ = 1
Using the formula given, we can simplify the integral:
a ∫01 θ1(1−θ)2dθ = 1
a (1!) (2!) / (1+2+1)! = 1
a (2/6) = 1
a = 3

Therefore, the value of a is 3.

b) To find E[Θ2], we need to use the formula for the expected value:
E[Θ2] = ∫01 θ2fΘ(θ)dθ
Substituting the given PDF and the value of a, we get:
E[Θ2] = ∫01 θ2(3θ(1−θ)2)dθ
Simplifying the integral using the formula given, we get:
E[Θ2] = 3 ∫01 θ3(1−θ)2dθ / 4
E[Θ2] = 3 (3!) (2!) / (3+2+1)! / 4
E[Θ2] = 3/20

Therefore, E[Θ2] is equal to 3/20.

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