Step-by-step explanation:
We have two cases for Ф,
1. Ф=1; it implies that Pr(Ф=1)=0.5, while y~N(1,α²)
2. Ф=2; it implies that Pr(Ф=2)=0.5, while y~N(2,α²)
Now,
For 1st case of α=2,
We have marginal probability density formula
p(y)=∑p(yIФ)p(Ф)
=p(yIФ=1)p(Ф=1)+p(yIФ=2)p(Ф=2)
=N(yI1,2²)(1/2)+N(yI2,2²)(1/2)
=(1/2)[N(yI1,2²)+N(yI2,2²)]
Now.
For Pr(Ф=1Iy=1) at α=2
We have,
=p(Ф=1Iy=1)
=[p(y=1,Ф=1)]/[p(y=1)]
=[p(y=1IФ=1)p(Ф=1)]/[p(y=1)]
={(1/[tex]\sqrt{2x-2}[/tex])exp[(-1/(2*2²))(1-1)²(1/2)]}/{(1/[tex]\sqrt{2x-2}[/tex])(1/2)[exp[(-1/(2*2²))(1-1)²]+exp[(-1/(2*2²))(1-2)²]}
=0.53 Answer
Now, to describe the changes in shape of Ф when α is increased and decreased:
The formula for posterioir density is p(ФIy)=p(yIФ)p(Ф)/p(y)
=exp[(-1/(2α²)(y-Ф)²]/{exp[(-1/(2α²)(y-1)²]+exp[(-1/(2α²))(y-2)²]}
Now at Ф=1 and solving the equation, we get
p(Ф=1Iy)=1 / {1+exp[(2y-3)/2α²]}
Similarly at Ф=1 and solving the equation, we get
p(Ф=2Iy)=1 / {1+exp[(2y-3)/2α²]}
Conclusion:
α² → ∞ ⇒p(ФIy) → p(Ф) = 1/2
α² → 0 ⇒ two cases
y > 3/2, α² → 0 ⇒p(Ф=2Iy) → 1
y < 3/2, α² → 0 ⇒p(Ф=1Iy) → 1
