Let Θ be a continuous random variable that represents the unknown bias (i.e., the probability of Heads) of a coin.
a) The prior PDF fΘ for the bias of a coin is of the form
fΘ(θ)=aθ9(1−θ), for θ∈[0,1],
where a is a normalizing constant. This indicates a prior belief that the bias Θ of the coin is
1. High
2. 10
3. 10
as rare as a unicorn sighting. Maybe you should reconsider using this coin for anything important.
To determine the normalizing constant, we need to integrate the prior PDF over its entire domain, which in this case is [0,1].
∫[0,1] aθ^9(1-θ) dθ
We can simplify this by splitting it into two separate integrals:
∫[0,1] aθ^9 dθ - ∫[0,1] aθ^10 dθ
Integrating the first term:
(a/10)θ^10 evaluated from 0 to 1
(a/10) - (a/10) * 0 = a/10
For the second term, we have:
(a/11)θ^11 evaluated from 0 to 1
(a/11) - (a/11) * 0 = a/11
To normalize the prior PDF, the integral of the PDF over its entire domain must equal 1. Therefore,
(a/10) + (a/11) = 1
To solve for a, we can multiply through by the least common denominator (10 * 11 = 110):
11a + 10a = 110
21a = 110
a = 110/21
Therefore, the normalizing constant a is 110/21.
To calculate the normalizing constant for the prior PDF, you need to integrate the PDF over its entire range and set it equal to 1.
The given prior PDF is: fΘ(θ) = aθ^9(1−θ), for θ ∈ [0,1].
To find the normalizing constant 'a', you need to evaluate the integral of fΘ(θ) over the range [0,1] and then solve for a.
∫[0,1] aθ^9(1−θ) dθ = 1
Integrating this expression:
a ∫[0,1] θ^9(1−θ) dθ = 1
Evaluate the integral:
a [ (θ^10/10 - θ^11/11) ] evaluated from θ=0 to θ=1 = 1
Now, evaluate the expression at θ=1 and θ=0:
a [ (1^10/10 - 1^11/11) ] - a [ (0^10/10 - 0^11/11) ] = 1
Simplifying this expression:
a [ (1/10 - 1/11) ] = 1
Now, solve for 'a':
a = 1 / [(1/10 - 1/11)]
Solving further:
a = 1 / [(11 - 10)/110]
a = 110
So, the normalizing constant 'a' for the prior PDF is 110.