Suppose that \Theta takes values in [0,1] and its PDF is of the form
f_{\Theta }(\theta )= a\theta (1-\theta )^2,\ \ \ \mbox{for}\ \theta \in [0,1],
where a is a normalizing constant.
Use the formula
\int _0^1 \theta ^\alpha (1-\theta )^{\beta }\, d\theta =\frac{\alpha !\, \beta !}{(\alpha +\beta +1)!}
to find the following:
a) a=\,
b) {\bf E}[\Theta ^2]=\,
a) To find the value of a, we need to find the normalizing constant that makes the probability density function (PDF) integrate to 1 over the interval [0,1].
We have the PDF:
f_Theta(theta) = a * theta * (1 - theta)^2 for theta in [0,1].
To find the value of a, we integrate the PDF over the interval [0,1]:
∫[0,1] (a * theta * (1 - theta)^2) d(theta).
Using the given formula, we can rewrite the integral as:
a * ∫[0,1] theta^1 * (1 - theta)^2 d(theta).
Using the formula, we have:
∫[0,1] theta^1 * (1 - theta)^2 d(theta) = (1!) * (2!) / (1 + 2 + 1)! = 2 / 24 = 1 / 12.
Therefore, a * (1 / 12) = 1.
Solving for a, we get:
a = 12.
Therefore, a = 12.
b) To find the expected value, E[Theta^2], we need to calculate the integral of theta^2 multiplied by the PDF over the interval [0,1].
Using the formula, we have:
E[Theta^2] = ∫[0,1] theta^2 * f_Theta(theta) d(theta)
= ∫[0,1] theta^2 * (12 * theta * (1 - theta)^2) d(theta)
= 12 * ∫[0,1] theta^3 * (1 - theta)^2 d(theta).
Using the formula, we have:
∫[0,1] theta^3 * (1 - theta)^2 d(theta) = (3!) * (2!) / (3 + 2 + 1)! = 6 / 120 = 1 / 20.
Therefore, E[Theta^2] = 12 * (1 / 20) = 3 / 5.
Therefore, E[Theta^2] = 3 / 5.
To find the value of a, we need to calculate the normalizing constant by integrating the given probability density function (PDF) over its entire range:
a)
∫₀¹ aθ(1 - θ)² dθ = 1
Using the given formula for integration, we have:
∫₀¹ θ(1 - θ)² dθ = a * ∫₀¹ θ² (1 - θ)² dθ = 1
Using the formula for integration:
∫₀¹ θ² (1 - θ)² dθ = a * [(2!)(2!)] / [(2 + 2 + 1)!]
= a * (2)(2) / 5!
= a * 4/120
= a/30
So, a/30 = 1
Solving for a:
a = 30
Therefore, a = 30.
b)
To find E[Θ²], we need to calculate the expected value of Θ². The expected value is given by:
E[Θ²] = ∫₀¹ θ² f_Θ(θ) dθ
Substituting the given PDF f_Θ(θ):
E[Θ²] = ∫₀¹ θ² (30θ(1 - θ)²) dθ
Using the linearity of integration, we can split this into two integrals:
E[Θ²] = 30 ∫₀¹ θ³ (1 - θ)² dθ
Using the formula for integration, we have:
E[Θ²] = 30 * [(3!)(2!)] / [(3 + 2 + 1)!]
= 30 * (6)(2) / 6!
= 30 * 12/720
= 360/720
= 1/2
Therefore, E[Θ²] = 1/2.