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.
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]=\,
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