problem 8

(a) Let $F$ be the CDF of the Beta distribution with parameters $a = 3$ , $b = 2$ .

Due to the properties of the CDF, $F$ must be 0 for $x \leq 0$ and 1 for $x \geq 1$ .

For $0 < x < 1$ :

\begin{matrix} \begin{matrix} F (x) & = \int_{0}^{x} f (t) dt = \int_{0}^{x} 12 t^{2} (1 - t) dt = 12 \int_{0}^{x} t^{2} dt - 12 \int_{0}^{x} t^{3} dt = 12 (\frac{x^{3}}{3} - \frac{x^{4}}{4}) = 4 x^{3} - 3 x^{4} \end{matrix} \end{matrix}

Then:

F (x) = {\begin{matrix} 0 & , for x \leq 0 x^{3} (4 - 3 x) & , for 0 < x < 1 \end{matrix}

(b)

\begin{matrix} \begin{matrix} P (0 < X < 1 ∕ 2) & = F (1 ∕ 2) - F (0) = {(\frac{1}{2})}^{3} (4 - \frac{3}{2}) - 0 = \frac{5}{16} \end{matrix} \end{matrix}

X

can be calculated by the definition of expectation:

\begin{matrix} \begin{matrix} E (X) & = \int_{0}^{1} xf (x) dx = \int_{0}^{1} 12 x^{3} (1 - x) dx = 12 \int_{0}^{1} x^{3} dx - 12 \int_{0}^{1} x^{4} dx = 12 (\frac{1}{4} - \frac{1}{5}) = \frac{3}{5} \end{matrix} \end{matrix}

By the definition of variance:

Var (X) = E (X^{2}) - {(EX)}^{2}

Let’s calculate the second moment of $X$ by LOTUS:

\begin{matrix} \begin{matrix} E (X^{2}) & = \int_{0}^{1} x^{2} f (x) dx = \int_{0}^{1} 12 x^{4} (1 - x) dx = 12 \int_{0}^{1} x^{4} dx - 12 \int_{0}^{1} x^{5} dx = 12 (\frac{1}{5} - \frac{1}{6}) = \frac{2}{5} \end{matrix} \end{matrix}

Substituting this value into the definition of variance:

Var (X) = \frac{2}{5} - {(\frac{3}{5})}^{2} = \frac{1}{25}