For the distribution function given by $\;$
$F \left(x\right) = \begin{cases}
0, & x < 0 \\
& \\
x^2, & 0 \leq x \leq 1 \\
& \\
1, & x > 1
\end{cases}$
find the density function. Also evaluate
- $P \left(0.5 < x < 0.75\right)$
- $P \left(x \leq 0.5\right)$
- $P \left(x > 0.75\right)$
Probability density function (p.d.f) $= f \left(x\right) = \dfrac{d}{dx} F \left(x\right)$
Now, $\dfrac{d}{dx} \left(x^2\right) = 2 x$ $\;$ and $\;$ $\dfrac{d}{dx} \left(1\right) = 0$
$\therefore \;$ $f \left(x\right) = \begin{cases}
2 x, & 0 \leq x \leq 1 \\
& \\
0, & \text{ elsewhere}
\end{cases}$
- $P \left(0.5 < x < 0.75\right) = \int \limits_{0.5}^{0.75} 2 \; x \; dx$
Now,
$\begin{aligned} \int \limits_{0.5}^{0.75} 2 \; x \; dx & = 2 \times \left[\dfrac{x^2}{2}\right]_{0.5}^{0.75} \\\\ & = \left(0.75\right)^2 - \left(0.5\right)^2 = 0.3125 \end{aligned}$
$\therefore \;$ $P \left(0.5 < x < 0.75\right) = 0.3125$
- $P \left(x \leq 0.5\right) = \int \limits_{0}^{0.5} 2 \; x \; dx$
Now,
$\begin{aligned} \int \limits_{0}^{0.5} 2 \; x \; dx & = 2 \times \left[\dfrac{x^2}{2}\right]_{0}^{0.5} \\\\ & = \left(0.5\right)^2 = 0.25 \end{aligned}$
$\therefore \;$ $P \left(x \leq 0.5\right) = 0.25$
- $P \left(x > 0.75\right) = \int \limits_{0.75}^{1} 2 \; x \; dx$
Now,
$\begin{aligned} \int \limits_{0.75}^{1} 2 \; x \; dx & = 2 \times \left[\dfrac{x^2}{2}\right]_{0.75}^{1} \\\\ & = \left(1\right)^2 - \left(0.75\right)^2 = 0.4375 \end{aligned}$
$\therefore \;$ $P \left(x > 0.75\right) = 0.4375$