# Finding the $cdf$ from a $pdf$ indicator function

Consider a random variable $X$ having $pdf$

$$f_X(x)=\frac{1}{2\sqrt{x}}I_{(0,1)}(x)$$

(a) Give the $cdf$ of $X$, expressing it as a piecewise function.

(b) Find the median of $X$’s distribution by solving $F_X(η_X)$ = $1\over2$ for $η_X$.

(c) Give the value of $P$($1\over{4}$ $\lt X \lt$ $3\over{4}$).

(d) Give the value of $E(X)$.

Attempted Solution:

(a) $\int\frac{1}{2\sqrt{x}}$ = $\sqrt{x}$ giving a $cdf$ of

$$F_X(x) = \begin{cases} {1} & \text{x \geq 1} \\ \sqrt{x} & \text{0 \lt x \lt 1} \\{0} & \text{x \leq 0}\end{cases}$$

(b) Setting $F_X(η_X)$ $=$ $1\over2$, we get $\sqrt{η_X}$ $=$ $1\over2$ and so $η_X$ = $1\over4$

(c) $P$($1\over{4}$ $\lt X \lt$ $3\over{4}$) = $\sqrt{3\over4} - \sqrt{1\over4} = .366$

(d) $E(X)$ = $\int_0^{1}$ $x$ $f_X(x)$ = $\int_0^{1}$ $x$ $1\over{2\sqrt{x}}$=$\int_0^{1}$ ${\sqrt{x}}\over{2}$ =$1\over3$$x^{3/2}$$|_0^1$ $=$ $1\over3$

Did I do these correctly?

• Looks fine to me. I cannot find any mistakes in it. – drhab Oct 9 '17 at 21:11
• Thanks for the confirmation. Just wanted to be sure. – Remy Oct 9 '17 at 21:20