Question

If the probability density function of a random variable X is given by 

𝑓(𝑥) = {2𝑘𝑥𝑒

−

𝑥2

, 𝑥 > 0

0,

𝑥

≤

Determine (i)k (ii)distribution function.

Answer 1

$$f(x)= \begin{cases} 2kxe^{-x^2} &x>0\\ 0 &x\le0 \end{cases}$$

(a)

$$\displaystyle\int_{-\infin}^{\infin}f(x)dx=1$$

$$\displaystyle\int_{0}^{\infin} 2kxe^{-x^2}dx=\lim\limits_{t\to \infin}\displaystyle\int_{0}^{t} 2kxe^{-x^2}dx$$

$$=k\lim\limits_{t\to \infin}[-e^{-x^2}]\begin{matrix} t \\ 0 \end{matrix}=k=1$$

$$k=1$$

$$f(x)= \begin{cases} 2xe^{-x^2} &x>0\\ 0 &x\le0 \end{cases}$$

(b)

$$F(x)=\displaystyle\int_{-\infin}^{x}f(y)dy$$

If $x\le0, F(x)=0.$

If $x>0$

$$F(x)=\displaystyle\int_{0}^{x}2ye^{-y^2} dy=[-e^{-y^2}]\begin{matrix} x \\ 0 \end{matrix}=1-e^{-x^2}$$

$$F(x)= \begin{cases} 0 &x\le0\\ 1-e^{-x^2} &x>0 \end{cases}$$