$X$ is a continuous random variable with probability density function
$$f(x) = \left\{\begin{array}{ll}6 \left(\sqrt{x} – x\right) & \textrm{for } 0 \leq x \leq 1\\
0 & \textrm{else} \end{array}\right.$$Find the distribution function of $X$.
Is it correct like that or have I missed anything / is it entirely wrong? :s
We have that $F(x)=\int_{-\infty}^{x} f(t) \, dt $
If $x<0 \,\,\,\ $ then $$F(x) = \int_{-\infty}^{x}f(t) \, dt = \int_{-\infty}^{x}0 \,dt = 0$$
If $0 \leq x \leq 1 \,\,\,\ $ then $$F(x) = \int_{-\infty}^{x} f(t) \, dt = \int_{-\infty}^{0} 0 \, dt + \int_{0}^{x} 6 \left(\sqrt{t}-t\right) \, dt = \int_{0}^{x} \left(6\sqrt{t}-6t\right) \, dt = \int_{0}^{x} \left(6t^{\frac{1}{2}}-6t\right) \, dt = \left[4t^{\frac{3}{2}} – 3t^2\right]_{0}^{x} = 4x^{\frac{3}{2}}-3x^2$$
Sooo the distribution function should be
$$F(x) = \left\{\begin{array}{ll}4x^{\frac{3}{2}}-3x^2 & \textrm{for } 0 \leq x \leq 1\\
0 & \textrm{for } x < 0\end{array}\right.$$
I hope it's done correct like that and maybe I did some unneccessary things anywhere?
Best Answer
Close. You missed that $F(x)=1$ for $1\leq x$.
$$F(x)=\begin{cases}0 &:& \qquad x\lt 0\\ 4x^{3/2}-3x^2&:&0\leqslant x<1\\1&:& 1\leqslant x\end{cases}$$