Probability — Question 4

Question 4 – Solution

Step 1: Normalize the density. $$1=\int_0^a kx\,dx =k\left[\frac{x^2}{2}\right]_0^a =\frac{ka^2}{2}.$$ Hence $$\boxed{k=\frac{2}{a^2}}.$$

Step 2: Build the CDF on the support. For $0\le x\le a$, $$\begin{align*} F(x)&=\int_0^x\frac{2t}{a^2}\,dt\\ &=\frac{2}{a^2}\left[\frac{t^2}{2}\right]_0^x =\frac{x^2}{a^2}. \end{align*}$$ Thus the complete CDF is $$F(x)= \begin{cases} 0,&x<0,\\[2pt] \dfrac{x^2}{a^2},&0\le x\le a,\\[4pt] 1,&x>a. \end{cases}$$

Step 3: Solve the quantile equation. The $p$-quantile $q_p$ satisfies $F(q_p)=p$: $$\frac{q_p^2}{a^2}=p.$$ Since $q_p\ge0$, $$\boxed{q_p=a\sqrt p},\qquad 0\le p\le1.$$