(a) \(F(x)=\begin{cases}0,&x\lt0,\\ \frac{2}{9}x^2+\frac{4}{9}x,&0\le x\lt1,\\ 1+\frac{1}{3}(x-2)^3,&1\le x\le2,\\ 1,&x\gt2.\end{cases}\)
(b) The median of \(X\) is \(\displaystyle \frac{-2+\sqrt{13}}{2}\).
(c) \(G(y)=P(Y\le y)=\begin{cases}0,&y\lt0,\\ \frac{2}{9}y^4+\frac{4}{9}y^2,&0\le y\lt1,\\ 1+\frac{1}{3}(y^2-2)^3,&1\le y\le\sqrt2,\\ 1,&y\gt\sqrt2.\end{cases}\)
(d) The median of \(Y\) is greater than the median of \(X\).
(a) The cumulative distribution function is \(F(x)=P(X\le x)\), so we integrate the density up to \(x\).
For \(0\le x\le1\),
\(F(x)=\int_0^x \frac{4}{9}(t+1)\,dt=\frac{4}{9}\left(\frac{x^2}{2}+x\right)=\frac{2}{9}x^2+\frac{4}{9}x.\)
For \(1\le x\le2\), first find the probability up to \(1\):
\(F(1)=\frac{2}{9}+\frac{4}{9}=\frac{2}{3}.\)
Then
\(F(x)=\frac{2}{3}+\int_1^x (t-2)^2\,dt=\frac{2}{3}+\left[\frac{(t-2)^3}{3}\right]_1^x.\)
Since \(\left[\frac{(t-2)^3}{3}\right]_1^x=\frac{(x-2)^3}{3}+\frac{1}{3}\), this gives
\(F(x)=1+\frac{1}{3}(x-2)^3.\)
Therefore
\(F(x)=\begin{cases}0,&x\lt0,\\ \frac{2}{9}x^2+\frac{4}{9}x,&0\le x\lt1,\\ 1+\frac{1}{3}(x-2)^3,&1\le x\le2,\\ 1,&x\gt2.\end{cases}\)
(b) Let the median of \(X\) be \(m\). Since \(F(1)=\frac{2}{3}\), the median lies in \(0\le x\lt1\). Hence
\(\frac{2}{9}m^2+\frac{4}{9}m=\frac{1}{2}.\)
Multiplying by \(18\),
\(4m^2+8m=9,\)
so
\(4m^2+8m-9=0.\)
Using the quadratic formula,
\(m=\frac{-8\pm\sqrt{8^2-4(4)(-9)}}{8}=\frac{-8\pm\sqrt{208}}{8}=\frac{-8\pm4\sqrt{13}}{8}.\)
The only root in \([0,1]\) is
\(m=\frac{-2+\sqrt{13}}{2}.\)
(c) Let \(G(y)=P(Y\le y)\). Since \(Y=\sqrt X\), and \(X\ge0\), we have
\(G(y)=P(\sqrt X\le y).\)
If \(y\lt0\), this probability is \(0\), because \(\sqrt X\) cannot be negative.
For \(y\ge0\), the inequality \(\sqrt X\le y\) is equivalent to \(X\le y^2\), so
\(G(y)=F(y^2).\)
Substituting \(x=y^2\) into the pieces of \(F\), and noting that the support of \(Y\) is \(0\le y\le\sqrt2\), gives
\(G(y)=\begin{cases}0,&y\lt0,\\ \frac{2}{9}y^4+\frac{4}{9}y^2,&0\le y\lt1,\\ 1+\frac{1}{3}(y^2-2)^3,&1\le y\le\sqrt2,\\ 1,&y\gt\sqrt2.\end{cases}\)
(d) Let the median of \(Y\) be \(m_Y\), and the median of \(X\) be \(m_X\). Since \(Y=\sqrt X\) is an increasing transformation, the median transforms in the same way:
\(m_Y=\sqrt{m_X}.\)
From part (b),
\(m_X=\frac{-2+\sqrt{13}}{2}\approx0.803.\)
Since \(0\lt m_X\lt1\), taking the square root makes it larger:
\(\sqrt{m_X}\gt m_X.\)
Therefore
\(m_Y\gt m_X\), so the median of \(Y\) is greater than the median of \(X\).
