here as we know from transformation method:
pdf of Y: f(y)=|dx/dy| *fx(g-1(y))
here X =-/+√y
|dx/dy| =1/(2√y)
hence f(y)=(1/(2√y))*(fx(√y))+fx(-√y)))
therefore for 0 <y<1
f(y)=(1/(2√y))*((2/9)*(√y+1)+(2/9)*(√y+1))
f(y)=(2/(9√y))*(√y+1)
for 1<y<4
f(y)=(1/(2√y))*((2/9)*(√y+1)+0)
f(y)=(1/(9√y))*(√y+1)
hence pdf f(y) =(2/(9√y))*(√y+1) for 0 <y<1
and f(y)=(1/(9√y))*(√y+1) for 1<y<4
otherwise f(y) =0
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