I have a question regarding probability transformations. Could someone tell me wether I am doing it good or not?
Consider $f_{X}(x)=3/2e^{-3x}+3e^{-6x}, x \geq 0$,
1) Calculate the pdf of $Y=e^{-X}$
I would say $y(x)=e^{-x} \implies ln(y)=-x \implies x(y)=-ln(y)$ From the transformation theorem, it follows that $f_{Y}(y)=f_{X}(x(y))\cdot|dx/dy|$
$f_{Y}(y)=3/2e^{-3(-ln (y))}+3e^{-6(-ln(y))}\cdot|-1/y|=3y^3/2+3y^5, 0<y<1$ Is this correct? And if not, could someone tell me where I made a mistake?
2) Let U be a uniform random variable on the interval $[0, 1]$. For which function $h: [0,1] \rightarrow [0,1]$ is $h(U)$ distributed as $Y$ (from part 1)?
I would say $h=F_{Y}^{-1}$. $F_{Y}(y)=\int_0^y 3v^3/2+3v^5 dv=y^4(4y^2+3)\cdot 1/8.$ I'm kinda stuck here, because I don't see the inverse of $F_Y$. Could anyone give me a hint here?
There is a minor error of arithmetic, you forgot to divide $\frac{3}{2}y^3$ by $y$.
After that is fixed, the cdf $F_Y$ is a quadratic in $y^3$ in the interesting part of the world, and finding the inverse can be done with the quadratic formula.