Let $X$ be a continuous uniform random variable on the interval $[0,3]$ and $Y = 2 \cdot X − X^2$
1) Compute the pdf of $Y $
2)check that it is a pdf
My answer is $\frac{1}{6 \sqrt{1-y}} I[-8,1]$ and otherwise $0$. I am not sure if this is right but if anyone could let me know where I went wrong or if its right that would be appreciated.
To calculate this used the quadratic formula and got
$$G(y)=\Pr(1-\sqrt{1-y}< x<1+\sqrt{1-y})$$since the bounds for X are 0 to 3 i did $$\textbf I_{[0,1+\sqrt{1-y}]} f(x)=\frac13$$
- then I got my answer to be $$\frac13(1+\sqrt{1-y})$$
- From there i took the derivative and got the final answer shown above -I got the bounds $-8$ to $1$ since $1$ is the highest value y could take and $-8$ since $x=-2$ gives you -8 when you do $Y=2(-2)-(-2)^2$
-Kinda new here so my apologies if its confusing