A random variable $X$ is drawn from a uniform distribution on interval $[0,1]$
$1)$ consider random variable $Y=1-(1-X)^2$, find it's probability distribution function $F_Y(y)$ and its probability density function $f_Y(y)$.
so far I have found (or tried) to get, could someone confirm?
$$F_Y(y)=1-\sqrt{1-y}, \ \ 0\leq y <1$$
and
$$f_Y(y)=F'_Y(y)=\frac{1}{2\sqrt{1-y}}, \ \ 0\leq y <1$$
Best Answer
Your solution is mostly okay.
1) Among other things, a cummulative distribution function must have an upper-infinite limit of one, so ensure that yours does too. (ie: add what happens when $y\geq 1$.)
2) The probability density function is the unsigned derivative of the cummulative distribution function. That doesn't change your answer, but should be noted for future reference.
Now, $X\sim\mathcal U[0..1]$ so $F_X(x)=x\mathbf 1_{0\leq x\lt 1}+\mathbf 1_{1\leq x}$ so...
$$\begin{align}F_Y(y)&=\mathsf P(1-(1-X)^2\leq y)\\[1ex]&=\mathsf P(X\leq 1-(1-y)^{1/2})\\[1ex]&=F_X(1-(1-y)^{1/2})\\[1ex]&=(1-(1-y)^{1/2})\mathbf 1_{0\leq y\lt 1}+\mathbf 1_{1\leq y}\\[1ex]&=\begin{cases}1-\surd(1-y)&:& 0\leq y\lt 1\\1&:& 1\leq y\\0&:& \text{otherwise}\end{cases}\\[5ex]f_Y(y)&=\lvert {F_Y}'(y)\rvert\\[1ex]&=\dfrac{1}{2(1-y)^{1/2}}\mathbf 1_{0\leq y\lt 1}\end{align}$$