$U \sim U(0,1)$ and let $X$ be the root of $3t^2 −2t^3 −U = 0.$Show that $X$ has p.d.f. $f(x)=6x(1−x)$, if $0\leq x\leq 1$

Question

Let $U \sim U(0,1)$ and let $X$ be the root of the equation $3t^2 −2t^3 −U = 0.$ Show that $X$ has p.d.f. $f(x)=6x(1−x)$, if $0\leq x\leq 1,\;=0$,otherwise.

I really don't know how to begin solving this problem.

I don't even know what the roots of $3t^2 −2t^3 −U = 0$ are

Help will be much appreciated.

Answer 1

First of all, for $U \in (0,1)$ there are three roots of $3t^2 -2t^3-U=0$ (two roots if $U = 0$ or $1$) so the correct definition of $X$ would be the unique root in $[0,1]$.

Now set $f(t) = 3t^2-2t^3-U$. Taking the derivative, it is easy to check that $f$ is increasing on $[0,1]$ and that $f(0) = -U$ and $f(1) = 1-U$ (proving that $X$ is well defined). Using this, we can show that for $x \in [0,1]$, we have $$\{X \leq x\} = \{U \leq 3x^2 - 2x^3\}.$$ Indeed, $X \leq x$ if and only if $f(X) \leq f(x)$ if and only if $0 \leq 3x^2 - 2x^3 - U$. It follows that $$\mathbb{P}(X\leq x) = \mathbb{P}(U \leq 3x^2 - 2x^3) = 3x^2 - 2x^3,$$ and the right-hand side is the cdf of the given distribution (just take the derivative).