Cumulative distribution (uniform distribution/continuous)

Question

Let $X$ be a continuous uniform random variable on the interval $[0,2]$.

How to determine the cumulative distribution function of $Y=X(2-X)$?

I got:

Since $X$ is uniformly distributed, the probability density function is $f_X(x)=\frac{1}{2}\boldsymbol{1}_{0<x<2}$

Now I tried to find the cumulative distribution function:

$F_Y(y)=\mathbb{P}(Y \leq x)=\mathbb{P}(X(2-X) \leq x)=\mathbb{P}((2X-X^2) \leq x)$

How to continue the transformation such that I get $\mathbb{P}(X \leq …)$?

I don't see how to rewrite $2X-X^2$ here.

Answer 1

Add $-1$ to both sides,$$2X-X^2-1=-(X-1)^2\le x-1\implies(X-1)^2\ge1-x$$When $x\ge1$, the inequality is always true and the c.d.f. of $Y$ is $1$. For $x<1$,$$P[(X-1)^2\ge1-x]\iff P(|X-1|\ge\sqrt{1-x})\\\iff P(X\le1-\sqrt{1-x})+P(X\ge1+\sqrt{1-x})$$For $x<0$, the first and second terms are both $0$ (Why?). So you only have to talk about $x\in[0,1)$. Can you take this forward?

Answer 2

You actually cannot, because $x \mapsto x(2-x)$ is not an injective function on $[0,2]$; each point in $(0,1)$ has two preimages. Thus you need to write the event $X(2-X) \leq x$ as $X \in I_1 \cup I_2$ where $I_1,I_2$ are appropriately chosen intervals. A graph will help you to do that.

Answer 3

Adding some sketches might help clarify the approach and motivate intuition/solving:

              Y
              |            Y=X(2-X)
              |Y=1           .  .
              |          .           .         pdf(X)
              |--------------------------------                       
              |   .                        .
              | .
            --0----------------1--------------2---- X
              |     |                   |
              |   X_lower            X_upper
              |

Solving $Y=X(2-X)$ for X we get: $X = \frac{2\pm \sqrt{4-4Y}}{2} = 1 \pm \sqrt{1-Y} $

Since we know that PDF of $X$ is $1/2$ for $X$ in $[0,2]$, the CDF is just the constant integral from 0 to X (lower portion) and X to 2 (upper portion) - see figure above for the sketch of Y(x):

CDF of lower portion is $1/2*(1-\sqrt{1-Y})$: Negative root is the lower solution for X.
CDF of upper portion is $1/2*(2-(1+\sqrt{1-Y}))$: Positive root is the upper solution for X.
Total is the sum = $1-\sqrt{1-Y}$
Therefore $P(Y \le y)=1-\sqrt{1-y}$

I hope this helps.