Knowing the distribution of $X$, what is the distribution function of $Y:= X(X-1)$?

Question

Let $X$ be exponentially distributed with parameter $1$ and let $Y := X(X-1)$.

Determine the distribution function $F_Y$ of $Y$.

I know that I need to calculate as follows: $ F_Y(t) = P[Y \leq t] = P[X(X-1) \leq t] = P[X^2-X \leq t]$.

I also know that $F_X(t) = 1-e^{-t}$ for $t \geq 0$, since $X$ is exponentially distributed with parameter $1$.

However, I don't know where to go from there. Thank you for your help!

Answer 1

$$ X(X-1) = X^2-X = X^2 - 2\cdot X\cdot \frac{1}{2} + \left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2 = \left(X-\frac{1}{2}\right)^2-\frac{1}{4} \Rightarrow $$

$$ \begin{aligned} F_Y(t) &= \Pr\left(Y \leq t\right) = \Pr\left(X(X-1) \leq t\right) = \Pr\left(\left(X-\frac{1}{2}\right)^2-\frac{1}{4} \leq t\right) = \Pr\left(\left(X-\frac{1}{2}\right)^2\leq t + \frac{1}{4}\right) = \\ &=\left\{ \begin{array}{rl} 0, & t < -\frac{1}{4} \\ \Pr\left(\left|X-\frac{1}{2}\right|\leq \sqrt{t + \frac{1}{4}}\right), & t \geq -\frac{1}{4} \end{array} \right. = \\ &=\left\{ \begin{array}{rl} 0, & t < -\frac{1}{4} \\ \Pr\left(\frac{1}{2} - \sqrt{t + \frac{1}{4}} \leq X \leq \frac{1}{2} + \sqrt{t + \frac{1}{4}}\right), & t \geq -\frac{1}{4} \end{array} \right. = \\ &=\left\{ \begin{array}{rl} 0, & t < -\frac{1}{4} \\ F_X\left(\frac{1}{2} + \sqrt{t + \frac{1}{4}}\right) - F_X\left(\frac{1}{2} - \sqrt{t + \frac{1}{4}}\right), & t \geq -\frac{1}{4} \end{array} \right. = \\ \end{aligned} $$

where $F_X(x)$ is a probability distribution function of $X$.