Non-standard question about random variables I am not sure which subbranch of mathematics this is, so I cannot give a precise tag. I am doing research, and this suddenly popped out of no where. So, please hear me out. 

$x$ is a variable that takes on random values. $x$ is said to 'stable' if $|x| \leq 1$. 
Here is an example: $x = \dfrac{cY+1}{2}$, where $Y \sim U(0,2)$. In this case, $c \in \mathbb{R}$ and $Y$ is a random variable. For what value of $c$ will $x$ become 'stable'? 
Since we know that $\dfrac{1}{2}cY \sim U(0, c)$, then $\dfrac{1}{2}cY + \dfrac{1}{2} \sim U(\frac{1}{2}, c+\frac{1}{2})$. We want $x$ to be stable so $|c+\frac{1}{2}| \leq 1$, therefore, $c \in [-\frac{3}{2}, \frac{1}{2}]$.

Here is my question: 
Let $Y \sim U(0,1)$. For what value of $c$ will make $x$ 'stable' if 
$$x = \dfrac{\sqrt{(Y-\frac{2}{c})^2-4}}{2}?$$

You want to find $c$ such that $$-1\le \dfrac{\sqrt{(Y-2/c)^2-4}}{2} \le 1$$ (and in particular is real) for $0 \le Y \le 1$.  Equivalently,
$$ 4 \le (Y - 2/c)^2  \le 8$$
for $0 \le Y \le 1$.  So either $2 \le Y - 2/c \le 2 \sqrt{2}$ in that interval, or $-2\sqrt{2} \le Y - 2/c \le -2$.  Now unfortunately, 
$2\sqrt{2} - 2 < 1$, so this can't ever be true for all $Y \in [0,1]$.
EDIT: If complex numbers are acceptable, then you just need $0 \le (Y - 2/c)^2 \le 8$, so $-2\sqrt{2} \le Y - 2/c \le 2 \sqrt{2}$ for $0 \le Y \le 1$, and thus either  $c \ge \sqrt{2}/2$ or $c \le -(2 + 4 \sqrt{2})/7$. 
