So i have a random vector $(X,Y)$ with probabilty density:
$$ p(x,y) = \begin{cases} e^{-x+y} & \quad \text{if } x>0 \text{ and } y<0\\ 0 & \quad \text{othervise} \end{cases} $$ So I have to define disstribution function of bigger of the roots of this equation:
$$t^2+Xt+Y$$ where $X,Y$ are random variables from the vector.
So the bigger of the roots is:
$$t=\frac{-X+\sqrt{X^2-4Y}}{2}$$
And now i don't know how to define the distribution function of this root.
Also i have to calculate the probability that both of the roots are lower than 1.
Any help would be appreciated.
Since $Y$ must be less than zero, we know that the bigger root $t^+$ must be greater than zero. So we are looking for the probability that $0< t^+ \leq 1$. We know that $t^+=1$ when $Y = -X-1$ and $t^+ = 0$ when $Y = 0$. So \begin{align} P(0<t^+\leq 1) &= P(-X-1 \leq Y < 0) \\ &= \int_0^\infty\int_{-x-1}^0e^{y-x}\;dy\;dx \\ &=\int_0^\infty e^{-x} - e^{-2x-1}\;dx \\ &= 1 - \frac{1}{2}e^{-1} \end{align} In general, $t^+\leq a$ for $a>0$ requires $Y \geq - aX-a^2$. So \begin{align} P(t^+\leq a) &=P(-aX-a^2 \leq Y < 0) \\ &= \int_0^\infty \int_{-ax-a^2}^0 e^{y-x}\;dy \;dx\\ &=\int_0^\infty e^{-x}-e^{-(a+1)x-a^2}\;dx \\ &= 1 - \frac{e^{-a^2}}{a+1} \end{align}