# Cumulative distribution function of quotient

Let's consider two independent variables $$X \sim Exp(4)$$ and $$Y \sim Exp(12)$$. I want to calculate $$P(\frac{X}{X-3Y} \le \frac19)$$

My work so far

$$P(\frac{X}{X-3Y} \le \frac19) = P(\frac{9X-X+3Y}{X-3Y} \le 0)= P(\frac{8X+3Y}{X-3Y} \le 0)=$$ $$= P(\{8X+3Y \ge 0\} \cap \{X-3Y\le0\}) + P(\{8X+3Y \le 0\} \cap \{X-3Y\ge0\})$$

I'm not sure what I should do next... Intuitevly I would just decompose $$P(\{8X+3Y \ge 0\} \cap \{X-3Y\le0\}) = P(\{8X+3Y \ge 0\}) \cdot P(\{X-3Y\le0\})$$ but thinking more it's not so obvious for me that these events are independent.

Am I going in the right direction ?

• You may have left out inequalities within some of the probabilities. P(X/(X-3Y) ≤ 1/9) = P((9X - X + 3Y)/(X - 3Y) ≤ ?) Feb 1, 2021 at 19:45
• Hint $\mathbb P(X \ge c Y) = \int_{-\infty}^\infty \int_{cy}^\infty f_{X,Y}(x,y) \ dx \ dy$. Feb 1, 2021 at 20:02
• Are 4 and 12 exponential rates or means? // Either way, note that denominator can often be very near $0.$ Feb 2, 2021 at 21:14
Because the support of $$X$$ and $$Y$$ is $$[0, \infty)$$, $$P(\{8X+3Y \le 0\}=0$$
$$P(\{8X+3Y \ge 0\} \cap \{X-3Y\le0\}) + P(\{8X+3Y \le 0\} \cap \{X-3Y\ge0\})= P(\{8X+3Y \ge 0\} \cap \{X-3Y\le0\}) = P(\{X-3Y\le0\}),$$
$$P(\{X-3Y\le0\}) = \int_0^{\infty}\int_0^{3y}f_y(y)f_x(x)dxdy= \\\int_0^{\infty}\int_0^{3y}12e^{-12y}4e^{-4x}dxdy = 0.5$$
Interpretation: Exp($$\lambda$$) represents the waiting times between Poisson($$\lambda$$)-distributed events. $$E_Y$$, the Poisson event associated with $$Y$$, is $$\frac{12}{4} = 3$$ times more frequent than $$E_X$$, the Poisson event associated with $$X$$. The probability that $$E_X$$ will occur once before $$E_Y$$ happens three times $$\big(P(\{X-3Y\le0\})\big)$$ is 50%, because the waiting times between different $$E_Y$$s are independent.