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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 ?

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  • $\begingroup$ You may have left out inequalities within some of the probabilities. P(X/(X-3Y) ≤ 1/9) = P((9X - X + 3Y)/(X - 3Y) ≤ ?) $\endgroup$
    – vanPelt2
    Feb 1, 2021 at 19:45
  • $\begingroup$ Jup thanks. Adjusted already $\endgroup$
    – Lucian
    Feb 1, 2021 at 19:48
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    $\begingroup$ Hint $\mathbb P(X \ge c Y) = \int_{-\infty}^\infty \int_{cy}^\infty f_{X,Y}(x,y) \ dx \ dy$. $\endgroup$
    – Gregory
    Feb 1, 2021 at 20:02
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    $\begingroup$ Hmmm.... I'm not so sure what exactly it simplify. Could you please explain this more comprehensively? $\endgroup$
    – Lucian
    Feb 1, 2021 at 20:35
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    $\begingroup$ Are 4 and 12 exponential rates or means? // Either way, note that denominator can often be very near $0.$ $\endgroup$
    – BruceET
    Feb 2, 2021 at 21:14

1 Answer 1

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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.

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