Correct way to calculate $\Pr \left[ {X > \frac{2}{Y},X > \frac{4}{Y}} \right]$ where $X,Y$ follow the exponential distribution?

Question

What would be the correct way to calculate $\Pr \left[ {X > \frac{2}{Y},X > \frac{4}{Y}} \right]$ where $X,Y$ are independent random variable that follow the exponential distribution. I end up with different results and a mind full ò doubt

1/ The first way $\Pr \left[ {X > \frac{2}{Y},X > \frac{4}{Y}} \right] = \Pr \left[ {X > \frac{2}{Y}} \right] \times \Pr \left[ {X > \frac{4}{Y}} \right]$

2/ The second way $\Pr \left[ {X > \frac{2}{Y},X > \frac{4}{Y}} \right] = \Pr \left[ {X > Max\left[ {\frac{2}{Y},\frac{4}{Y}} \right]} \right] = \Pr \left[ {X > \frac{4}{Y}} \right]$

3/ The third way: use order statistics to find the pdf of $Z = Max\left[ {\frac{2}{Y},\frac{4}{Y}} \right] = {F_{\frac{2}{Y}}}\left( z \right){F_{\frac{4}{Y}}}\left( z \right)$ and then invoking the law of total probability $\Pr \left[ {X > \frac{2}{Y},X > \frac{4}{Y}} \right] = \Pr \left[ {X > Z} \right] = \int\limits_{x = 0}^{x = \infty } {\Pr \left( {X > x} \right){f_Z}\left( x \right)dx}$

I am thinking that I am having some misconception in my head since I am an engineer and not a mathematician.

P/S: As a though exercise and extension to the original problem:

In case their is a slight changes with the introduction of the variable $Z$ which is also an independent random variable follow the exponential distribution as $\Pr \left[ {X > \frac{2}{Y},X > \frac{4}{Z}} \right]$ what would be the best approach selected from 3 of the above way ?

Please help me clear my doubt, thank you very much

Answer 1

Since $\mathbb P(Y>0)=1$, it is clear that $\mathbb P\left(\frac 4Y>\frac 2Y\right)=1$ so $\left\{X>\frac 2Y\right\}\cap\left\{X>\frac 4Y\right\} = \left\{X>\frac 4Y\right\}$. Now, for independent continuous random variables $X$ and $Y$ with densities $f_X$ and $f_Y$, the density of the product $Z=XY$ is $$ f_Z(z) = \int_{\mathbb R} f_X(x) f_Y(z/x)\frac 1{|x|}\ \mathsf dx. $$ Supposing $X\sim\mathsf{Expo}(\lambda)$ and $Y\sim\mathsf{Expo}(\mu)$, we find that $$ f_Z(z) = \int_0^\infty \lambda e^{-\lambda x}\mu e^{-\mu\frac zx}\frac 1x\ \mathsf dx = 2\lambda\mu K_0(2\sqrt{\lambda z\mu}), $$ where $K_0$ denotes the modified Bessel function of the second kind. We may then compute $$ \mathbb P\left(X>\frac 4Y\right) = \mathbb P(Z>4) = \int_4^\infty f_Z(z)\ \mathsf dz = 4\sqrt{\lambda\mu}K_1(4\sqrt{\lambda\mu}). $$ More specifically, $K_n(z)$ satisfies the differential equation $$ z^2y'' +zy' -(z^2+n^2)y = 0. $$

Answer 2

As said the second way is correct.

You cant do the first way since the events $\{X>\frac{2}{Y}\}$ and $\{X>\frac{4}{Y}\}$ are not $\textit{Independent}$.