Knowing that both $x \sim \mathcal{N}(0,1)$ and $y \sim \mathcal{N}(0,1)$ ($x,y$ independent from each other), I want to compute

$$\mathbb{P}(3x > -y > x \land x > 0 \land y < 0)$$

I'm aware that if two events are independent, I can take the two probabilities and multiply. E.g., if I were only interested in

$$ \mathbb{P}(x > 0 \land y < 0) $$

I would simply compute $\mathbb{P}(x > 0) \mathbb{P}(y < 0)$.

However, in the problem described above, the condition $3x > -y > x$ is clearly not independent from the two other conditions $x >0 \land y <0$. I think that I should re-formulate the condition to $3x > -y \land -y > x$ ... but that still leaves the issue that the conditions are somewhat dependent from each other.

How can I approach such a problem?

  • 1
    $\begingroup$ The duplicate addresses the same problem, changing only minor details: given a joint density $f_{X,Y}$ and inequalities determining an event, how does one go about finding the probability of the event? $\endgroup$ – whuber Nov 29 '14 at 14:58

Since $X$ and $Y$ are independent Gaussians, they are also jointly normaly distributed. So, one can just take the pdf of a bivariate standard normal distribution and adjust the integrals according to the given conditions. With $Z = -Y$ and $f(x,z)$ denoting the bivariate normal distribution, I thought about the problem like this:

$ \quad P(3x > -y > x \land x > 0 \land y < 0) \\ =P(3x > z > x) ; \ \ x,z>0 \\ =P(3x > z) - P(x > z); \ \ x,z>0 \\ =P(3x - z > 0) - P(x - z > 0); \ \ x,z>0 \\ = \int_0^{\infty} \int_0^{3x} f(x,z)dzdx - \int_0^{\infty} \int_0^{x} f(x,z)dzdx.$

Since the analytical solution at this point wasn't easy, I have to admit that I just used WolframAlpha. As a result I got:

$\quad \int_0^{\infty} \int_0^{3x} f(x,z)dzdx - \int_0^{\infty} \int_0^{x} f(x,z)dzdx \\ = \frac{tan^{-1}(3)}{2\pi} - \frac{1}{8} \approx 0.073792. $

  • 4
    $\begingroup$ This is correct, as one can see by noting that (a) the joint distribution has a circular symmetry and (b) the region of integration is a sector bounded by lines at angles $-\pi/4$ and $\arctan(-3)$. Therefore the probability is whatever fraction of a full circle this sector occupies, which obviously is $(-\pi/4 - \arctan(-3))/(2\pi)$. That's equivalent to your answer--but is much more easily obtained! $\endgroup$ – whuber Nov 29 '14 at 15:06
  • $\begingroup$ Thank you for this comment. There will hopefully a time when I can solve such problems just like this! $\endgroup$ – random_guy Nov 29 '14 at 15:15
  • $\begingroup$ @whuber, can you please elaborate which picture you have in mind when to say that "the region of integration is a sector bounded by lines at angles", and how you come up with $-\pi/4$ and $\arctan(-3)$ ? $\endgroup$ – bonifaz Nov 29 '14 at 16:04
  • 3
    $\begingroup$ @bonifaz The criteria $x\gt 0$ and $y \lt 0$ place this region in the fourth quadrant. Its boundaries are portions of the lines $3x=-y$, with slope $-3$ (and therefore at an angle of $\arctan(-3)$), and $-y=x$, with slope $-1$ (at an angle of $\arctan(-1)$). For any circularly symmetric distribution the probability of this event will be $\frac{1}{2\pi}$ times the angular difference $\arctan(-1)-\arctan(-3)$. $\endgroup$ – whuber Nov 29 '14 at 16:40

$\mathbb{P}(3x>-y>x \ \& \ x>0 \ \& \ y<0) = \mathbb{P}(3x>-y>x >0)$ Maybe this will help you on start.

Then, let's divide by x ( that is greater than 0)

$ \mathbb{P}(3x>-y>x >0) = \mathbb{P}(3>-\frac{y}{x}>1, x>0)$$

  • 1
    $\begingroup$ Have you solved it? I thought my solution must be correct but I think with your approach one has only to look at cdf of a Cauchy distribution. But then, I got two times my result. But I don't know where the difference is. $\endgroup$ – random_guy Nov 29 '14 at 13:51
  • 1
    $\begingroup$ @random It looks like you might not have accounted for the restrictions $x\gt 0, y\lt 0$. They cause $-y/x$ not to have a Cauchy distribution--which is obvious when you consider that $-y/x$ is necessarily positive, whereas half the probability of a Cauchy distribution is for negative values. $\endgroup$ – whuber Nov 29 '14 at 16:47
  • $\begingroup$ Yes of course, it just came to my mind when I saw this ratio and then I was probably too keen to go this way to do well. I also thought about squaring everything and using two Chi-squared distributed rv's instead to get the cdf of the F-distribution. $\endgroup$ – random_guy Nov 29 '14 at 17:08

Not the answer you're looking for? Browse other questions tagged or ask your own question.