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Note: this is based on this post at physics stack exchange, but I thought it would be good to cross-post the final question here as well since it's pretty mathematical in nature.

Given the following inequality

$$ (z_2 \cos \theta - z_1)^2 + (z_2 \sin \theta)^2 - a^2 \geq 0 $$

where $z_1$, $z_2$, and $\theta$ are independent random variables, and $a$ is a positive real number. The values of $z_1$ and $z_2$ follow a Gaussian distribution such that

$$P(z) = \frac{1}{\sqrt{2\sigma^2 \pi}}\exp{\left(-\frac{z^2}{2\sigma^2}\right)} $$

and the value of $\sigma$ is known. $\theta$ has a uniform distribution given by

$$P_{phase}(\theta_i) = 1/2\pi; \quad \theta_i\in[0,2\pi]$$

What is the probability that the initial inequality is true?

I feel like there should be some way to evaluate this by somehow integrating $z_1$ and $z_2$ from $-\infty$ to $+\infty$ and $\theta$ from $0$ to $2\pi$, but I can't figure out how to put it in terms of the combined probability of all three variables making that inequality true.

I could of course evaluate this numerically by doing a brute-force Monte Carlo approach, but I feel like there should be some kind of closed-form integral or something, even if that integral ultimately has to be evaluated numerically.

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  • $\begingroup$ are they all independent? $\endgroup$ – Markoff Chainz Jan 27 '17 at 18:18
  • $\begingroup$ Yes, $z_1$, $z_2$, and $\theta$ are all independent. Let me clarify that with an edit in the main post. $\endgroup$ – derio Jan 27 '17 at 18:47
  • $\begingroup$ Why, of course there should be a way via integrating, only it would not be from $−\infty$ to $+\infty$. See, given a fixed $\theta$, what would be the territory in $(z_1,z_2)$ where the inequality is true? $\endgroup$ – Ivan Neretin Jan 27 '17 at 18:59
  • $\begingroup$ OK, I think you're trying to look at the problem in a slightly different way. I agree that for any fixed $\theta$ that there will be some domain in $(z_1,z_2)$ that the inequality is true, and then you could do that for all values of $\theta$, but how exactly would you find that domain in $(z_1,z_2)$? $\endgroup$ – derio Jan 27 '17 at 19:11
  • $\begingroup$ The domain is simple: it is bounded by some quadratic curve in $(z_1,z_2)$. The integral is probably not that simple, but that's another story. $\endgroup$ – Ivan Neretin Jan 27 '17 at 21:34

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