# Order statistics of absolute value of bivariate normal distribution

Suppose $X_1$ and $X_2$ are bivariate normal and let $|X|_{(1)}$ and $|X|_{(2)}$ be the ordered version of their absolute value. I am interesting in finding the following probabilities or some bounds on it. \begin{align*} Pr(|X|_{(1)} < c_1, |X|_{(2)} < c_2) \end{align*} and \begin{align*} Pr(|X_1| < c_1, |X|_{(1)} < c_2) \end{align*} Can anyone familier with this give some hint or some references? Thank you very much. Hanna

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## 1 Answer

For $c_1 < c_2$, the event $\left\{|X|_{(1)} < c_1, |X|_{(2)} < c_2\right\}$ is the union of the two events \begin{align} A &= \left\{-c_1 < X_{1} < c_1, -c_2 < X_{2} < c_2\right\}\\ B &= \left\{-c_2 < X_{1} < c_2, -c_1 < X_{2} < c_1\right\} \end{align} whose intersection is the event $A \cap B = \left\{-c_1 < X_{1} < c_1, -c_1 < X_{2} < c_1\right\}$. The probabilities of all three events can be expressed in terms of the bivariate cumulative normal distribution function, and then we can use $$P\left\{|X|_{(1)} < c_1, |X|_{(2)} < c_2\right\} = P(A\cup B) = P(A) +P(B) - PA\cap B).$$

For $c_1 > c_2$, the event $\left\{|X|_{(1)} < c_1, |X|_{(2)} < c_2\right\}$ is the same as the event $A\cap B$.

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 Thank you very much, Dilip. Hanna. – user13154 Oct 8 '12 at 13:54