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The following is a derivation of a density from a paper I am currently studying. Sorry for the bad quality, it is quite an old paper. I need to clarify that $R$ has the standard exponential density in $(0,\infty)$, $U$ is uniform on $(0,1)$ and they are independent. The population correlation coefficient $\rho$ is a constant of course. $X$ and $Y$ come from the standard bivariate normal distribution, hence the trigonometric representation, but this plays no role here, I believe.

What I do not understand is how the author reaches these conclusions for positive or negative $t$. It seems to me that the division by a negative number and the nonnegativity of $R$ are not properly taken into account. I could be mistaken of course so I would appreciate some advice. Thank you.

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    $\begingroup$ @Xi'an Thank you for your comment. This representation is derived from the fact that $XY = \left[ (X+Y)^2 -(X-Y)^2 \right]/4$ with $X-Y$ and $X+Y$ independent. Since the sum has variance $2(1+\rho)$ and the difference $2(1-\rho)$, then $XY$ has the same distribution as $$\frac{1}{2}\left((1+\rho)Z_1^2 -(1-\rho) Z_2 ^2 \right)$$ where $Z_i$ now has the standard normal distribution. Then the result follows by putting $Z_1=\sqrt{-2\log(U_1})\cos(2\pi U_2)$ and $Z_2=\sqrt{-2\log(U_1)} \sin(2\pi U_2)$, the Box-Muller transform and uitlizing that $- \log(U)$ has the standard exponential distribution $\endgroup$ – JohnK Oct 23 '15 at 20:59
  • $\begingroup$ @Xi'an No problem. Would you say that the subsequent steps are correct, then? $\endgroup$ – JohnK Oct 23 '15 at 21:08
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I may also be mistaken but see no difficulty with the decomposition.

When $t\ge 0$, \begin{align*} P(R(\cos(\pi U)&+\varrho)\ge t,\cos(\pi U)+\varrho\ge 0)\\ &+P(R(\cos(\pi U)+\varrho)\ge t,\cos(\pi U)+\varrho\le 0)\\ &=P(R(\cos(\pi U)+\varrho)\ge t,\cos(\pi U)+\varrho\ge 0) \end{align*} since the second term is zero, $R$ being multiplied there by a negative term. Thus $$ P(XY\ge t) = P(R(\cos(\pi U)+\varrho)\ge t,\cos(\pi U)+\varrho\ge 0)\quad\qquad \\ \ \ \ = P(R(\cos(\pi U)+\varrho)\ge t,U\le\cos^{-1}(-\varrho)/\pi)\\ = \int_0^{\cos^{-1}(-\varrho)/\pi} P(R(\cos(\pi U)+\varrho)\ge t)\,\text{d}u $$ seems to be correct.

When $t\le 0$, since $$R(\cos(\pi U)+\varrho)\ge t$$ is always true when $\cos(\pi U)+\varrho\ge 0$, \begin{align*} P(R(\cos(\pi U)&+\varrho)\ge t,\cos(\pi U)+\varrho\ge 0)\\ &+P(R(\cos(\pi U)+\varrho)\ge t,\cos(\pi U)+\varrho\le 0)\\ =P(\cos(\pi U)&+\varrho\ge 0)\\ &+P(R(-\cos(\pi U)-\varrho)\le -t,\cos(\pi U)+\varrho\le 0)\\ =P(\cos(\pi U)&+\varrho\ge 0)\\ &+P\left\{R\le t\big/(\cos(\pi U)+\varrho),U\ge\cos^{-1}(-\varrho)/\pi\right\}\\ =P(\cos(\pi U)&+\varrho\ge 0)\\ &+\int_{\cos^{-1}(-\varrho)/\pi}^1 P\left\{R\le t\big/(\cos(\pi u)+\varrho\right\} \end{align*} so this seems to be correct as well.

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    $\begingroup$ I think my mistake was in inverting the cosine, didn't switch the inequalities. Thank you very much for your answer. $\endgroup$ – JohnK Oct 23 '15 at 21:31

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