A possible mistake in a conditional probability derivation 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.
 
 A: 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.
