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So assume $Y$ and $X$ are exponentially distributed with parameters $y_1$, and $x_1$ respecitively. assume c is a constant.

I am having huge trouble to understand the integration of the following expression.

$P(Y<c/u(X))$

$=\int_{t}^{\infty}f_X(x)\int_{0}^{c/u(x)}f_Y(y)dydx +\int_{0}^{t}f_X(x)\int_{c/u(x)}^{\infty}f_Y(y)dydx$

where t is the cross-point that $u(x)$ change sign


here $c/u(x)$ is given by the plot below, t is the point crossing the zero: enter image description here

Confusion:

I don't understand the second integration of the second term "$\int_{c/u(x)}^{\infty}f_Y(y)dydx$", this isn't right because Y an X only defined for y>0, and x>0. So it's the first quadrant in this plot.

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    $\begingroup$ The RHS is most bizarre. You might want to indicate the source. $\endgroup$ – Did Jul 9 '14 at 20:54
  • $\begingroup$ @Did, do you think this is correct, because I think the second term is not right. $\endgroup$ – kou Jul 9 '14 at 23:25
  • $\begingroup$ Quote: "You might want to indicate the source." $\endgroup$ – Did Jul 9 '14 at 23:39
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    $\begingroup$ @Did, the source comes from an paper:math.stackexchange.com/questions/859663/… $\endgroup$ – kou Jul 10 '14 at 0:04
  • $\begingroup$ I didn't understand the yellow part of math.stackexchange.com/questions/859663/…, I ask the author, and he gave me his skiped steps as this question, which I also don't understand. $\endgroup$ – kou Jul 10 '14 at 0:06
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Actually, one is not interested in $P(Y<c/u(X))$ but in $P(Yu(X)\lt c)$, and this is $$P(Yu(X)\lt c)=\int_{t}^{\infty}f_X(x)F_Y(c/u(x))dx +\int_{0}^{t}f_X(x)dx.$$ Thus, the factor $$ \int_{c/u(x)}^{\infty}f_Y(y)dy $$ in the second part of the RHS is not useful. Note that, if $x\lt t$, $u(x)\lt0$ hence $c/u(x)\lt0$ and, since $f_Y(y)=0$ for every $y\lt0$, $$ \int_{c/u(x)}^{\infty}f_Y(y)dy=\int_0^{\infty}f_Y(y)dy=1. $$ And it happens that this factor is rightfully replaced by $1$ in the second line of (75) in the paper you linked to.

Note finally that $$P(Y<c/u(X))=P(Yu(X)\lt c,X\gt t), $$ while $$P(Yu(X)\lt c)=P(Yu(X)\lt c,X\gt t)+P(X\lt t),$$ hence replacing the latter by the former was misleading.

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  • $\begingroup$ Thank you very much. You are very brilliant! Very helpful! $\endgroup$ – kou Jul 16 '14 at 18:57
  • $\begingroup$ I agree your answer, and I understand all of the above except the last two equations in your answer. The part where I still don't get is why this term is included: "$$\int_{0}^{t}f_X(x)dx$$" $\endgroup$ – kou Jul 16 '14 at 20:26
  • $\begingroup$ Because, if $X\lt t$ then $u(X)\lt0$ hence $Yu(X)\lt c$ for every positive $c$ (remember that $Y\geqslant0$, always). $\endgroup$ – Did Jul 16 '14 at 20:28

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