# [Probability]need help to understand the following expression

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:

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.

• The RHS is most bizarre. You might want to indicate the source. – Did Jul 9 '14 at 20:54
• @Did, do you think this is correct, because I think the second term is not right. – kou Jul 9 '14 at 23:25
• Quote: "You might want to indicate the source." – Did Jul 9 '14 at 23:39
• @Did, the source comes from an paper:math.stackexchange.com/questions/859663/… – kou Jul 10 '14 at 0:04
• 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. – kou Jul 10 '14 at 0:06

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.
• 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: "$$\int_{0}^{t}f_X(x)dx$$" – kou Jul 16 '14 at 20:26
• Because, if $X\lt t$ then $u(X)\lt0$ hence $Yu(X)\lt c$ for every positive $c$ (remember that $Y\geqslant0$, always). – Did Jul 16 '14 at 20:28