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Let $Z_1$ and $Z_2$ denotes two dependent random variables defined as \begin{align} Z_1&=\frac{XY}{aX+bY+c}\\ Z_2&=\frac{XY}{uX+vY+w} \end{align} where $X$ and $Y$ are independent exponentially distributed random variables with parameters $2\sigma_x^2$ and $2\sigma_y^2$ respectively. Further more $a$, $b$, $c$, $u$, $v$, and $w$ are finite non-zero positive real constants.

I'm evaluating \begin{align} Pr\{\min(Z_1,Z_2)\leq\eta\} &=\underbrace{\Pr\{Z_1\leq\eta,Z_1\leq Z_2\}}_{P_1}+\underbrace{\Pr\{Z_2\leq\eta,Z_2\leq Z_1\}}_{P_2}\\ \end{align} Both $P_1$ and $P_2$ will follow the same approach of evaluation. So evaluating $P_1$ as \begin{align} P_1 &= \Pr\{Z_1\leq\eta,Z_1\leq Z_2\}\\ &=\Pr\bigg\{\frac{XY}{aX+bY+c}\leq\eta,\frac{XY}{aX+bY+c}\leq\frac{XY}{uX+vY+w}\bigg\}\\ &=\Pr\bigg\{X\leq\frac{bY\eta+c\eta}{Y(1-a\eta)},X\leq\frac{(b-u)Y+(c-w)}{v-a}\bigg\}\\ &=\int_{y=0}^{\infty}\Pr\bigg\{X\leq\frac{by\eta+c\eta}{y(1-a\eta)},X\leq\frac{(b-u)y+(c-w)}{v-a}\bigg\}f_Y(y)dy\tag{1} \end{align}

How can I expand $(1)$ for further evaluation?

This is what I've tried

Defining event $A: X\leq\frac{by\eta+c\eta}{y(1-a\eta)}$ and $B:X\leq\frac{(b-u)y+(c-w)}{v-a}$ and using the identity given in $(2)$ \begin{equation} \Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)\tag{2} \end{equation} Hence form $(1)$ and $(2)$ we get \begin{align} P_1=\int_{y=0}^{\infty}\bigg[&\Pr\bigg\{X\leq\frac{by\eta+c\eta}{y(1-a\eta)}\bigg\}+\Pr\bigg\{X\leq\frac{(b-u)y+(c-w)}{v-a}\bigg\}\\ &-\Pr\bigg\{X\leq\frac{by\eta+c\eta}{y(1-a\eta)}\cap X\leq\frac{(b-u)y+(c-w)}{v-a}\bigg\}\bigg]f_Y(y)dy \end{align} Expanding integral \begin{align} P_1&=\int_{y=0}^{\infty}\Pr\bigg\{X\leq\frac{by\eta+c\eta}{y(1-a\eta)}\bigg\}f_Y(y)dy+\int_{y=0}^{\infty}\Pr\bigg\{X\leq\frac{(b-u)y+(c-w)}{v-a}\bigg\}f_Y(y)dy\\ &-\int_{y=0}^{\infty}\Pr\bigg\{X\leq\frac{by\eta+c\eta}{y(1-a\eta)}\cap X\leq\frac{(b-u)y+(c-w)}{v-a}\bigg\}f_Y(y)dy \end{align} Rewriting in terms of cdf \begin{align} P_1&=\underbrace{\int_{y=0}^{\infty}F_X\bigg(\frac{by\eta+c\eta}{y(1-a\eta)}\bigg)f_Y(y)dy}_{I_1}+\underbrace{\int_{y=0}^{\infty}F_X\bigg(\frac{(b-u)y+(c-w)}{v-a}\bigg)f_Y(y)dy}_{I_2} \\&-\underbrace{\int_{y=0}^{\infty}F_{XX}\bigg\{\frac{by\eta+c\eta}{y(1-a\eta)}\cap \frac{(b-u)y+(c-w)}{v-a}\bigg\}f_Y(y)dy}_{I_3}\tag{3} \end{align}

Evaluating integrals $I_1$, $I_2$ are simple, but not sure about $I_3$.


Further update The integrand in $I_3$ can be written as \begin{align} I_3 &= \int_{y=0}^{\infty}F_{XX}\bigg\{\frac{by\eta+c\eta}{y(1-a\eta)}\cap \frac{(b-u)y+(c-w)}{v-a}\bigg\}f_Y(y)dy\\ &=\int_{y=0}^{\infty}F_{X}\bigg\{\min{\bigg(\frac{by\eta+c\eta}{y(1-a\eta)},\frac{(b-u)y+(c-w)}{v-a}}\bigg)\bigg\}f_Y(y)dy\\ &=\int_{y=0}^{\infty}\int_{x=0}^{\min{\big(\frac{by\eta+c\eta}{y(1-a\eta)},\frac{(b-u)y+(c-w)}{v-a}}\big)}f_X(x)dxf_Y(y)dy \end{align}

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