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Consider two random variables $X$ and $Y$. I would like to derive the PDF and CDF of $$Z=\frac{X}{\max(X,Y)}.$$ The direct method would be to derive distributions for the numerator and denominator and then the ratio. But I am trying to do this slightly differently.

The way I see it there are two cases: $Z_1=X/X$ and $Z_2 = X/Y$. For $Z_1$ the CDF is $$ F_{Z_1}(z) = P(X/X \le z) = \begin{cases} 0 ,& \text{if } z < 1\\ 1, & z \ge 1. \end{cases} $$ For $Z_2 = X/Y$ the CDF is less straightforward but assume the PDF $f_{Z_2}(z)$ is known. Then $$ F_{Z_2}(z) = \int_{-\infty}^{z}f_{Z_2}(x)dx. $$ How can I stitch these pieces together into a CDF and PDF for the original $Z$?

My guess: $$ F_Z(z) = \begin{cases} F_{Z_2}(z) ,& \text{if } z < 1\\ 1, & z \ge 1 \end{cases} $$ There is likely a discontinuity at $z=1$ but we differentiate to obtain $$ f_Z(z) = \begin{cases} f_{Z_2}(z) ,& \text{if } z < 1\\ \delta(z-1), & z \ge 1 \end{cases} $$

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  • $\begingroup$ Are $X$ and $Y$ independent? $\endgroup$ May 16, 2020 at 2:12

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I'll assume that you know how to calculate $P(X\leq zY)$ for any $z$...

Following your logic, $$ Z=\begin{cases} 1, & X>Y\\ \frac{X}{Y}, & X\leq Y. \end{cases} $$ Then using conditional probability, for any $z\in\mathbb{R}$, $$ P(Z\leq z)=P(Z\leq z|X>Y)P(X>Y)+P(Z\leq z|X\leq Y)P(X\leq Y)\quad (1) $$ For the first part in the right-hand side of (1) we have that $$ P(Z\leq z|X>Y) = P(1\leq z|X>Y) = P(1\leq z) = \mathbb{1}_{[1,\infty)}(z). $$ For the second part of (1) we have that \begin{align*} P(Z\leq z|X\leq Y) &= P(X\leq zY|X\leq Y)\\ &=\begin{cases} \frac{P(X\leq zY)}{P(X\leq Y)}, & z\leq 1\\ 1, & z>1. \end{cases} \end{align*}

Then you just add all the pieces together.

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