# Cauchy CDF derivation from standard normal?

I'm studying Probability, from the book "Introduction to probability" by Joseph K. Blitzstein and Jessica Hwang page 294 talks about Cauchy CDF, it says:

Let $X$ and $Y \sim N(0,1)$ (Standard Normal) and let $T = \frac{X}{Y}$. The distribution of $T$ is called Cauchy Distribution.

The CDF of $T$ is:

$$F_T(t) = P(T \le t) = P\Big(\frac{X}{Y} \le t\Big) = P\Big(\frac{X}{|Y|} \le t\Big)$$

since the r.v.s $\frac{X}{Y}$ and $\frac{X}{|Y|}$ are identically distributed by the symmetry of the standard Normal distribution.

I have few questions:

1. Random variable has some similarity with function which we can manipulate it for example if I have function in general(like a high school math function) ex. $f(x) = x+2$ and another $g(x) = x+2$ then if I do $\frac{f(x)}{g(x)} =1$ I get $1$. lets talk about the r.v.s $X$ and $Y$, if it is identical shouldn't I get $\frac{X}{Y} = 1$?

2. Why $P(\frac{X}{Y} \le t) = P(\frac{X}{|Y|} \le t)$ ? I know symmetry of standard normal talks about if $Z$ has standard normal distribution then $-Z$ and $Z$ has the same distribution but in this case it is the absolute value of $Y$, what does that mean? When we talk about manipulating r.v.s ex. $Y = X-1$ it is to minus all the support of $X$ by $1$ to get distribution of $Y$ but what about the r.v.s divide by another r.v.s? Do we think of it as a function that already crystallised into a number? Or how we think about it? I really have no clue about what this is about.

Please give me detail and step by step answer. I'm a newbie.

1. You are right, a real random variable is a function from a set $\Omega$ to $\mathbb{R}$. Therefore, if $X(\omega) = Y(\omega) \neq 0$, then $X/Y(\omega) = 1$. Here, something else is assumed: it is assumed that $X$ and $Y$ are identically distributed with normal distribution $\mathcal{N}(0,1)$, but we don't have $X=Y$! This may be interpreted in terms of probabilities: $$\int_A f_X(x)\, dx = P(X(\omega)\in A) = P(Y(\omega)\in A) = \int_A f_Y(x)\, dx$$ for any given subset $A$ of $\mathbb{R}$, which does not imply that $P(X/Y(\omega)\in A) = P(1\in A)$. In other words, if two dices have the same probability distribution, there is no reason that the ratio of results is always equal to one. The ratio of results is a new random variable with its own probability distribution. In the case of two dices, we have $$\frac{X(\omega)}{Y(\omega)} \in \left\lbrace 1, \frac{1}{2},\frac{1}{3},\dots, \frac{6}{5}\right\rbrace ,$$ where $X(\omega) \in \left\lbrace 1, 2,\dots, 6\right\rbrace$ and $Y(\omega) \in \left\lbrace 1, 2,\dots, 6\right\rbrace$ denote the result of each dice throw.
2. The probability $P( X/Y\leqslant t)$ may be viewed as \begin{aligned} &P(\lbrace X/Y\leqslant t, Y>0 \rbrace\cup\lbrace X/Y\leqslant t, Y<0 \rbrace) \\ =\; &P(X/|Y|\leqslant t, Y>0) + P(-X/|Y|\leqslant t, Y<0)\\ =\; &P(X/|Y|\leqslant t, Y>0) + P(X/|Y|\leqslant t, Y<0)\\ =\; &P(X/|Y|\leqslant t) \end{aligned} by symmetry of the standard normal distribution, independence and the fact that $P(Y=0)=0$.
• @John (1) $X/|Y|$ is not assumed independent of $Y$, but the event $X/|Y|\leq t$ is independent from the event $Y>0$. (2) $-X/|Y|$ has the same distribution as $X/|Y|$ since $-X$ has the same distribution as $X$, and $X$, $Y$ are independent. Commented Sep 17, 2017 at 20:21