If $X_{1}, X_{2}$ is a r.s. from $N(0,\sigma^2)$ (Scale Family) show that $\frac{X_1}{X_2}\sim \operatorname{Cauchy}(0,1)$

Question

If $X_{1}, X_{2}$ is a random sample from $N(0,\sigma^2)$ (Scale Family) show that $\frac{X_1}{X_2}\sim \operatorname{Cauchy}(0,1)$.

Here, I have been trying to use the CDF $F_{\frac{X_1}{X_2}}\left( y_{1} \right)=P\left( \frac{X_1}{X_2}\leq y_{1} \right)$. But I am confused in how to proceed.

Note that $X_1/X_2 \le y$ is the same as $$ \Big[ X_1\le yX_2\ &\ X_2>0 \Big] \text{ or } \Big[ X_1\ge yX_2\ &\ X_2<0\Big] $$ (this neglects the case where $X_2=0,$ but that has zero probability). That tells you what part of the plane to integrate over. — Michael Hardy, Feb 18 '21 at 18:37

score 0 · Answer 1 · answered Feb 18 '21 at 18:57

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This is a ratio distribution, where without loss of generality $\sigma=1$ so$$f_{X_1/X_2}(z):=\int_{\Bbb R}|x_2|\tfrac{1}{2\pi}e^{-(1+z^2)x_2^2/2}dx_2=\tfrac{1}{\pi}\int_0^\infty x_2e^{-(1+z^2)x_2^2/2}dx_2=\tfrac{1}{\pi(1+z^2)}.$$

answered Feb 18 '21 at 18:57

J.G.

115,835

Thank you, however, I still don't understand why you assume $\sigma$=1. Can you explain? – Alexis Sandoval Feb 18 '21 at 19:12
@AlexisSandoval Because $\tfrac{X_1}{X_2}=\tfrac{X_1/\sigma}{X_2/\sigma}$ is a ratio of independent $N(0,,1)$ variables. – J.G. Feb 18 '21 at 19:13

If $X_{1}, X_{2}$ is a r.s. from $N(0,\sigma^2)$ (Scale Family) show that $\frac{X_1}{X_2}\sim \operatorname{Cauchy}(0,1)$

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