Quantile function, bivariate joint density.

Question

Consider random variables $X,Y$ with joint density function $$ f(x, y)=\frac{1}{2 \pi}\left(1+x^2+y^2\right)^{-3 / 2} $$ I want to find the quantile function of $|Y|$. I have learned how to find the quantile function when considering a single random variable, but I am in doubt how to do it for a bivariate function. Since I am considering $|Y|$, I believe that I should first find the marginal density function, $$ f_Y(y)=\int_{-\infty}^{+\infty} f(x, y) d x $$ and then calculate $$ Q_{|Y|}(p)=\inf \left\{y \mid F_Y(|y|)=p\right\} $$ where $F_Y=\int_{-\infty}^y f_Y(t) d t$ Is this the correct procedure?

Answer 1

First you find the marginal density: $f_Y(y)=\int_{\mathbb{R}}f_{X,Y}(x,y)dx=(\pi (y^2+1))^{-1}$; for reference this the density of a standard Cauchy distribution. Then you find the cumulative distribution function of $|Y|$ (note the rv is s.t. $P(Y=y)=0,\,\forall y$): we obtain $F_{|Y|}(y)=F_Y(y)-F_Y(-y)=$$\frac{1}{\pi}(\arctan(y)-\arctan(-y)),\,y \geq 0$. The quantile function is defined through $F_{|Y|}$: by definition $Q_{|Y|}(u)=\inf\{y\geq 0:F_{|Y|}(y)\geq u\}$. We can invert the cumulative distribution directly with this property: $\arctan(-y)=-\arctan(y)$. So $$\frac{2}{\pi}\arctan(y)=u\implies Q_{|Y|}(u)=\tan(u\pi/2)$$