Show that $V=\frac{Z_1}{\sqrt{(Z^2_1 + Z^2_2)/2}}$ has pdf $f(v) = 1 / (\pi \sqrt{2-v^2}),-\sqrt2<v<\sqrt2$

Question

Let $Z_1, Z_2$ have independent standard normal distributions, $N(0,1)$.

If the random variable in the numerator did not also appear in the denominator this would be a t distribution. Should start with:

$$F(v) = P(\frac{Z_1}{\sqrt{(Z^2_1 + Z^2_2)/2}} \leqq v)$$

or

$$F(v) = P(\frac{Z_1}{\sqrt{X^2(2)/2}} \leqq v)$$

or is there a better way?

Answer 1

$$E(g(V)) = \frac{1}{2\pi}\int_{\mathbb{R^2}}g(\frac{x}{\sqrt{(x^2 + y^2)/2}})\exp(-\frac{x^2 + y^2}{2})dxdy$$

by $x = r\sin \theta$ and $y = r\cos\theta$, this becomes

$$\frac{1}{2\pi} \int_0^{2\pi}\int_0^\infty g(\sqrt{2}\sin\theta) \exp(-\frac{r^2}{2})r d\theta dr = \frac{1}{2\pi} \int_0^{2\pi} g(\sqrt{2}\sin\theta)d\theta = \frac{1}{\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} g(\sqrt{2}\sin\theta)d\theta$$

by $v = \sqrt{2}\sin\theta$, it becomes

$$\frac{1}{\pi} \int_{-\sqrt{2}}^{\sqrt{2}}g(v)\dfrac{1}{\sqrt{2-v^2}}dv$$

So we get the density of $V$ as desired

Answer 2

This distribution belongs to the family of distributions for sample correlation coefficients. With the change of variables, $$ \begin{eqnarray*} X&{}={}&Z_1\\ &&\\ Y&{}={}&\sqrt{Z_1^2{}+{}Z_2^2}\,\,, \end{eqnarray*} $$

one finds

$$ F_R\left(\nu\right){}={}P\left(\dfrac{Z_1}{\sqrt{Z_1^2{}+{}Z_2^2}} < \dfrac{\nu}{\sqrt{2}}\right){}={}P\left(\dfrac{X}{Y} < \dfrac{\nu}{\sqrt{2}}\right)\\ \,\\{}={}\dfrac{1}{\pi}\int\limits^{\infty}_{0}\left(\,\,\,\int\limits^{y\nu /\sqrt{2}}_{-\infty}\dfrac{{\textbf{1}}_{ \left\{|x| < |y|\right\}}}{\sqrt{y^2-x^2}}\,\text{d}x\right)e^{-\frac{1}{2} y^2}y\,\text{d}y\,\,.\\ $$

Therefore, by appealing to Leibniz's rule (using the constant parameter $\nu$), the pdf of the ratio is given as

$$ \begin{eqnarray*} f_{R}(\nu){}={}\dfrac{{\text{d}}F_R}{{\text{d}}\nu}&{}={}&\dfrac{{\text{d}}}{{\text{d}}\nu}\left(\dfrac{1}{\pi}\int\limits^{\infty}_{0}\left(\,\,\,\int\limits^{y\nu /\sqrt{2}}_{-\infty}\dfrac{{\textbf{1}}_{ \left\{|x| < |y|\right\}}}{\sqrt{y^2-x^2}}\,\text{d}x\right)e^{-\frac{1}{2} y^2}y\,\text{d}y\right)\newline &{}={}&\dfrac{1}{\pi}\int\limits^{\infty}_{0}\left(\,\,\,\dfrac{{\text{d}}}{{\text{d}}\nu}\int\limits^{y\nu /\sqrt{2}}_{-\infty}\dfrac{{\textbf{1}}_{ \left\{|x| < |y|\right\}}}{\sqrt{y^2-x^2}}\,\text{d}x\right)e^{-\frac{1}{2} y^2}y\,\text{d}y\newline &{}={}&\dfrac{1}{\pi}\int\limits^{\infty}_{0}\left(\,\,\,\dfrac{y}{\sqrt{2}}\dfrac{{\textbf{1}}_{ \left\{|y\nu /\sqrt{2}| < |y|\right\}}}{\sqrt{y^2-({y\nu /\sqrt{2}})^2}}\,\text{d}x\right)e^{-\frac{1}{2} y^2}y\,\text{d}y\newline &{}={}&{\textbf{1}}_{ \left\{|\nu| < \sqrt{2}\right\}}\dfrac{1}{\pi}\int\limits^{\infty}_{0}\dfrac{e^{-\frac{1}{2} y^2}y}{\sqrt{2-\nu^2}}\,\text{d}y{}={}\dfrac{1}{\pi}\dfrac{{\textbf{1}}_{ \left\{|\nu| < \sqrt{2}\right\}}}{\sqrt{2-\nu^2}}\,. \end{eqnarray*} $$