If $X$ and $Y$ are independent standard normal random variables, and $U=X+Y$, $V=\frac{X}{Y}$ then $f_V(v) = \frac{1}{\pi (1+v^2)}$

Question

If $X$ and $Y$ are independent standard normal random variables, and $U=X+Y$, $V=\frac{X}{Y}$ prove that $$f_V(v) = \frac{1}{\pi (1+v^2)}$$

Solution:

I know that $f_{xy}(x,y)=\frac{e^{-\frac{x^2+y^2}{2}}}{2\pi}$

Also, $$u=x+y \ \ \ \ \ \ AND \ \ \ \ \ \ v = \frac xy$$

Then $\frac{uv}{1+v}=x$ and $\frac{u}{v+1}=y$

Then the Jacobian would be $J=\frac{-u}{(v+1)^2}$

So:

$$f_{u,v}=f_{xy}(h^{-1}(u,v),g^{-1}(u,v))*|J|=$$ $$\frac{e^{-\frac{u^2(1+v^2)}{2(1+v)^2}}}{2\pi}*\frac{u}{(v+1)^2}$$

My question now is, how can I prove that: $$f_V(v)=\int^{\infty}_{-\infty}\frac{e^{-\frac{u^2(1+v^2)}{2(1+v)^2}}}{2\pi}*\frac{u}{(v+1)^2}du=\frac{1}{\pi(1+v^2)}$$

Answer 1

Please note that the support of $u$ is $(- \infty, \infty)$.

So, $|J| = -\dfrac{u}{(v+1)^2}$ when $u \in (-\infty,0]$

Your integral should be,

$ \displaystyle f_V(v)=\int^{\infty}_{-\infty}\frac{e^{-\frac{u^2(1+v^2)}{2(1+v)^2}}}{2\pi} \cdot \frac{|u|}{(v+1)^2}du$

Substituting $u^2 = w, u du = \frac 12 dw$ and using symmetry,

$ \displaystyle f_V(v) = \frac{1}{2\pi(1+v)^2} \int^{\infty}_{0} e^{-\frac{w(1+v^2)}{2(1+v)^2}}dw = \frac{1}{\pi(1+v^2)}$