(a) Show that $F_Y(y)=\tan^{-1}(y)$
I tried to use the usual method but quickly realized that it doesn't work. I have been working other angles for hours and now I'm just stuck.
$F_Y=P(Y\leq y)$ $=\int\limits_{-\infty}^{y}f_Y(t)dt$ $=\int\limits_{-\infty}^{y}\frac{1}{1+t^2}dt$ $=[\tan^{-1}(t)]_{-\infty}^{y}$ $=\tan^{-1}(y)+\frac{\pi}{2}$
Obviously, this solution doesn't work. I was thinking that there must be a transformation, $Y=g(X)$ on a standard Cauchy r.v., $X \sim$ Cauchy(0,1) with pdf $\;f_X(x)=\frac{1}{\pi(1+x^2)}$, so that I can get the pdf of $Y$ as stated in the problem.
EDIT 1: I figured if I could get limits on $y$ such that $\int f_Y=1$ then I'd have something. So I got
$1=\int\limits_{-\infty}^{\alpha}\frac{1}{1+t^2}dt$ $=[\tan^{-1}(t)]_{-\infty}^{\alpha}$ $=\tan^{-1}(\alpha)+\frac{\pi}{2}$
$\Rightarrow \alpha=\tan(1-\frac{\pi}{2})$
So know I think I can say that $f_Y(y)=\frac{1}{1+y^2},\;-\infty < y \leq \tan(1-\frac{\pi}{2})$
But now what...
EDIT 2: By suggestions, I made the transformation $Y=g(x)=\frac{X}{\pi}$.
$\Rightarrow g^{-1}(y)=y\pi\;$ and $\;|J|=\pi$ so that
$f_Y=f_X(g^{-1}(y))|J|=\frac{1}{1+(\frac{y}{1/\pi})^2}$
But I'm still not sure how to get to $\;F_Y(y)=\tan^{-1}(y)$
EDIT 3: I'm pretty sure that I am somehow supposed to use the theorem stating that if $X$ is a continuous r.v. with cdf $F_X$, then a r.v. $Y=F_X$ is uniformly distributed on $(0,1)$, ie. $Y\sim Uniform(0,1)$. (probability integral transformation)
So if $X \sim$ Cauchy$(0,1)$ and $Y=F_X=\frac{1}{\pi}\tan^{-1}(x)+\frac{1}{2}$, then $g^{-1}(y)=\tan(\pi(y-\frac{1}{2}))$. This inverse function is so similar to the value of the upper limit that makes $\int_{-\infty}^{\tan(\pi(1-\frac{1}{2})}\frac{1}{1+y^2}=1$, that I feel the answer must be close.
Seems the density function should be normalized by $\pi$.