How to find a distribution of the following characteristic function \begin{align} g(t)=\frac{1+t^2}{1+a^2t^2}, \end{align} for $a>1$.
Since, \begin{align} \lim_{t \to \infty}g(t)=\frac{1}{a^2}. \end{align} This means that the distribution is neither continuos nor discrete.
I was also able to show that \begin{align} F(0)-F(0^{-})=\frac{1}{a^2} \end{align} which means that there is a mass at zero.
However, I am not able to show anything more about distribution $F$.
Thanks ;)
$$\frac{1+t^2}{1+a^2 t^2} = \frac{1}{a^2} 1 + (1-\frac{1}{a^2}) \frac{1}{1+a^2 t^2}.$$
That is, we are looking at the convex combination of ${\bf 1}$ and \frac{1}{1+a^2 t^2}$.
Now $1$ is the characteristic function of the random variable equal to $0$ with probability $1$.
Let $X\sim \mbox{Exp}(1/a)$ and let $Y$ be independent of $X$ equal to $1$ or $-1$ with probability $\frac12$ each. The characteristic function of $Z=XY$ is (I'm skipping the calculation):
$$\frac{1}{1+a^2 t^2}.$$
This is known as centered Laplace distribution. Simply ``symmetrized" exponential. Denote this distribution by $\mbox{Laplace}(a)$. Note that it is equal in distribution to $a\times \mbox{Laplace}(1)$.
Back to our original equation.
Let $W\sim \mbox{Bern}(1-\frac{1}{a^2})$, independent of $Z$, it follows that the answer to our question is
$$(1-W) \times 0 + W Z=WZ \sim a\times \mbox{Bern}(1-\frac{1}{a^2})\times \mbox{Laplace}(1).$$