MGF relevant to modified 2nd kind Bessel

Question

Given the moment-generating function $$ m_{0}(t)=\frac{1}{\sqrt{1-t^2}}\,\text{ for }t<1, $$ which corresponds to a distribution with density $$ f(u) = \frac{1}{\pi}K_{0}(\frac{u}{\pi }) $$ where $K_0$ is the modified Bessel function of the second kind of order zero, what is the moment-generating function $m(t)$ for the distribution with density $$ f(u)=\frac{1}{\pi}K_{0}\left(\frac{|u|}{\pi }\right)\;? $$ While I have derived bounds as $m_{0}(t)\leq m(t) \leq 2m_{0}(t)$, I would appreciate an exact expression or a more rigorous evaluation.

Answer 1

For $$f(u)=\frac{1}{\pi}K_{0}\left(\frac{|u|}{\pi }\right),$$ we have $$M(t):=\int_{-\infty}^\infty e^{tu}f(u)\,du=\frac\pi{\sqrt{1-\pi^2 t^2}} \tag{1}\label{1}$$ for $t\in(-1/\pi,1/\pi)$. In particular, $M(0)=\pi\ne1$, so that your $f$ is not a probability density.

To get \eqref{1}, first note that $$K_0(x)=\int_0^\infty dz\,e^{-x\cosh z}$$ for real $x>0$. So, for $t\in(-1/\pi,1/\pi)$,
$$N(t):=\frac1\pi\int_0^\infty du\,e^{tu}K_0\Big(\frac u\pi\Big) \\ =\frac1\pi\int_0^\infty dz\,\int_0^\infty du\,\exp\Big(\Big(t-\frac{\cosh z}\pi\Big)u\Big) =\int_0^\infty \frac{dz}{\cosh z-\pi t}.$$ So, for $t\in(-1/\pi,1/\pi)$, using the substitution $e^z=v$, we have $$M(t)=N(t)+N(-t)=\int_0^\infty dz\,\Big(\frac1{\cosh z-\pi t}+\frac1{\cosh z+\pi t}\Big)=\frac\pi{\sqrt{1-\pi^2 t^2}}.\quad\Box$$

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Following the lines of the reasoning above, it is easy to see that $M(t)=\infty$ for real $t$ with $|t|\ge1/\pi$.