Which distribution has the moment-generating function $\frac{\pi t}{\sin \pi t}$

Question

The distribution $F(x) = e^{-e^{-x}}$ has moment-generating function $M_F(t) = \Gamma(1-t)$. From this it follows that the distribution of $X-Y$ for independently $F$-distributed $X,Y$ has the moment-generating function $$ M_{X-Y}(t) = \frac{\pi t}{\sin \pi t} \text{,} $$ and the characteristic function $$ \phi_{X-Y}(t) = M_{X-Y}(it) = \frac{\pi t}{\sinh \pi t} \text{.} $$

I'd now like to find the distribution function of $X-Y$, but so far without success. I've tried reverse transforming $\phi_{X-Y}$, but my integration skills are a bit rusty. Does anyone see an easier way? ...

Answer 1

To compute the distribution of $Z=X-Y$, one can start from the PDFs $f_X$ and $f_Y$ of $X$ and $Y$. Since $X$ and $Y$ are independent and $Z=X-Y$ means $X=Y+Z$, the PDF $f_Z$ of $Z$ is such that, for every $z$, $$ f_Z(z)=\int_\mathbb R f_X(y+z)f_Y(y)\mathrm dy. $$ Here, $f_X=f_Y=f$ is the derivative of $F:x\mapsto\exp(-\mathrm e^{-x})$ hence $f(x)=\mathrm e^{-x}\exp(-\mathrm e^{-x})$ and $$ f_Z(z)=\int_\mathbb R \mathrm e^{-y-z}\exp(-\mathrm e^{-y-z})\mathrm e^{-y}\exp(-\mathrm e^{-y})\mathrm dy=\int_{t\geqslant0} \mathrm e^{-z}t\exp(-(1+\mathrm e^{-z})t)\mathrm dt, $$ thanks to the change of variable $t=\exp(-y)$. Now, using the change of variable $s=t(1+\mathrm e^{-z})$, $$ f_Z(z)=\frac{\mathrm e^{-z}}{(1+\mathrm e^{-z})^2}\int_{s\geqslant0} s\mathrm e^{-s}\mathrm ds, $$ hence, for every real number $z$, $$ f_Z(z)=\frac{\mathrm e^{-z}}{(1+\mathrm e^{-z})^2}=\frac{\mathrm e^{z}}{(1+\mathrm e^{z})^2}=\frac1{4\cosh^2(z/2)}. $$

Answer 2

One can invert the characteristic function ... using an inverse fourier transform ... to obtain the pdf. Here is the code to do this in Mathematica:

f = InverseFourierTransform[(Pi*t)/Sinh[Pi*t], t, x, FourierParameters -> {1,1}] 

which produces the output:

(1/4)*(Sech[x/2])^2

which is the pdf, with domain of support on the real line.