Where to find the characteristic function of the generalized arcsine distribution?

Question

Well, the title says it all. I need the characteristic function of the (generalized) arcsine distribution. I desperately searched the internet for it but haven't found anything. Is there some standard reference?

Thanks!

Answer 1

I recently had to derive this, myself. I followed this progression: derive the characteristic function for the Arcsine distribution, then employ change of variables to convert the generalized Arcsine distribution into the regular one, keeping track of parameters to identify the characteristic function.

$\varphi_{X}(t)$ for Arcsine

First consider the definition of the characteristic function as applied to the random variable $X \sim \mathrm{Arcsine}$: \begin{align} \varphi_{X}(t) &= \mathbb{E}\left[ e^{itX} \right] = \int_{-\infty}^{\infty} e^{i t x} \, f_{X}(x) \, \mathrm{d}x , \end{align} which according to Wikipedia [1] has p.d.f given by \begin{align} f_{X}(x) &= \dfrac{1}{\pi} \dfrac{1}{\sqrt{x(1-x)}} , \quad 0 < x < 1. \end{align} Therefore, the characteristic function of $X$ is obtained via \begin{align} \varphi_{X}(t) = \dfrac{1}{\pi} \int_{0}^{1} \dfrac{e^{itx}}{ \sqrt{x (1-x)} } \, \mathrm{d}x. \end{align}

To determine the process of evaluating this integral, I had to discover the following change of variable steps: $x \rightarrow u^{2}$, $u \rightarrow \sin \theta$, and $\theta \rightarrow \phi/2$. Combining these into one step, make the substitution $x = \sin^{2} \frac{\theta}{2}$. Employing the change of variable results in \begin{equation} \mathrm{d}x = \left| \dfrac{\partial}{\partial \theta} T(\theta) \right| \, \mathrm{d}\theta \end{equation} with \begin{equation} T(\theta) = \sin^{2}\frac{\theta}{2} . \end{equation} Therefore, the Jacobian of $T(\theta)$ is evaluated to be \begin{align} \left| \dfrac{\partial}{\partial \theta} T(\theta) \right| &= \left| \dfrac{\partial}{\partial \theta} \sin^{2}\frac{\theta}{2} \right| \\ &= \left| 2 \sin\frac{\theta}{2} \cos\frac{\theta}{2} \frac{1}{2} \right| \\ &= \sin\frac{\theta}{2} \cos\frac{\theta}{2} \end{align} The result is that \begin{equation} \mathrm{d}x = \sin\frac{\theta}{2} \cos\frac{\theta}{2} \mathrm{d}\theta. \end{equation} In light of this and by observing that $0 < \theta < \pi$, the new form for the characteristic function is given by \begin{align} \varphi_{X}(t) &= \dfrac{1}{\pi} \int_{0}^{\pi} \dfrac{\sin\frac{\theta}{2} \cos\frac{\theta}{2} e^{it \sin^{2}\frac{\theta}{2}}}{\sqrt{\sin^{2}\frac{\theta}{2} (1 - \sin^{2}\frac{\theta}{2}) } } \, \mathrm{d}\theta \\ &= \dfrac{1}{\pi} \int_{0}^{\pi} \dfrac{\sin\frac{\theta}{2} \cos\frac{\theta}{2} e^{it \sin^{2}\frac{\theta}{2}}}{ \sin\frac{\theta}{2} \cos\frac{\theta}{2} } \, \mathrm{d}\theta \\ &= \dfrac{1}{\pi} \int_{0}^{\pi} e^{it \sin^{2}\frac{\theta}{2}} \, \mathrm{d}\theta. \end{align}

The final step required to evaluate the integral is to employ the half-angle identity: \begin{align} \varphi_{X}(t) &= \dfrac{1}{\pi} \int_{0}^{\pi} e^{it \sin^{2}\frac{\theta}{2}} \, \mathrm{d}\theta \\ &= \dfrac{1}{\pi} \int_{0}^{\pi} e^{it \frac{1}{2} \left(1 - \cos \theta \right) } \, \mathrm{d}\theta \\ &= e^{it/2} \dfrac{1}{\pi} \int_{0}^{\pi} e^{-\frac{it}{2} \cos\theta } \, \mathrm{d}\theta. \end{align} At this point, the above is recognized as an integral representation for the zero$^{\text{th}}$ order Bessel function of the first kind, $J_{0}$ [2]. Thus, one obtains \begin{equation} \varphi_{X}(t) = e^{\frac{it}{2}} J_{0} \left(\frac{t}{2}\right) . \end{equation} This can be written in terms of the confluent hypergeometric function via [3] (but not done here, since I find Bessel functions easier to parse than hypergeometric functions).

$\varphi_{X}(t)$ for Generalized Arcsine

To obtain $\varphi_{X}(t)$ for the Generalized Arcsine distribution, first consider its p.d.f [1]: \begin{equation} f_{X}(x;a,b) = \dfrac{1}{\pi} \dfrac{1}{\sqrt{(x - a)(b - x)}} \, \mathrm{d}x , \end{equation} where the parameters $a$ and $b$ define the domain for $X$ such that $a < x < b$. This can be converted to the p.d.f. for the regular Arcsine distribution by making the change of variables $ x = (b-a)z + a$, resulting in $0 < z < 1$ and \begin{align} f_{Z}(z) &= \dfrac{b-a}{\pi} \dfrac{1}{\sqrt{\left(\big[(b-a)z + a \big] - a\right) \left(b - \big[(b-a)z + a\big] \right) }} \\ &= \dfrac{1}{\pi} \dfrac{1}{\sqrt{(b-a)(z) (b-a)(1-z)}} \\ &= \dfrac{1}{(b-a)\pi} \dfrac{1}{\sqrt{z (1-z)}} , \end{align} which is recognized as the p.d.f of the regular Arcsine distribution as claimed. Therefore, the characteristic function for the Generalized Arcsine distribution may be obtained via \begin{align} \varphi_{X}(t) &= \dfrac{1}{\pi} \int_{a}^{b} \dfrac{e^{itx}}{\sqrt{(x-a)(b-x)}} \, \mathrm{d}x \\ &= \dfrac{1}{\pi} \int_{0}^{1} \dfrac{ e^{it[(b-a)z+a]} }{ \sqrt{z (1 - z)} } \, \mathrm{d}z \\ &= \dfrac{e^{ita}}{\pi} \int_{0}^{1} \dfrac{ e^{it(b-a)z} }{ \sqrt{z (1 - z)} } \, \mathrm{d}z , \end{align} where the $(b-a)$ term in the denominator of the p.d.f after change of variables and coefficient in the differential $\mathrm{d}x = (b-a) \mathrm{d}z$ cancel. At this point, the expression is the same as above except an extra $(b-a)$ multiplied onto the Fourier transform variable $t$. Thus, for $X \sim \mathrm{Arcsine}(a,b)$, \begin{equation} \varphi_{X}(t) = e^{\frac{1}{2} it(b+a)} J_{0}\left( \frac{1}{2} t(b-a) \right). \end{equation}

[1] : https://en.wikipedia.org/wiki/Arcsine_distribution
[2] : https://dlmf.nist.gov/10.9.E2
[3] : https://dlmf.nist.gov/10.16.E5

Answer 2

This is 8.5 years too late, but just incase anyone else is interested, I've got a rather more straightforward derivation of the characteristic function for the generalized arcsine distribution.

My derivation:

Start from a distribution of uniform probability on the unit circle, i.e. $\{(x,y) \in \mathbb{R}^2: x^2 + y^2 = r^2 \}$. It turns out that the arcsine distribution with support on $[-r,r]$ can be obtained by marginalizing out the y dependence, i.e. projecting onto the x-axis.

We can consider each point on the circle to contribute $\delta(x-rcos(\theta))$. Then averaging over all possible values of $\theta$, we've got:

$\frac{1}{2\pi}\int^{2\pi}_0\delta(x-rcos(\theta))d\theta = \frac{1}{\pi\sqrt{r^2-x^2}} = arcsine(-r,r)$

Well, we know the characteristic function of a delta function is just a complex exponential, so we can proceed from the integral form of the arcsine distribution above to get the following (See Equation 71):

$\varphi_{X}(t) = \frac{1}{2\pi}\int^{2\pi}_0e^{itrcos(\theta) }d\theta = J_0(rt)$

Generalizing to $arcsine(a,b)$ is easy from here: simply set $r = (b - a)/2$ and account for the offset by multiplying by $e^{it\frac{b+a}{2}}$. So all told, the characteristic function is:

$\varphi_{X}(t) = e^{it\frac{b+a}{2}}J_0(\frac{b-a}{2}t)$