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Show that a mixed Poisson with an inverse Gaussian mixing distribution is the same as a Poisson-ETNB distribution with $r=-0.5$.

$f(x)=\left(\frac{\theta}{2 \pi x^3}\right)^{1 / 2} \exp \left[-\frac{\theta}{2 x}\left(\frac{x-\mu}{\mu}\right)^2\right]$

$$\int_{0}^{\infty}\left(\frac{\theta}{2 \pi x^3}\right)^{1 / 2} \exp \left[-\frac{\theta}{2 x}\left(\frac{x-\mu}{\mu}\right)^2\right] \, dx =1$$

$$\int_{0}^{\infty} \exp(xt)\left(\frac{\theta}{2 \pi x^3}\right)^{1 / 2} \exp \left[-\frac{\theta}{2 x}\left(\frac{x-\mu}{\mu}\right)^2\right]\, dx =\int_{0}^{\infty} \left(\frac{\theta}{2 \pi x^3}\right)^{1 / 2} \exp \left[-\frac{\theta}{2 x}\left(\frac{x-\mu}{\mu}\right)^2+xt\right]\, dx \\= \int_{0}^{\infty} \left(\frac{\theta}{2 \pi x^3}\right)^{1 / 2} \exp \left[-\frac{\theta}{2 x}\left(\frac{x-\mu}{\mu}\right)^2+\frac{2x^2 \mu^2 t}{2x\mu^2}\right]\, dx$$

Notice that $$-\frac{\theta}{2 x}\left(\frac{x-\mu}{\mu}\right)^2+\frac{2x^2 \mu t}{2x\mu^2} = \frac{-\theta x^2+2\mu\theta x-\mu^2\theta+2x^2\mu^2 t}{2x \mu^2}$$

Notice further that $$-\theta x^2+2\mu\theta x-\mu^2\theta+2x^2\mu^2 t = (-\theta+2\mu^2 t)x^2 +2\mu\theta x-\mu^2 \theta \\= (-\theta +2\mu^2 t)\left(x^2 + \frac{\mu^2 \theta}{ \theta-2\mu^2 t}\right)+2\mu\theta x \\ = (-\theta +2\mu^2 t)\left(x^2 -2 \left(\frac{ \theta}{ \theta-2\mu^2 t}\right)^{1/2} \mu x+ \frac{\mu^2 \theta}{ \theta-2\mu^2 t}\right)+2\mu\theta x - 2 \left(\theta ( \theta-2\mu^2 t)\right)^{1/2} \mu x$$

Let $\Lambda := \left(\frac{ \theta}{ \theta-2\mu^2 t}\right)^{1/2} \mu$ and $C:=2\mu\theta - 2 \left(\theta ( \theta-2\mu^2 t)\right)^{1/2} \mu $

And so we will have

$$\frac{(-\theta+2\mu^2 t)\left(x - \Lambda \right)^2+Cx}{2 x \mu^2} = \frac{(-\theta+2\mu^2 t)\left(x - \Lambda \right)^2}{2 x \frac{\mu^2 \theta}{\theta - 2\mu^2 t}}\frac{\theta}{(\theta-2\mu^2 t)} +\frac{C}{2\mu^2} \\= -\frac{\theta\left(x - \Lambda \right)^2}{2 x \Lambda^2} +\frac{C}{2\mu^2}$$

Since we know that $$\int_{0}^{\infty}\left(\frac{\theta}{2 \pi x^3}\right)^{1 / 2} \exp \left[-\frac{\theta}{2 x}\left(\frac{x-\mu}{\mu}\right)^2\right] \, dx =1$$

And so we will have

$$M_X(t) = \exp\left(\frac{\mu \theta - (\theta (\theta -2\mu^2 t))^{1/2}\mu}{\mu^2}\right)=\exp \left(\frac{\theta}{\mu}\left(1 - \frac{1}{\theta} (\theta^2 -2\theta\mu^2 t)^{1/2}\right)\right) =\exp \left(\frac{\theta}{\mu} \left(1-\sqrt{1-2\frac{\mu^2}{\theta}t}\right)\right).$$

And now if we take $1/n$-th power to $M_X(t)$, then we will have

$$M_X(t)^{1/n} =\exp \left(\frac{\theta/n^2}{\mu/n} \left(1-\sqrt{1-2\frac{\mu^2}{\theta}t}\right)\right) \\ = \exp \left(\frac{\theta/n^2}{\mu/n} \left(1-\sqrt{1-2\frac{\mu^2 /n^2}{\theta/n^2}t}\right)\right) \\=\exp \left(\frac{\theta/n^2}{\mu/n} \left(1-\sqrt{1-2\frac{(\mu/n)^2 }{\theta/n^2}t}\right)\right) $$

And so it represents a inverse Gaussian distribution with $\theta/n^2$ and $\mu/n$ as parameters.

Now our mixed distribution has the following pgf:

$$P(z) = M_X(z-1) = \exp \left(\frac{\theta}{\mu} \left(1-\sqrt{1-2\frac{\mu^2}{\theta}(z-1)}\right)\right)$$

And since our inverse Gaussian distribution is infinitely divisible, we will have

$$\lambda = -\ln P(0) = \frac{\theta}{\mu} \left(\sqrt{1+2\frac{\mu^2}{\theta}}-1\right)$$

And so

$$P_2(z) = 1 + \frac{\ln P(z)}{\lambda} \\= \frac{\frac{\theta}{\mu}\left(\sqrt{1+2\frac{\mu^2}{\theta}}-1\right)+\frac{\theta}{\mu}\left(1-\sqrt{1-2\frac{\mu^2}{\theta}(z-1)}\right)}{\frac{\theta}{\mu}\left(\sqrt{1+2\frac{\mu^2}{\theta}}-1\right)} \\= \frac{\sqrt{1-2\frac{\mu^2}{\theta}(z-1)} - \sqrt{1+2\frac{\mu^2}{\theta}}}{1-\sqrt{1+2\frac{\mu^2}{\theta}}}$$

which is a probability generating function with $\beta = \frac{2\mu^2}{\theta}$ and $r= -1/2.$