Derivation of a function missing one step in Nandi & Aich, 1994

39 Views Asked by At

In Nandi and Aich, (1994) Sankhya, vol. 56, p. 129-136 (accessible here), there is a density function derived. However, there is one step where I cannot figure out how it was obtained. On bottom of p. 131, the authors go from (reported as in the paper):

$$ g(t') = \frac{1}{\sqrt{2 \pi}} \sum_{i=0}^\infty \frac{ q_i ((1+\rho)\sigma^2)^{-(m+i-1)} }{ 2^{m+i-1} \Gamma(m+i-1)} \cdot \int_0^\infty \exp\left(-\frac{1}{2}\left((t'\sqrt{z} - \delta)^2+ \frac{z}{(1+\rho)\sigma^2} \right) \right) z^{m+i-3/2} \; \mathrm{d}\,z $$

(which is checked to be ok) to this expression

$$ = \exp\left(-\frac{\delta^2}{2}\right) \sqrt{\frac{(1+\rho)\sigma^2}{\pi}} \sum_{i=0}^\infty \sum_{j=0}^\infty \frac{q_i}{\Gamma(m+i-1)} \; \frac{(\sqrt{2} \delta t')^j}{j!} \cdot \frac{ ((1+\rho)\sigma^2)^{j/2} \Gamma(m+i+(j-1)/2) }{ (1+(1+\rho)\sigma^2 t'^2)^{m+i+(j-1)/2} } $$

for $-\infty<t'<\infty$. The parameters $\delta$, $\sigma$, $\rho$ and $z$ are all real numbers, with $\sigma > 0$, $z>0$, $-1\le \rho \le 1$. $m$ is an integer greater than 2.

Can you help filling the gap? Also, there might be an error in the second equation because when I evaluate both functions with the same values for the various variables ($m$, $z$, $\delta$, $\rho$, $\sigma$, $t'$), I get two different results...

Note. In case this is relevant, here is the definition of $q_i$ (and $\sum_{i=0}^\infty q_i = 1$):

$$\dbinom{-(m-1)/2}{i} \left(\frac{2\rho}{1-\rho}\right)^i \left(\frac{1+\rho}{1-\rho}\right)^{(m-1)/2} $$

1

There are 1 best solutions below

1
On BEST ANSWER

This must be an application of $$\int_0^\infty z^{a-1}e^{b\sqrt{z}-cz}\,dz=\sum_{j=0}^\infty\frac{b^j}{j!}\frac{\Gamma(a+j/2)}{c^{a+j/2}}\qquad(a,c>0)$$ (shown using the power series of $e^{b\sqrt{z}}$ and termwise integration).