1
$\begingroup$

I encountered a problem to get the expectation with respect to an inverse gamma distribution. My problem can be simplified as to solve the following integral: $$ \int_{0}^{\infty}x^{-a}\exp\left(-\left[\frac{b}{x} + \frac{c}{\,\sqrt{\, x\,}\,}\right]\right)\,\mathrm{d}x, \quad\mbox{where}\quad a,b,c \in \mathbb{R}\quad\mbox{and}\quad a > 1\,,\ b > 0. $$ It is basically something like $E\left[c_{0}/\sqrt{\, X\,}\right]$ if $X$ follows inverse gamma distribution.

Is there an analytical form or approximation for above the integral? Thanks in advance!

$\endgroup$
0
0
$\begingroup$

Not a complete answer, but for what it's worth: via Mathematica I found that your integral $I$ can be expressed in the terms of the Kummer confluent hypergeometric function, ${}_1F_1$: $$I=b^{-a} \left(b \Gamma (a-1) \, _1F_1\left(a-1;\frac{1}{2};\frac{c^2}{4 b}\right)-\sqrt{b} c \Gamma \left(a-\frac{1}{2}\right) \, _1F_1\left(a-\frac{1}{2};\frac{3}{2};\frac{c^2}{4 b}\right)\right)\,.$$ For certain specific values of $a$ and $b$ (integer, half-integer, quarter-integer), this can be simplified considerably, in terms of error and Bessel fn.'s especially. E.g., for $a=3/2$ and $b=2$ the expression simplifies to $\sqrt{\frac{\pi }{2}} e^{\frac{c^2}{8}} \text{erfc}\left(\frac{c}{2 \sqrt{2}}\right)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.