Does any one know approximations for the lower incomplete gamma function $\gamma(a,bx)$. The problem is this: I want to find the quantile function for the CDF of the gamma distribution. The CDF of the gamma distribution is $$F(x)=\frac{\gamma(a,bx)}{\Gamma(a)}=\frac{b^a}{\Gamma(a)}\sum_{m=0}^{\infty} \frac{(-b)^{m}}{m!} \frac{x^{m+a}}{m+a}.$$

The problem is to find $x$ (the quantile function) such that $F(x)= \epsilon$, $\epsilon$ is a very small value. For a small value of $F(x)$, $x$ is also very small.


Since $x$ is small, I approximate $F(x)$ using a single term in the above summation, that is,

$$ F(x) \approx \frac{b^a}{\Gamma(a)} \frac{x^{a}}{a}=\epsilon,$$

which gives $x \approx \frac{1}{b} (\epsilon~\Gamma(a)~a)^{\frac{1}{a}}$. Are there any other better approximations than this? Please help.


If I correctly understand, you want an approximation of $x$ such that $F(x)=\epsilon$ and you consider that $x_0=\frac{1}{b} (\epsilon~\Gamma(a)~a)^{\frac{1}{a}}$ is a legitimate approximation when $\epsilon$ is small.

So, to refine the approximation, you could use one step of Newton method writing $$x_1=x_0-\frac{F(x_0)-\epsilon}{F'(x_0)}$$ This should give you $$x_{n+1}=x_n+x_n e^{b x_n}(b x_n)^{-a} (\Gamma (a,b x_n)+(\epsilon -1) \Gamma (a))$$

Another solution could be to use the first two terms and write $$\frac{x^a}{a}=\frac{\epsilon \Gamma(a)}{b^a}+\frac{b x^{a+1}}{a+1}$$ that is to say $$x_{n+1}=\Big(\frac{a\epsilon \Gamma(a)}{b^a}+\frac{ab x_n^{a+1}}{a+1}\Big)^{\frac{1}{a}}$$ If you start with $x=0$ in the rhs, you have your approximation $x_0$. Then, one interation gives $$x_1=\left(a \left(\epsilon b^{-a} \Gamma (a)+\frac{b \left(\left(a \epsilon b^{-a} \Gamma (a)\right)^{\frac{1}{a}}\right)^{a+1}}{a+1}\right)\right)^{\frac{1}{a}}$$

  • $\begingroup$ Thank you for this beautiful insight Dr. Claude. May I ask you couple of more questions? 1. I am not clear how do you arrive at the recursive equation $x_{n+1}$, from the first two terms? Pardon me for my ignorance. 2. Your approximation is really helpful to me. For simplicity, can I say $x _{1} \approx \alpha x_{0},$ where $\alpha=1,2 or 3$. Thanks in advance. $\endgroup$ – Oliver Aug 2 '14 at 21:54
  • $\begingroup$ In the second part, I used the first two terms of the expansion. I rewrote putting the first in the lhs a,d the second in the rhs in order to arrive to something looking as a recurrence relation. $\endgroup$ – Claude Leibovici Aug 3 '14 at 2:46
  • $\begingroup$ Thank you for your reply. I am still not convinced with the recurrence relation.In the two term approximation, the $x$ assumes the same value on both sides. How can we take the $x$ (on the L.H.S) as $x_{n+1}$ and the $x$ (on the R.H.S) as $x_{n}$? I am clueless. Thanks. $\endgroup$ – Oliver Aug 3 '14 at 12:44
  • $\begingroup$ I like the second way of approximation, because that does not involve the incomplete gamma functions! $\endgroup$ – Oliver Aug 3 '14 at 12:52

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