I need to estimate the following quantity $$ K(m,y) \equiv \left(\frac e m \right)^m \times \Gamma(m,(1+y)m), $$ as $m \to \infty$, where $\Gamma(a,z)$ is the incomplete gamma function.

My question is, is there a valid expansion of the incomplete Gamma function in the range when $y \approx m^{-1/2}$ that can give us a good estimation of $K(m,y)$?

Further details:

As I need to estimate $K(m,y)$ for all $y > 0$, I looked up for uniform expansions of incomplete Gamma functions. The most promising formula that I can find is give by R. B. Paris: $$ \Gamma(a,z)= z^{a-1/2}e^{-z} \left(\sqrt{\frac \pi 2}\cdot e^{\chi^2/2} \mathrm{erfc}\left(\frac{\chi}{\sqrt{2}}\right)+O(z^{-1/2})\right) $$ where $\chi = (z-a)/\sqrt{z}$, and $\mathrm{erfc}$ is the complementary error function.

So in my case, I have $$ K(m,y) = \left(\frac{1+y}{e^y} \right)^m \frac{1}{\sqrt{(1+y)m}} \left(\sqrt{\frac \pi 2}\cdot e^{\chi^2/2} \mathrm{erfc}\left(\frac{\chi}{\sqrt{2}}\right)+O(m^{-1/2})\right). $$ with $\chi = y \sqrt{m}/\sqrt{1+y}$.

When $y > m^{-1/2+\epsilon}$ or $y < m^{-1/2-\epsilon}$, i.e., when $\chi \to \infty$ or $\chi \to 0$, the above estimation is good, because we can expand $\mathrm{erfc}(\chi)$ easily.

But when $\chi$ is bounded from above and below, all I can say is that $\mathrm{erfc}(\chi) = O(1)$, which implies that $$ K(m, y) = O(m^{-1/2}). $$ This is not good enough for my application. I would like to get the first order approximation of $K(m,y)$ for $y$ in this range.

Even more details:

I am actually trying is to upper bound the following $$ H(n,m) = \int_0^\infty e^{-y(n-m+1)} (1+y)^{n-m} K(m+1, y) \mathrm dy \qquad (m \to \infty), $$ where $n > m$ and $n, m$ are both integers.

If I use $K(m+1,y) = O(m^{-1/2})$, I will only get $H(n,m) = O(m(n-m))^{-1/2}$. But again I actually want to get the first order approximation of $H(m,n)$.

  • 2
    $\begingroup$ Could you clarify what you mean by "not good enough"? $\endgroup$ – Antonio Vargas Feb 11 '17 at 9:35
  • $\begingroup$ @AntonioVargas I added more background of the question. $\endgroup$ – ablmf Feb 11 '17 at 15:40
  • $\begingroup$ When y=m^(-1/2), Mathematica shows a limit of 0.3976897/sqrt(m). $\endgroup$ – user210229 Feb 14 '17 at 14:35
  • $\begingroup$ It appears that your difficulty is in approximating the value of the error function in a simple way. Have a look at AS 7.125, for example. $\endgroup$ – user26872 Feb 18 '17 at 21:18

From Stirling's approximation: $$m\Gamma(m)=\Gamma(m+1)\approx\left(\frac me\right)^m\sqrt{2\pi m}$$ one can write: $$K(m,y)=\left(\frac e m \right)^m \times \Gamma(m,(1+y)m)\approx\sqrt{\frac{2\pi}m}\times\frac{\Gamma(m,(1+y)m)}{\Gamma(m)}$$ and since $\Gamma(m,x)+\gamma(m,x)=\Gamma(m)$ then $$K(m,y)\approx\sqrt{\frac{2\pi}m}\left(1-\frac{\gamma(m,(1+y)m)}{\Gamma(m)}\right)$$ So you only need an approximation of regularized Gamma function: $$P(m,x):=\frac{\gamma(m,x)}{\Gamma(m)}$$ Note that $P(m,x)$ is the CDF of gamma distribution whose scale parameter is $1$. Therefore if a random variable, $x$, has a gamma distribution with shape parameter $m$, then: $$\mathbf{P}(x\ge m(1+y))=1-\frac{\gamma(m,(1+y)m)}{\Gamma(m)}$$ The mean and variance of a gamma distribution with unit scale are both $m$. For large $m$ the gamma distribution approaches to a normal distribution. So from a statistical point of view, you can consider a random variable $z$ with normal distribution $\mathcal N(m,m)$. Then your function becomes approximately proportional to the probability of $\mathbf P(z\ge m+my)$ or in other words: $$z\sim\mathcal N(m,m)\rightarrow K(m,y)\approx\sqrt{\frac{2\pi}m}\mathbf P\left(\frac{z-\mu}{\sigma}\ge \sigma y\right)$$ Hence $$K(m,y)\approx\sqrt{\frac{\pi}{2m}}\left(1-\text{erf}\left(y\sqrt{\frac m2} \right)\right)$$

  • $\begingroup$ I need to think a bit if this helps. But thanks a lot. :) $\endgroup$ – ablmf Feb 20 '17 at 10:02

For large $m$ with $1+\frac{k}{m}\approx 1$ and $m!\approx \left(\frac{m}{e}\right)^m\sqrt{2\pi m}$:

$\displaystyle \left(\frac{e}{m}\right)^m\Gamma(m,(1+y)m)= \left(\frac{e}{m}\right)^m (m-1)!- \left(\frac{e}{m}\right)^m \sum\limits_{k=0}^\infty\frac{(-1)^k ((1+y)m)^{k+m}}{k!(k+m)}$$\displaystyle \approx \frac{\sqrt{2\pi}}{\sqrt{m}}-\frac{(1+y)^m}{m}e^{-ym}$


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