Let $a>1$. I need to approximate (for n large) $$ \Gamma(n,n/a) \approx f(n) $$

Approximations I found are for the case which the second argument is fixed. Is there any simple formula for this asymptotic?

  • $\begingroup$ Stirling's approximation of $n!$ will do ... See en.wikipedia.org/wiki/… $\endgroup$ Feb 25 at 1:42
  • $\begingroup$ Is there any relation between factorial and the incomplete gamma function? I think I need this relation to use Stirling's approximation. $\endgroup$
    – Rostam22
    Feb 25 at 1:56
  • 1
    $\begingroup$ For Laplace's method to apply, we need to find the turning point where the largest contribution will occur. In your case this turning point is within the limits (you do need to check this ... luckily $a>1$) ... so this method will give the asymptotic expansion. $\endgroup$ Feb 25 at 2:01
  • $\begingroup$ Have a look at arxiv.org/pdf/1408.0674.pdf $\endgroup$ Feb 25 at 9:25

If $0<\alpha<1$ and $\alpha$ is bounded away from $1$ (e.g., $1 - n^{ - 1/2} \gg \alpha$), then $$ \Gamma (n,\alpha n) \sim \Gamma (n) + \alpha ^n \frac{{n^{n - 1} e^{ - \alpha n} }}{{1-\alpha}}\left( {1 - \frac{\alpha }{{n(1-\alpha)^2 }} + \frac{{\alpha (2\alpha + 1)}}{{n^2 (1-\alpha )^4 }} - \cdots } \right) $$ as $n\to +\infty$ (cf. http://dlmf.nist.gov/8.11.E6). If $\alpha$ is close to $1$, one may use \begin{align*} \frac{{\Gamma (n,n - \beta \sqrt n )}}{{\Gamma (n)}} \sim 1 & - \frac{1}{2}\operatorname{erfc}\left( {\frac{\beta }{{\sqrt 2 }}} \right) \\ &+ \frac{1}{{\sqrt {2\pi n} }}\exp \left( { - \frac{{\beta ^2 }}{2}} \right)\left( {\frac{{\beta ^2 - 1}}{3} - \frac{{\beta (2\beta ^4 - 11\beta ^2 + 3)}}{{36\sqrt n }} + \cdots } \right) \end{align*} as $n\to +\infty$ with $0 < \beta \ll n^{1/6}$. Here $\operatorname{erfc}$ denotes the complementary error function (cf. https://doi.org/10.1007/s00365-018-9445-3).


$\Gamma(n, n/a)$ and $\Gamma(n)$ will become very close (in the sense that their ratio converges to $1$) as $n$ grow. The following computation provides a more precise error bound for their ratio:

I will write $\alpha = 1/a$ so that $0 < \alpha < 1$. Then

$$ \Gamma(n, \alpha n) = \Gamma(n) \biggl[ 1 - \frac{1}{\Gamma(n)} \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \biggr]. $$

Since $t \mapsto t^{n-1}e^{-t}$ is increasing on $0 \leq t \leq n-1$, for large $n$ we have

$$ \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \leq (\alpha n)^n e^{-\alpha n}. $$

So by the Stirling's approximation,

$$ \frac{1}{\Gamma(n)} \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \leq \frac{(\alpha n)^n e^{-\alpha n}}{\sqrt{2\pi} \, n^{n-\frac{1}{2}} e^{-n}} = \frac{1}{\sqrt{2\pi}} n^{1/2} (\alpha e^{1-\alpha})^n . $$

Similarly, by noting that

$$ \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \geq \int_{\alpha n-1}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \geq (\alpha n - 1)^{n-1} e^{-\alpha n+1}, $$

we have

$$ \frac{1}{\Gamma(n)} \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \geq c_{\alpha} n^{-1/2} (\alpha e^{1-\alpha})^{n} $$

for some constant $c_{\alpha} \in (0, 1)$ depending only on $\alpha$. This tells that

$$ \Gamma(n, \alpha n) = \Gamma(n) \left(1 - n^{\Theta(1)}(\alpha e^{1-\alpha})^n \right), $$

where $\Theta(1)$ represents a bounded sequence in $n$.

Remarks. A less precise asymptotic formula $\Gamma(n, \alpha n) = \Gamma(n)(1 - o(1))$ can be obtained by applying the central limit theorem to

$$ \frac{\Gamma(n, \alpha n)}{\Gamma(n)} = \mathbf{P}(\tau_1 + \cdots + \tau_n \geq \alpha n), $$

where $\tau_k$'s are independent $\operatorname{Exp}(1)$ variables.


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.