I am trying to find a tight upper bound the following expression \begin{align} \frac{ \Gamma \left( x; a\right)}{\Gamma(x)} \end{align} where $\Gamma \left( x; a\right)$ is incomplet Gamma function \begin{align} \Gamma \left( x; a\right)= \int_{a}^\infty t^{x-1} e^{-t} dt. \end{align}

Also, assume $x \ge 1/2$. Clearly a trivial upper bound is \begin{align} \frac{ \Gamma \left( x; a\right)}{\Gamma(x)} \le 1 \end{align}

However, I suspect the following bound should be true

\begin{align} \frac{ \Gamma \left( x; a\right)}{\Gamma(x)} \le O( e^{-a} ) \end{align}

  • $\begingroup$ You've given a lower bound on $x$. Do you have any bounds on $a$? $\endgroup$ – Eric Towers Apr 12 '16 at 14:20
  • $\begingroup$ @EricTowers I want a general bound that holds for any $a$ and $x$ without any assumption on $x$ and $a$. In fact, I would like to vary $a$ not $x$. $\endgroup$ – Boby Apr 12 '16 at 14:22
  • $\begingroup$ Then this is hopeless. For $x<0$, this ratio is unbounded. $\endgroup$ – Eric Towers Apr 12 '16 at 14:27
  • $\begingroup$ @EricTowers Still with assumption $x>1/2$. $\endgroup$ – Boby Apr 12 '16 at 14:29

$$\int_{a}^{+\infty}t^{x-1}e^{-t}\,dt = e^{-a}\int_{0}^{+\infty}(t+a)^{x-1}e^{-t}\,dt\leq e^{-a}a^{x-1}\int_{0}^{+\infty}\exp\left(\frac{t(x-1)}{a}-t\right)\,dt$$ hence a simple upper bound is: $$ \Gamma(x;a)\leq \frac{a^x e^{-a}}{a+1-x}.$$ That approximation can be refined, producing a continued fraction representation for the incomplete $\Gamma$ function: $$ \Gamma(x;a) = \frac{a^x e^{-a}}{a+\frac{1-x}{1+\frac{1}{a+\frac{2-x}{\ldots}}}}$$ Have also a look at this paper.

  • $\begingroup$ ?? that's not what is asked $\endgroup$ – reuns Apr 7 '16 at 16:37
  • $\begingroup$ If $a+1-x<0$ then the bound is negative, right? This doesn't seem to be correct. $\endgroup$ – Boby Apr 7 '16 at 16:43
  • $\begingroup$ Obviously it is an approximation that holds for large $a$s. $\endgroup$ – Jack D'Aurizio Apr 7 '16 at 16:47

Your final inequality seems to be false. For instance, $\dfrac{\Gamma(2;a)}{\Gamma(2)} = (a+1)\mathrm{e}^{-a}$ and $\dfrac{\Gamma(3;a)}{\Gamma(3)} = \dfrac{1}{2}(a^2 + 2a + 2)\mathrm{e}^{-a}$. The bound $\dfrac{\Gamma(x;a)}{\Gamma(x)} \leq O(a^{x-1}\mathrm{e}^{-a})$ seems possible, but I haven't taken the time to find a proof.

  • $\begingroup$ Thank you. In case you are curios, please do take the time. $\endgroup$ – Boby Apr 12 '16 at 14:28
  • $\begingroup$ For example how would you compute $\lim_{n \to \infty } \frac{\Gamma(n;n)}{\Gamma(n)}$ ? $\endgroup$ – Boby Apr 12 '16 at 17:23
  • 1
    $\begingroup$ @Boby: that limit is $\frac{1}{2}$ by a famous probabilistic argument: you are computing the probability that a limit random variable (fulfilling the hyphotesis of the central limit theorem) takes a value that is less than its expected value. $\endgroup$ – Jack D'Aurizio Apr 13 '16 at 15:56
  • $\begingroup$ @JackD'Aurizio Thank you. What aboutn $\lim_{n \to \infty} \frac{\Gamma(n,\lambda n)}{\Gamma(n)}$. Is it $0$ if $\lambda>1$ and $1$ if $\lambda<1$ ? $\endgroup$ – Boby Apr 13 '16 at 15:59
  • $\begingroup$ @Boby: yes, you're right. In such cases it is enough to use some concentration inequality for the Gamma distribution, whose mean and variance are known. You may also replace the Gamma distribution with a normal distribution and prove the same: the Central Limit Theorem allow us to do so. $\endgroup$ – Jack D'Aurizio Apr 13 '16 at 16:07

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