# Monotonicity of the regularized incomplete gamma function

In the theory of chi-squared distribution in statistics, for the random variable $X$ following $\chi^2 (k)$, the probability that $X$ is lower than its expactation is given by $$P(X\le k) = \frac{\gamma\left(\frac{k}2,\frac{k}2\right)}{\Gamma\left(\frac{k}2\right)}.$$ Here $$\Gamma(a):=\int_0^\infty{t^{a-1}e^{-t}dt}~~(a>0)$$ is Euler's gamma function and $$\gamma(a,x):=\int_0^x{t^{a-1}e^{-t}dt}~~(a>0,x\ge0)$$ is the incomplete lower gamma function. By the central limit theorem, we know that $P(X\le k)$ converges to $\frac12$ as $k$ tends to $\infty$.

Concerning its monotonicity, I could prove that $$\frac{\gamma(x+1,x+1)}{\Gamma(x+1)}<\frac{\gamma(x,x)}{\Gamma(x)}$$ holds for $x>0$ and find some papers regarding this. In fact, a numerical method supports that $\frac{\gamma(x,x)}{\Gamma(x)}$ decreases for all $x>0$, however, there seems to be few literature on this. My question is whether there is a way to prove this analytically.


By straightforward manipulations, the problem can be restated as follows: Show that \begin{equation*} G_a:=\int_1^\infty g_a(x)\,dx \tag{1} \end{equation*} is increasing in $a>0$, where $g_a(x):=a^a x^{a-1} e^{-a x} /\Ga(a)$.

The key is the following construction: consider \begin{equation*} r_{a,b}(x):=\frac{g_b(x)}{h_{a,b}(x)}, \end{equation*} where \begin{equation*} h_{a,b}(x):=g_a(\la_{a,b}(x))\la'_{a,b}(x),\quad \la_{a,b}(x):=1+\frac{b}{a}\,(x-1)-\frac12\, \Big(\frac ba - 1\Big) \ln x, \end{equation*} and $\la'_{a,b}(x)$ is the derivative of $\la_{a,b}(x)$ in $x$; here everywhere $0<a\le b$ and $x>1$. Note that $\la'_{a,b}(x)\ge\la'_{a,b}(1)= (a + b)/(2 a)>0$, $\la_{a,b}(1)=1$, and $\la_{a,b}(x)\to\infty$ as $x\to\infty$. So, by (1), $G_a=\int_1^\infty h_{a,b}(x)\,dx$. Also, $r_{a,a}(x)=1$, and hence \begin{equation*} \frac{G_b-G_a}{b-a}=\int_1^\infty h_{a,b}(x)\frac{r_{a,b}(x)-r_{a,a}(x)}{b-a}\,dx, \end{equation*} which yields \begin{equation*} G'_a=\int_1^\infty g_a(x)\rho(x)\,dx, \end{equation*} where $G'_a$ is the derivative of $G_a$ in $a$, \begin{equation*} \rho(x):=\rho_a(x):=\frac{\partial r_{a,b}(x)}{\partial b}\Big|_{b=a} =-\frac{1}{2 a x}+\frac{(a x+a-1) \ln x}{2 a x}+\ln a-\psi(a)+\frac{1}{x}-1, \end{equation*} and $\psi=\Ga'/\Ga$.

It suffices to show that $\rho(x)>0$. Writing $x=1+u$ (so that $u>0$), we have \begin{equation*} \rho'(x)2 a (1 + u)^2=a u - (a-1) \ln(1 + u)>a u - a \ln(1 + u)>0, \end{equation*} so that $\rho$ is increasing.

It remains to show that $\rho(1)>0$. In view of the Gauss formula \begin{equation*} \psi(a)=\int_0^\infty\Big(\frac{e^{-z}}{z}-\frac{e^{-az}}{1-e^{-z}}\Big)\,dz \end{equation*} and the elementary formula \begin{equation*} \ln a=\int_0^\infty\frac{e^{-z}-e^{-az}}{z}\,dz \end{equation*} (see e.g. Theorem 1.6.1 and formula 1.6.1, respectively, in Andrews--Askey--Roy), we see that \begin{equation*} \rho(1)=-\frac1{2a}+\ln a-\psi(a) = \int_0^\infty\Big(-\frac12-\frac1{z}+\frac1{1-e^{-z}}\Big)\,e^{-az}\,dz = \int_0^\infty R(z)\,e^{-az}\,dz, \end{equation*} where \begin{equation*} R(z):=\frac{e^z (z-2)+z+2}{2 \left(e^z-1\right) z} \end{equation*} is easily shown to be $>0$ for $z>0$. This completes the proof of the conjecture.

Added: Let me explain the idea of the key construction. We needed to show that $G_b>G_a$ when $0<a\le b$, with $G_a$ as in (1). In the following picture, one can see the graphs $\{(x,g_a(x))\colon 1<x<5\}$ (red) and $\{(x,g_b(x))\colon 1<x<5\}$ (green) for $a=1,b=3$.

The functions $g_a$ and $g_b$ are not directly comparable on the interval $(1,\infty)$: none of them majorizes the other one on this entire interval. The idea is then to rearrange the mass under the graph of $g_a$ so as to make a direct comparison possible. This is done by using the transformation/change of integration variable $x\mapsto\la_{a,b}(x)$, $dx\to\la'_{a,b}(x)\,dx$. The resulting function $h_{a,b}$ (whose graph is orange in the picture) is "majorized in the limit as $b\downarrow a$" by $g_b$ on the entire interval $(1,\infty)$, and the mass under the graph of $h_{a,b}$ over the interval $(1,\infty)$ is the same as that under the graph of $g_a$.

Using transportation of mass to prove inequalities is an old and very broadly used idea. One of earlier examples in this area is using the Steiner symmetrization (where segments are moved to a central position) to prove isoperimetric-type inequalities. Optimal transport by definition produces inequalities; the recent bible in this area is "Optimal Transport" by C. Villani with 846 further references, including ones to the use of transportation techniques by Marton and Talagrand to obtain concentration-of-measure inequalities.

However, I have not seen use of the mass transportation idea in contexts similar to the one here. While the idea is, in principle, very simple, it took me many failed attempts and a great deal of tying up loose ends to finally get a working transformation $\la_{a,b}$.

• Thank a lot for your admiring proof. It seems so elaborate and fantastic. But I still wonder what the motivation of this proof is. Is it a general technique available for other inequalities? Or are there any further references to recommend for me? – Ramanasa May 22 '18 at 10:00
• @Ramanasa : I have added a description of the idea, including mentioning of transportation of mass/transformation techiniques. I do think that this approach can be used in other similar situations. – Iosif Pinelis May 22 '18 at 16:57
• Upon reading the question, I knew it would be you who could solve it. :-) I have not gone through your proof in details yet. But would you be so kind as to list some references in your answer if possible of other examples of using the idea of transportation of mass to prove inequalities? Thank you, Iosif. – Hans May 22 '18 at 17:32
• Thank you Hans for your kind comment. I have added remarks on the use of transportation techniques to obtain inequalities. – Iosif Pinelis May 22 '18 at 19:05
• Again I appreciate your efforts and the idea with such a nice description. Both of the motivating idea and the delecate technique are very impressive! I learned a lot. Thank you:D – Ramanasa May 23 '18 at 1:58