Inequality for incomplete Gamma Function

I am trying to find if the following inequality is correct or not

\begin{align} f_a(x) \le f_b(x), \forall x\in \mathbb{R} \end{align} for $0<a\le b$, where \begin{align} f_a(x)= \frac{\gamma \left( \frac{1}{a}, \frac{|x|^a}{2} \right)}{\Gamma \left( \frac{1}{a} \right)}, \end{align}

and where the gamma functions are defined as follows \begin{align} \Gamma\left(x \right)= \int_0^\infty t^{x-1} e^{-t} dt, \\ \gamma(x,s) = \int_0^s t^{x-1}\,e^{-t}\,dt. \end{align}

I tried some simulation and this inequality seems to hold. However, I can not show it.

This inequality can also be seen as a monotonicity result in terms of a variable $a$ for $f_a(x)$.

Simulation, results seem to suggest the inequality is true, see the figure below (on the figure x-axis is $x$). Thanks.

• It might be easier to look at $g_a(x) = f_{1/a}(x)$ and show the inverse inequality. That makes the Gamma and incomplete Gamma much more manageable. Then since $\frac{1}{\Gamma(a)}$ decreases, your inequality is equivalent to showing $\gamma(a,\frac{|x|^{1/a}}{2})$ is a decreasing function, and that shouldn't be too hard to prove. (Decreasing in $a$.) – user335907 Mar 9 '17 at 18:12
• But $\gamma(a,\frac{|x|^{1/a}}{2})$ isn't necessarily decreasing; this is just one possible way of approaching the question. Looking at it, getting my bearings about it, it might actually be slightly increasing, but increasing slow enough so that the inverse Gamma kills its growth. Then all you have to do is obtain a nice bound for $\gamma(a,\frac{|x|^{1/a}}{2})$, that it's $O(b^a)$ should be enough. – user335907 Mar 9 '17 at 18:17
• @james.nixon Thank you. But don't see how the bound on $\gamma(a, \frac{|a|^{1/a}}{2})$ will show this? Can you give more details on this? – Lisa Mar 9 '17 at 19:41
• Hmm, well, it will give a partial answer, if $\gamma(a,\frac{|x|^{1/a}}{2}) = O(b^a)$ then since $\frac{1}{\Gamma(a)}$ goes to zero faster, it will kill its growth and make it tend to $0$. On the fact that it will monotonically tend to zero is a little more difficult. Since $\frac{1}{\Gamma(a)}$ is monotonically decreasing, we only need to show that $\gamma(a,\frac{|x|^{1/a}}{2})$ is monotonically increasing and $O(b^a)$ (which isn't hard to show). You might need a clever argument for the monotonicity part, but the exponential bound isn't hard. – user335907 Mar 9 '17 at 22:02
• If $\gamma(a,\frac{|x|^{1/a}}{2})$ wobbles though, I haven't a clue how to approach this. – user335907 Mar 9 '17 at 22:03

We may assume WLOG $x>0$ and consider that:
$$f_a(x) = \frac{\int_{0}^{x^a/2}t^{1/a-1}e^{-t}\,dt}{\int_{0}^{+\infty}t^{1/a-1}e^{-t}\,dt}=\frac{\int_{0}^{x}e^{-t^a/2}\,dt}{\int_{0}^{+\infty}e^{-t^a/2}\,dt}$$ so the inequality $f_a\leq f_b$ boils down to: $$\int_{0}^{x}e^{-t^a/2} dt \int_{0}^{+\infty} e^{-s^b/2}\,ds \leq \int_{0}^{x} e^{-s^b/2}\,ds \int_{0}^{+\infty}e^{-t^a/2}\,dt$$ or to: $$\int_{0}^{x}\int_{0}^{+\infty}\exp\left(-\frac{t^a+s^b}{2}\right)\,ds\,dt \leq \int_{0}^{x}\int_{0}^{+\infty}\exp\left(-\frac{t^a+s^b}{2}\right)\,dt\,ds$$ that is trivial due to the monotonicity of the argument of the exponential and the transformation $(s,t)\mapsto (t,s)$.
• I know this is an older question, but I don't really follow the last step. For example, if we do $(s,t) \to (t,s)$ then the inequality becomes \begin{align} \int_0^x \int_0^\infty \exp(-\frac{s^a+t^b}{2}) dt ds \le \int_0^x \int_0^\infty \exp(-\frac{t^a+s^b}{2}) dt ds \end{align} Is this correct? But how do I use monotonicity now? – Lisa Jul 26 '17 at 19:14