# Estimation of a gamma function-like integral

A random variable $$X$$ has a pdf: $$f(x) = \frac{1}{k!} \cdot x^k \cdot e^{-x}$$

Prove that $$P(0 \frac{k}{k+1}$$

There are no conditions about $$k$$, so it can be non-integer.

I did these steps:

$$P(0

How can the second integral be estimated to show the required inequality?

• the property is probably true for all $k > 0$ but you need to replace the factorial by the Gamma function (I'm not sure for $-1 < k < 0$) Commented Jul 19 at 12:51
• Two directions I see. 1. The distribution is log-concave. Can we perhaps bound the log by an affine function / the proba by an exponential? 2. Integration by parts Commented Jul 19 at 12:55
• Look up the incomplete gamma functions.
– Gary
Commented Jul 19 at 13:06
• @GuillaumeDehaene, yes, assume that $k>0$ Commented Jul 19 at 13:25
• Have you tried to take advantage of Markov's inequality ? Commented Jul 20 at 5:03

This is probably not the simplest answer. We need to show that $$\frac{1}{{k!}}\int_{2k + 2}^{ + \infty } {x^k {\rm e}^{ - x} {\rm d}x} < \frac{1}{{k + 1}}$$ for $$k>-1$$. (I simply define $$k!$$ by $$\Gamma(k+1)$$ for non-integer $$k$$.) We perform the change of variables from $$x$$ to $$t$$ via $$x = 2(k + 1){\rm e}^t$$ to obtain $$\frac{1}{{k!}}\int_{2k + 2}^{ + \infty } {x^k {\rm e}^{ - x} {\rm d}x} = \frac{{(2{\rm e}^{ - 1} )^{k + 1} }}{{k!}}(k + 1)^{k + 1} {\rm e}^{ - (k + 1)} \int_0^{ + \infty } {\exp \left( { - (k + 1)(2{\rm e}^t - t - 2)} \right){\rm d}t} .$$ By $$(5.6.1)$$, we have $$(k + 1)^{k + 1} {\rm e}^{ - (k + 1)} < \frac{{(k + 1)!}}{{\sqrt {2\pi (k + 1)} }}$$ for any $$k>-1$$. By Maclaurin series expansion, we can see that $$2{\rm e}^t - t - 2 \ge t$$ for any $$t\ge 0$$. Therefore, $$\tag{1} \frac{1}{{k!}}\int_{2k + 2}^{ + \infty } {x^k {\rm e}^{ - x} {\rm d}x} < (2{\rm e}^{ - 1} )^{k + 1} \frac{{\sqrt {k + 1} }}{{\sqrt {2\pi } }}\int_0^{ + \infty } {\exp \left( { - (k + 1)t} \right){\rm d}t} = \frac{{(2{\rm e}^{ - 1} )^{k + 1} }}{{\sqrt {2\pi (k + 1)} }}$$ for any $$k>-1$$. I leave it as an exercise to show that $$\frac{{(2{\rm e}^{ - 1} )^{k + 1} }}{{\sqrt {2\pi (k + 1)} }} < \frac{1}{{k + 1}},$$ for any $$k>-1$$. Let me add that by Stirling's formula and the Laplace method, it can be shown that $$\frac{1}{{k!}}\int_{2k + 2}^{ + \infty } {x^k {\rm e}^{ - x} {\rm d}x} \sim \frac{{(2{\rm e}^{ - 1} )^{k + 1} }}{{\sqrt {2\pi (k + 1)} }}$$ as $$k\to+\infty$$, whence the bound $$(1)$$ is sharp.