1 Introduction and main result
Recently, there have been many papers about the ratio of gamma functions in the literature; see [1‐9]. Some of the papers use statistical methods. Gurland [10] has given an inequality satisfied by the gamma function, using the so-called Cramér-Rao lower bound for the variance of unbiased estimators. Olkin [11] has given an extension of Gurland’s inequality. Gokhale [12] has given another inequality, which used an analogue of the Cramér-Rao lower bound derived by Rao [13]. Rao gave a stronger version of Wallis’ formula [14]. We, inspired by the above papers, give an inequality concerning the gamma function. Applications of the inequality are also given. We first recall some definitions, notation, and well-known results in statistical theory, which will be used in this paper.
A normal distribution \(N(\mu,\sigma^{2})\) is described by the probability density function, When a random variable X is distributed normally with mean μ and variance \(\sigma^{2}\), we write \(X\sim N(\mu,\sigma^{2})\).
$$ p(x)=\frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}},\quad x\in \mathbb{R}. $$
(1.1)
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Suppose that \(x_{1},x_{2},\ldots,x_{n}\) is a sample from a population with a distribution function \(F_{\theta}(x)\) (\(\theta\in\Omega\)). Let \(\hat{g}=\hat{g}(x_{1},x_{2},\ldots,x_{n})\) be an estimator of a parametric function \(g(\theta)\). If \(E(\hat{g})=g(\theta)\) for all values of parameter \(\theta\in\Omega\), we call \(\hat{g}\) an unbiased estimator of \(g(\theta)\).
Consider an estimation of \(g(\theta)\) based on a sample \(x_{1},x_{2},\ldots,x_{n}\) from some member of a family of distribution functions \(F_{\theta}(x)\), \(\theta\in\Omega\), where Ω is the parameter space. An unbiased estimator \(\hat{g}(x_{1},x_{2},\ldots,x_{n})\) of \(g(\theta)\) is UMVUE, if \(\forall\theta\in\Omega\), for any other unbiased estimator \(\tilde{g}\).
$$ \operatorname{var}_{\theta}(\hat{g}(x_{1},x_{2}, \ldots,x_{n})\leq \operatorname{var}_{\theta}\bigl(\tilde{g}(x_{1},x_{2}, \ldots,x_{n})\bigr), $$
(1.2)
Euler’s gamma function Γ is defined for \(x>0\) by If \(x_{1},x_{2},\ldots,x_{n}\) is a sample from a population with distribution \(N(\mu,\sigma^{2})\), then is the UMVUE of σ.
$$ \Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}\,dt. $$
(1.3)
$$ \hat{\sigma}=\frac{\Gamma (\frac {n}{2} )}{ \sqrt{2}\Gamma (\frac{n+1}{2} )}\sqrt{\sum _{i=1}^{n}x_{i}^{2}} $$
(1.4)
The main result of this paper is the following theorem.
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Theorem 1.1
Suppose that
\(n_{k}\) (\(k=1,2,\ldots,m\)) are nonnegative integers and
\(\lambda_{k}\in\mathbb{R}\), such that
\(0\leq\lambda_{k}\leq1\), \(\sum_{k=1}^{m}\lambda_{k}=1\). Then we have
where
\(n=\sum_{k=1}^{m}n_{k}\).
$$ \frac{n\Gamma^{2}(\frac{n}{2})}{2\Gamma^{2}(\frac{n+1}{2})}-1\leq\sum_{k=1}^{m} \lambda_{k}^{2} \biggl(\frac{n_{k}\Gamma^{2}(\frac{n_{k}}{2})}{ 2\Gamma^{2}(\frac{n_{k}+1}{2})}-1 \biggr), $$
(1.5)
2 Proof of the main result
In this section, we use statistical methods to prove the theorem.
Proof
Let \(x_{11}, x_{12}, \ldots, x_{1n_{1}}, x_{21}, x_{22}, \ldots, x_{2n_{2}}, x_{m1}, x_{m2}, \ldots, x_{mn_{m}}\) be a random sample from a normal distribution \(X\sim N(\mu,\sigma^{2})\). From (1.4), it is known that is the UMVUE of σ, where \(n=\sum_{i=1}^{m}n_{i}\).
$$ \hat{\sigma}=\frac{\Gamma (\frac {n}{2} )}{ \sqrt{2}\Gamma (\frac{n+1}{2} )}\sqrt{\sum _{i=1}^{m}\sum_{j=1}^{n_{i}}x_{ij}^{2}} $$
(2.1)
For any \(x_{k1}, x_{k2}, \ldots, x_{kn_{k}}\), \(1\leq k\leq m\)
is an unbiased estimate of σ.
$$ \frac{\Gamma(\frac{n_{k}}{2})}{ \sqrt{2}\Gamma(\frac{n_{k}+1}{2})}\sqrt{\sum_{k=1}^{n_{k}}x_{ki}^{2}} $$
(2.2)
Using (2.2), we construct a new unbiased estimate of σ, i.e., where \(0\leq\lambda_{k}\leq1\), \(\sum_{k=1}^{m}\lambda_{k}=1\).
$$ \hat{\sigma}_{1}=\sum_{k=1}^{m} \frac{\lambda_{k}\Gamma(\frac{n_{k}}{2})}{ \sqrt{2}\Gamma(\frac{n_{k}+1}{2})}\sqrt{\sum_{k=1}^{n_{k}}x_{ki}^{2}}, $$
(2.3)
Due to the definition of the UMVUE, the following inequality holds: After some simple computations, we can obtain and Substituting (2.5) and (2.6) into (2.4) gives (1.5). Thus, we complete the proof. □
$$ \operatorname{Var}\hat{\sigma}_{1}\geq \operatorname{Var} \hat{\sigma}. $$
(2.4)
$$ D(\hat{\sigma})= \biggl(\frac{n\Gamma^{2}(\frac{n}{2})}{2\Gamma^{2}(\frac {n+1}{2})}-1 \biggr) \sigma^{2} $$
(2.5)
$$ D(\hat{\sigma}_{1})=\sum_{k=1}^{m} \lambda_{k}^{2} \biggl(\frac {n_{k}\Gamma^{2}(\frac{n_{k}}{2})}{2\Gamma^{2}(\frac{n_{k}+1}{2})}-1 \biggr) \sigma^{2}. $$
(2.6)
3 Some applications of Theorem 1.1
In this section, we show some applications of the main result of this paper. First, we give the following inequalities, which include the gamma function and a trigonometric function.
Theorem 3.1
Suppose
n
is any positive integer, and
\(\theta\in\mathbb{R}\), then
$$ \frac{\Gamma^{2}(2n+1)}{\Gamma^{2}(2n+\frac{1}{2})}\leq n\sin^{2}\theta+\bigl(2- \sin^{2}\theta\bigr)\frac{\Gamma^{2}(n+1)}{\Gamma ^{2}(n+\frac{1}{2})}. $$
(3.1)
Proof
Letting \(m=2\) and \(\lambda_{1}=\sin^{2}\theta\), \(\lambda _{2}=\cos^{2}\theta\) in (1.5) gives Letting \(n_{1}=n_{2}=a\) in (3.2), one obtains Employing the following trigonometric formula: (3.3) becomes Replacing 2θ by θ and a by n, we obtain (3.1). Thus, we finish the proof. □
$$\begin{aligned}& \frac{(n_{1}+n_{2})\Gamma^{2}(\frac{n_{1}+n_{2}}{2})}{2\Gamma ^{2}(\frac{n_{1}+n_{2}+1}{2})}-1 \\& \quad \leq\sin^{4}\theta \biggl(\frac{n_{1}\Gamma^{2}(\frac{n_{1}}{2})}{2\Gamma ^{2}(\frac{n_{1}+1}{2})}-1 \biggr)+ \cos^{4}\theta \biggl(\frac{n_{2}\Gamma^{2}(\frac{n_{2}}{2})}{2\Gamma ^{2}(\frac{n_{2}+1}{2})}-1 \biggr). \end{aligned}$$
(3.2)
$$ \frac{\Gamma^{2}(2a+1)}{\Gamma^{2}(2a+\frac{1}{2})}- 2\bigl(\sin^{4} \theta+\cos^{4}\theta\bigr)\frac{\Gamma^{2}(a+1)}{\Gamma^{2}(a+\frac {1}{2})}\leq2a \bigl(1- \sin^{4}\theta-\cos^{4}\theta \bigr). $$
(3.3)
$$ \sin^{4}\theta+\cos^{4}\theta=1- \frac{\sin^{2}2\theta}{2}, $$
(3.4)
$$ \frac{\Gamma^{2}(2a+1)}{\Gamma^{2}(2a+\frac{1}{2})}\leq a\sin^{2}(2\theta)+\bigl(2- \sin^{2}(2\theta)\bigr)\frac{\Gamma^{2}(a+1)}{\Gamma ^{2}(a+\frac{1}{2})}. $$
(3.5)
In [15], Gurland gave the following estimator of π: Mortici [1] gave the refinements of Gurland’s formula for π: Using (3.1), we can get the following similar result.
$$ \frac{4n+3}{ (2n+1 )^{2}} \biggl(\frac{ (2n )!!}{ (2n-1 )!!} \biggr)^{2}< \pi. $$
(3.6)
$$ \biggl(\frac{n+\frac{1}{4}}{n^{2}+\frac{1}{2}n+\frac{3}{32}}+\frac {9}{2\text{,}048n^{5}}-\frac{45}{8\text{,}192n^{6}} \biggr) \biggl(\frac{ (2n )!!}{ (2n-1 )!!} \biggr)^{2}< \pi. $$
(3.7)
Corollary 3.2
Suppose
n
is any nonnegative integer, then
$$ \frac{1}{n} \biggl[ \biggl(\frac{(4n)!!}{(4n-1)!!} \biggr)^{2}- \biggl(\frac {(2n)!!}{(2n-1)!!} \biggr)^{2} \biggr]< \pi. $$
(3.8)
Proof
Letting \(\theta=\frac{\pi}{2}\) in (3.1) gives We have See, e.g., [1]. So Because the equality in (3.11) cannot hold, we get (3.8). □
$$ \frac{\Gamma^{2}(2n+1)}{\Gamma^{2}(2n+\frac{1}{2})}\leq n+\frac{\Gamma^{2}(n+1)}{\Gamma^{2}(n+\frac{1}{2})}. $$
(3.9)
$$ \frac{ (2q )!!}{ (2q-1 )!!}=\sqrt{\pi}\frac{\Gamma (q+1 )}{\Gamma (q+\frac{1}{2} )}. $$
(3.10)
$$ \frac{1}{n} \biggl[ \biggl(\frac{(4n)!!}{(4n-1)!!} \biggr)^{2}- \biggl(\frac {(2n)!!}{(2n-1)!!} \biggr)^{2} \biggr]\leq \pi. $$
(3.11)
Now we give an inequality involving combinational coefficients \({n\choose m}\), defined by
$${n\choose m }=\frac{n!}{m!(n-m)!}. $$
Theorem 3.3
Suppose
m, w
are any positive integers, then
$$\begin{aligned}& \frac{1}{m} \biggl[\frac{(2mw)!!}{(2mw-1)!!} \biggr]^{2} +\frac{1}{2^{2m}} \biggl[w\pi- \biggl(\frac{ (2w )!!}{ (2w-1 )!!} \biggr)^{2} \biggr] \left[{2m\choose m}-1 \right] \\& \quad\leq w\pi. \end{aligned}$$
(3.12)
Proof
Since Letting \(n_{k}=2w\), \(\lambda_{k}={m\choose k}\frac{1}{2^{m}}\) in (1.5), then \(n=\sum_{k=1}^{m}n_{k}=2mw\). We get Using the inequality of (3.10) and after some simple derivations, we have Substituting into (3.15) one obtains (3.12). □
$$ \sum_{k=1}^{m}{m\choose k} \frac{1}{2^{m}}=1. $$
(3.13)
$$ mw\frac{\Gamma^{2}(mw)}{\Gamma^{2}(mw+\frac{1}{2})}-1\leq\sum_{k=1}^{m} \biggl({m\choose k}\frac{1}{2^{m}} \biggr)^{2} \biggl(w \frac{\Gamma ^{2}(w)}{\Gamma^{2}(w+\frac{1}{2})}-1 \biggr). $$
(3.14)
$$\begin{aligned}& \Biggl[\frac{1}{m} \biggl(\frac{(2mw)!!}{(2mw-1)!!} \biggr)^{2} -\sum_{k=1}^{m} \biggl({m\choose k}\frac{(2w)!!}{(2w-1)!!}\frac{1}{2^{m}} \biggr)^{2} \Biggr]\frac{1}{\pi} \\& \quad\leq w \Biggl[1-\sum_{k=1}^{m} \biggl({m\choose k}\frac{1}{2^{m}} \biggr)^{2} \Biggr]. \end{aligned}$$
(3.15)
$$ \sum_{k=0}^{m}{m\choose k}^{2}={2m\choose m} $$
(3.16)
The special case \(w=1\) of (3.12) results in
$$ \frac{1}{m} \biggl[\frac{(2m)!!}{(2m-1)!!} \biggr]^{2} +\frac{\pi-4}{2^{2m}} \left[{2m\choose m}-1 \right]\leq\pi. $$
(3.17)
Finally, we give the following double inequality for π.
Theorem 3.4
Let
p, d
are positive integers, then
$$\begin{aligned}& \frac{1}{2pd} \biggl[ \biggl(\frac{ (4pd+2p )!!}{ (4pd+2p-1 )!!} \biggr)^{2}- \biggl(\frac{ (2p )!!}{ (2p-1 )!!} \biggr)^{2} \biggr] \\& \quad< \pi < \frac{4d}{ (2p+1 ) [ (2d+1 )^{2} (\frac { (4pd+2p+2d-1 )!!}{ (4pd+2p+2d )!!} )^{2}- (\frac{(2p-1)!!}{(2p)!!} )^{2} ]}. \end{aligned}$$
(3.18)
Proof
Letting \(m=2d+1\) and \(n_{1}=n_{2}=\cdots=n_{m}=2p+1\) in (1.5) gives Using (3.10), the inequality (3.19) can be written in the equivalent form Letting \(\lambda_{1}=\lambda_{2}=\cdots=\lambda_{m}=\frac{1}{2d+1}\) in (3.20) gives Similarly, letting \(m=2d+1\), \(n_{1}=n_{2}=\cdots=n_{m}=2p\), and \(\lambda_{1}=\lambda_{2}=\cdots=\lambda_{m}=\frac{1}{2d+1}\) in (1.5), one can obtain
$$\begin{aligned}& \frac{(2p+1)(2d+1)}{2}\frac{\Gamma^{2}(2pd+p+d+\frac{1}{2})}{\Gamma ^{2}(2pd+p+d+1)}-1 \\& \quad\leq\sum_{k=1}^{2d+1} \lambda_{k}^{2} \biggl[\frac{2p+1}{2}\frac {\Gamma^{2}(\frac{2p+1}{2})}{ \Gamma^{2}(\frac{2p+2}{2})}-1 \biggr]. \end{aligned}$$
(3.19)
$$ \pi< \frac{2 (1-\sum_{k=1}^{2d+1}\lambda_{k}^{2} )}{(2p+1) [(2d+1) (\frac{(4pd+2p+2d-1)!!}{(4pd+2p+2d)!!} )^{2}-\sum_{k=1}^{2d+1}\lambda_{k}^{2} (\frac{(2p-1)!!}{(2p)!!} )^{2} ]}. $$
(3.20)
$$ \pi< \frac{4d}{(2p+1) [(2d+1)^{2} (\frac {(4pd+2p+2d-1)!!}{(4pd+2p+2d)!!} )^{2}- (\frac{(2p-1)!!}{(2p)!!} )^{2} ]}. $$
(3.21)
$$ \frac{1}{2pd} \biggl[ \biggl(\frac{(4pd+2p)!!}{(4pd+2p-1)!!} \biggr)^{2}- \biggl(\frac{(2p)!!}{(2p-1)!!} \biggr)^{2} \biggr]< \pi. $$
(3.22)
The special case \(p=1\) of (3.18) results in
$$ \frac{1}{2d} \biggl[ \biggl(\frac{ (4d+2 )!!}{ (4d+1 )!!} \biggr)^{2}-4 \biggr]< \pi < \frac{4d}{3 [ (2d+1 )^{2} (\frac{ (6d+1 )!!}{ (6d+2 )!!} )^{2}-\frac{1}{4} ]}. $$
(3.23)
Acknowledgements
The authors would like to thank two anonymous referees for many helpful comments and suggestions. We acknowledge support by the National Natural Science Foundation (grant 11271057) of China.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors read and approved the final manuscript.