What Is the Difference between Gamma and Gaussian Distributions? ()

Xiao-Li Hu

School of Electrical Engineering and Computer Science, University of Newcastle, Newcastle, Australia.

**DOI: **10.4236/am.2013.42043
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School of Electrical Engineering and Computer Science, University of Newcastle, Newcastle, Australia.

An inequality describing the difference between Gamma and Gaussian distributions is derived. The asymptotic bound is much better than by existing uniform bound from Berry-Esseen inequality.

Keywords

Gamma Distribution; Gaussian Distribution; Berry-Esseen Inequality; Characteristic Function

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Hu, X. (2013) What Is the Difference between Gamma and Gaussian Distributions?. *Applied Mathematics*, **4**, 285-289. doi: 10.4236/am.2013.42043.

1. Introduction

1.1. Problem

We first introduce some notations. Denote Gamma distribution function as

(1)

for and, where is the Gamma function, i.e.,

Assume for. The density of chisquare distributed random variable with degrees of freedom is

It is well-known that the random variable can be interpreted by with independent and identically distributed (i.i.d.) random variables where denotes the standard Gaussian distribution. The mean and variance of is respectively

Then, by simple change of variable we find

(2)

On the other side, by the Berry-Esseen inequality to

, it is easy to find a bound

such that

(3)

where is the standard Gaussian distribution function, i.e.,

(4)

Then, by Equations (2) and (3) it follows

(5)

which describes the distance between Gamma and Gaussian distributions. The purpose of this paper is to derive asymptotic sharper bound in Equation (5), which much improves the constant by directly using Berry-Esseen inequality. The main framework of analysis is based on Gil-Pelaez formula (essentially equivalent to Levy inversion formula), which represents distribution function of a random variable by its characteristic function.

The main result of this paper is as following.

Theorem 1.1 A relation of the Gamma distribution (1) and Gaussian distribution (4) is given by

(6)

where

with and for any.

Clearly, as. Thus, the asymptotical bound is

as. To check the tightness of the limit value of, we plot in Figure 1 the multiplication

for, where the straight line is the limit value. From this experiment it seems that

is the best constant. The tendency of the theoretical formula is plotted for in Figure 2, which also shows the tendency to the limit value

. The slow trend is due to that some upper bounds formulated over interval have been weakly estimated, e.g., the third and fourth terms of.

1.2. Comparison to the Bound Derived by Berry-Esseen Inequality

Let be a sequence of independent identically distributed random variables with EX_{1} = 0

Figure 1. Experiment.

Figure 2. Trend of.

and finite third absolute moment. Denote

By classic Berry-Esseen inequality, there exists a finite positive number such that

(7)

The best upper bound is found in [1] in 2009. The bound is improved in [2] at some angle in a slight different form as

(8)

with

The inequality (8) will be sharper than Equation (7) for.

Now let us derive the constant in (5) by applying Berry-Esseen inequality to. It is difficult to calculate the exact value of third absolute moment of the random variable. Thus, it is approximated as

by using Matlab to integrate over interval divided equivalently 100,000 subinterval for its half value.

By Equation (7) with we have

and by Equation (8) we have

Hence, the best constant in Equation (5) by applying Berry-Esseen inequality is. Obviously, the limit bound

found in this paper for chi-square distribution is much better.

The technical reason is that the Berry-Esseen inequality deals with general i.i.d. random sequences without exact information of the distribution.

2. Proof of Main Result

Before to prove the main result, we first list a few lemmas and introduce some facts of characteristic function theory.

2.1. Some Lemmas

Lemma 2.1 For a complex number satisfying,

Proof First show that

By Taylor’s expansion and noting, we have

Together with

the assertion follows.

Lemma 2.2 For a real number satisfying,

where is the imaginary unit and

Clearly,

Proof. By Taylor expansion for complex function, for we have

where is shown above. By further noting the two alternating real series above, it follows the upper bound.

We cite below a well-known inequality [3] as a lemma.

Lemma 2.3 The tail probability of the standard normal distribution satisfies

for.

2.2. Characteristic Function

Let us recall, see e.g., [4], the definition and some basic facts of characteristic function (CF), which provides another way to describe the distribution function of a random variable. The characteristic function of a random variable is defined by

where is the imaginary unit, and is the argument of the function. Clearly, the CF for random variable with real numbers and is

Another basic quality is

for with and independent to each other.

It is well-known that the CF of standard Gaussian is

(9)

and the CF of chi-square distributed variable is

Thus, the CF for is

(10)

The CF is actually an inverse Fourier transformation of density function. Therefore, distribution function can be expressed by CF directly, e.g., Levy inversion formula. We use another slightly simpler formula. For a univariate random variable, if is a continuity point of its distribution, then

(11)

which is called Gil-Pelaez formula, see, e.g., page 168 of [4].

2.3. Proof of Main Result

We are now in a position to prove the main result.

Proof of Theorem 1.1 First analyze CF of given by Equation (10). Denote. For, i.e., , by Lemma 2.2,

(12)

where

Clearly,

To make sure for somedenote. Then, it is easy to see that

(13)

for. Hence, by Equations (12) and (13) and Lemma 2.1,

(14)

for.

Now let us consider the difference between and, i.e., the CF (9) of Gaussian distribution, over the interval. By Equation (14)

Note that

it follows

(15)

Similarly,

(16)

Below let us analyze the residual integrals over the interval. By Lemma 2.3,

(17)

Similarly,

(18)

It is somewhat difficult to analyze the residual integral over for. We divide it into two subintervals as following:

where.

Observe that decreases on interval and for, we have

where

The fact is used in above formula. Thus,

(19)

For the other interval, we proceed as

(20)

By Equations (19) and (20)

(21)

Similarly,

(22)

By Equation (15), Equation (17), Equation (21) and Equation (16), Equation (18), Equation (22)

where

In view of Formula (11) , the formula to be proved follows directly.

Conflicts of Interest

The authors declare no conflicts of interest.

[1] | I. S. Tyurin, “On the Accuracy of the Gaussian Approximation,” Doklady Mathematics, Vol. 80, No. 3, 2009, pp. 840-843. doi:10.1134/S1064562409060155 |

[2] | V. Koroleva and I. Shevtsova, “An Improvement of the Berry-Essen in Equality with Application to Possion and Mixed Poison Random Sums,” Scandinavian Actuarial Journal, Vol. 2012, No. 2, 2012, pp. 81-105. doi:10.1080/03461238.2010.485370 |

[3] | R. D. Gordon, “Values of Mills’ Ratio of Area to Bounding Ordinate and of the Normal Probability Integral for Large Values of the Argument,” The Annals of Mathematical Statistics, Vol. 12, No. 3, 1941, pp. 364-366. doi:10.1214/aoms/1177731721 |

[4] | K. L. Chung, “A Course in Probability Theory,” 3rd Edition, Probability and Mathematical Statistics, Academic, New York, 2001. |

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