# 2.2 The gamma and chi-square distributions

 Page 1 / 1
This course is a short series of lectures on Introductory Statistics. Topics covered are listed in the Table of Contents. The notes were prepared by EwaPaszek and Marek Kimmel. The development of this course has been supported by NSF 0203396 grant.

## Gamma and chi-square distributions

In the (approximate) Poisson process with mean $\lambda$ , we have seen that the waiting time until the first change has an exponential distribution . Let now W denote the waiting time until the $\alpha$ th change occurs and let find the distribution of W . The distribution function of W ,when $w\ge 0$ is given by

$\begin{array}{l}F\left(w\right)=P\left(W\le w\right)=1-P\left(W>w\right)=1-P\left(fewer_than_\alpha _changes_occur_in_\left[0,w\right]\right)\\ =1-\sum _{k=0}^{\alpha -1}\frac{{\left(\lambda w\right)}^{k}{e}^{-\lambda w}}{k!},\end{array}$

since the number of changes in the interval $\left[0,w\right]$ has a Poisson distribution with mean $\lambda w$ . Because W is a continuous-type random variable, $F\text{'}\left(w\right)$ is equal to the p.d.f. of W whenever this derivative exists. We have, provided w >0, that

$\begin{array}{l}F\text{'}\left(w\right)=\lambda {e}^{-\lambda w}-{e}^{-\lambda w}\sum _{k=1}^{\alpha -1}\left[\frac{k{\left(\lambda w\right)}^{k-1}\lambda }{k!}-\frac{{\left(\lambda w\right)}^{k}\lambda }{k!}\right]=\lambda {e}^{-\lambda w}-{e}^{-\lambda w}\left[\lambda -\frac{\lambda {\left(\lambda w\right)}^{\alpha -1}}{\left(\alpha -1\right)!}\right]\\ =\frac{\lambda {\left(\lambda w\right)}^{\alpha -1}}{\left(\alpha -1\right)!}{e}^{-\lambda w}.\end{array}$

## Gamma distribution

The gamma function is defined by $\Gamma \left(t\right)=\underset{0}{\overset{\infty }{\int }}{y}^{t-1}{e}^{-y}dy,0

This integral is positive for $0 , because the integrand id positive. Values of it are often given in a table of integrals. If $t>1$ , integration of gamma fnction of t by parts yields

$\Gamma \left(t\right)={\left[-{y}^{t-1}{e}^{-y}\right]}_{0}^{\infty }+\underset{0}{\overset{\infty }{\int }}\left(t-1\right){y}^{t-2}{e}^{-y}dy=\left(t-1\right)\underset{0}{\overset{\infty }{\int }}{y}^{t-2}{e}^{-y}dy=\left(t-1\right)\Gamma \left(t-1\right).$

Let $\Gamma \left(6\right)=5\Gamma \left(5\right)$ and $\Gamma \left(3\right)=2\Gamma \left(2\right)=\left(2\right)\left(1\right)\Gamma \left(1\right)$ . Whenever $t=n$ , a positive integer, we have, be repeated application of $\Gamma \left(t\right)=\left(t-1\right)\Gamma \left(t-1\right)$ , that $\Gamma \left(n\right)=\left(n-1\right)\Gamma \left(n-1\right)=\left(n-1\right)\left(n-2\right)...\left(2\right)\left(1\right)\Gamma \left(1\right).$

However, $\Gamma \left(1\right)=\underset{0}{\overset{\infty }{\int }}{e}^{-y}dy=1.$

Thus when n is a positive integer, we have that $\Gamma \left(n\right)=\left(n-1\right)!$ ; and, for this reason, the gamma is called the generalized factorial .

Incidentally, $\Gamma \left(1\right)$ corresponds to 0!, and we have noted that $\Gamma \left(1\right)=1$ , which is consistent with earlier discussions.

## Summarizing

The random variable x has a gamma distribution if its p.d.f. is defined by

$f\left(x\right)=\frac{1}{\Gamma \left(\alpha \right){\theta }^{\alpha }}{x}^{\alpha -1}{e}^{-x/\theta },0\le x<\infty .$

Hence, w , the waiting time until the $\alpha$ th change in a Poisson process, has a gamma distribution with parameters $\alpha$ and $\theta =1/\lambda$ .

Function $f\left(x\right)$ actually has the properties of a p.d.f., because $f\left(x\right)\ge 0$ and

$\underset{-\infty }{\overset{\infty }{\int }}f\left(x\right)dx=\underset{0}{\overset{\infty }{\int }}\frac{{x}^{\alpha -1}{e}^{-x/\theta }}{\Gamma \left(\alpha \right){\theta }^{\alpha }}dx,$ which, by the change of variables $y=x/\theta$ equals

$\underset{0}{\overset{\infty }{\int }}\frac{{\left(\theta y\right)}^{\alpha -1}{e}^{-y}}{\Gamma \left(\alpha \right){\theta }^{\alpha }}\theta dy=\frac{1}{\Gamma \left(\alpha \right)}\underset{0}{\overset{\infty }{\int }}{y}^{\alpha -1}{e}^{-y}dy=\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha \right)}=1.$

The mean and variance are: $\mu =\alpha \theta$ and ${\sigma }^{2}=\alpha {\theta }^{2}$ .

Suppose that an average of 30 customers per hour arrive at a shop in accordance with Poisson process. That is, if a minute is our unit, then $\lambda =1/2$ . What is the probability that the shopkeeper will wait more than 5 minutes before both of the first two customers arrive? If X denotes the waiting time in minutes until the second customer arrives, then X has a gamma distribution with $\alpha =2,\theta =1/\lambda =2.$ Hence,

$p\left(X>5\right)=\underset{5}{\overset{\infty }{\int }}\frac{{x}^{2-1}{e}^{-x/2}}{\Gamma \left(2\right){2}^{2}}dx=\underset{5}{\overset{\infty }{\int }}\frac{x{e}^{-x/2}}{4}dx=\frac{1}{4}{\left[\left(-2\right)x{e}^{-x/2}-4{e}^{-x/2}\right]}_{5}^{\infty }=\frac{7}{2}{e}^{-5/2}=0.287.$

We could also have used equation with $\lambda =1/\theta$ , because $\alpha$ is an integer $P\left(X>x\right)=\sum _{k=0}^{\alpha -1}\frac{{\left(x/\theta \right)}^{k}{e}^{-x/\theta }}{k!}.$ Thus, with x =5, $\alpha$ =2, and $\theta =2$ , this is equal to

$P\left(X>x\right)=\sum _{k=0}^{2-1}\frac{{\left(5/2\right)}^{k}{e}^{-5/2}}{k!}={e}^{-5/2}\left(1+\frac{5}{2}\right)=\left(\frac{7}{2}\right){e}^{-5/2}.$

## Chi-square distribution

Let now consider the special case of the gamma distribution that plays an important role in statistics.

Let X have a gamma distribution with $\theta =2$ and $\alpha =r/2$ , where r is a positive integer. If the p.d.f. of X is
$f\left(x\right)=\frac{1}{\Gamma \left(r/2\right){2}^{r/2}}{x}^{r/2-1}{e}^{-x/2},0\le x<\infty .$
We say that X has chi-square distribution with r degrees of freedom, which we abbreviate by saying is ${\chi }^{2}\left(r\right)$ .

The mean and the variance of this chi-square distributions are

$\mu =\alpha \theta =\left(\frac{r}{2}\right)2=r$ and ${\sigma }^{2}=\alpha {\theta }^{2}=\left(\frac{r}{2}\right){2}^{2}=2r.$

That is, the mean equals the number of degrees of freedom and the variance equals twice the number of degrees of freedom.

In the fugure 2 the graphs of chi-square p.d.f. for r =2,3,5, and 8 are given.

the relationship between the mean $\mu =r$ , and the point at which the p.d.f. obtains its maximum.

Because the chi-square distribution is so important in applications, tables have been prepared giving the values of the distribution function for selected value of r and x ,

$F\left(x\right)=\underset{0}{\overset{x}{\int }}\frac{1}{\Gamma \left(r/2\right){2}^{r/2}}{w}^{r/2-1}{e}^{-w/2}dw.$

Let X have a chi-square distribution with r =5 degrees of freedom. Then, using tabularized values,

$P\left(1.145\le X\le 12.83\right)=F\left(12.83\right)-F\left(1.145\right)=0.975-0.050=0.925$

and $P\left(X>15.09\right)=1-F\left(15.09\right)=1-0.99=0.01.$

If X is ${\chi }^{2}\left(7\right)$ , two constants, a and b , such that $P\left(a , are a =1.690 and b =16.01.

Other constants a and b can be found, this above are only restricted in choices by the limited table.

Probabilities like that in Example 4 are so important in statistical applications that one uses special symbols for a and b . Let $\alpha$ be a positive probability (that is usually less than 0.5) and let X have a chi-square distribution with r degrees of freedom. Then ${\chi }_{\alpha }^{2}\left(r\right)$ is a number such that $P\left[X\ge {\chi }_{\alpha }^{2}\left(r\right)\right]=\alpha$

That is, ${\chi }_{\alpha }^{2}\left(r\right)$ is the 100(1- $\alpha$ ) percentile (or upper 100a percent point) of the chi-square distribution with r degrees of freedom. Then the 100 $\alpha$ percentile is the number ${\chi }_{1-\alpha }^{2}\left(r\right)$ such that $P\left[X\le {\chi }_{1-\alpha }^{2}\left(r\right)\right]=\alpha$ . This is, the probability to the right of ${\chi }_{1-\alpha }^{2}\left(r\right)$ is 1- $\alpha$ . SEE fugure 3 .

Let X have a chi-square distribution with seven degrees of freedom. Then, using tabularized values, ${\chi }_{0.05}^{2}\left(7\right)=14.07$ and ${\chi }_{0.95}^{2}\left(7\right)=2.167.$ These are the points that are indicated on Figure 3.

what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
Got questions? Join the online conversation and get instant answers!