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THE GAMMA AND CHI-SQUARE DISTRIBUTIONS

Module by: Ewa Paszek. E-mail the author

Summary: Gamma Distribution and Chi-Square Distribution

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GAMMA AND CHI-SQUARE DISTRIBUTIONS

In the (approximate) Poisson process with mean λ λ , we have seen that the waiting time until the first change has an exponential distribution. Let now W denote the waiting time until the α α th change occurs and let find the distribution of W. The distribution function of W ,when w0 w0 is given by

F( w )=P( Ww )=1P( W>w )=1P( fewer_than_α_changes_occur_in_[ 0,w ] ) =1 k=0 α1 ( λw ) k e λw k! , F( w )=P( Ww )=1P( W>w )=1P( fewer_than_α_changes_occur_in_[ 0,w ] ) =1 k=0 α1 ( λw ) k e λw k! ,

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

F'( w )=λ e λw e λw k=1 α1 [ k ( λw ) k1 λ k! ( λw ) k λ k! ] =λ e λw e λw [ λ λ ( λw ) α1 ( α1 )! ] = λ ( λw ) α1 ( α1 )! e λw . F'( w )=λ e λw e λw k=1 α1 [ k ( λw ) k1 λ k! ( λw ) k λ k! ] =λ e λw e λw [ λ λ ( λw ) α1 ( α1 )! ] = λ ( λw ) α1 ( α1 )! e λw .

Gamma Distribution

Definition 1:
1. If w<0 w<0 , then F( w )=0 F( w )=0 and F'( w )=0 F'( w )=0 , a p.d.f. of this form is said to be one of the gamma type, and the random variable W is said to have the gamma distribution.
2. The gamma function is defined by Γ( t )= 0 y t1 e y dy ,0<t. Γ( t )= 0 y t1 e y dy ,0<t.

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

Γ( t )= [ y t1 e y ] 0 + 0 ( t1 ) y t2 e y dy =( t1 ) 0 y t2 e y dy= ( t1 )Γ( t1 ). Γ( t )= [ y t1 e y ] 0 + 0 ( t1 ) y t2 e y dy =( t1 ) 0 y t2 e y dy= ( t1 )Γ( t1 ).

Example 1

Let Γ( 6 )=5Γ( 5 ) Γ( 6 )=5Γ( 5 ) and Γ( 3 )=2Γ( 2 )=( 2 )( 1 )Γ( 1 ) Γ( 3 )=2Γ( 2 )=( 2 )( 1 )Γ( 1 ) . Whenever t=n t=n , a positive integer, we have, be repeated application of Γ( t )=( t1 )Γ( t1 ) Γ( t )=( t1 )Γ( t1 ) , that Γ( n )=( n1 )Γ( n1 )=( n1 )( n2 )...( 2 )( 1 )Γ( 1 ). Γ( n )=( n1 )Γ( n1 )=( n1 )( n2 )...( 2 )( 1 )Γ( 1 ).

However, Γ( 1 )= 0 e y dy=1 . Γ( 1 )= 0 e y dy=1 .

Thus when n is a positive integer, we have that Γ( n )=( n1 )! Γ( n )=( n1 )! ; and, for this reason, the gamma is called the generalized factorial.

Incidentally, Γ( 1 ) Γ( 1 ) corresponds to 0!, and we have noted that Γ( 1 )=1 Γ( 1 )=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( x )= 1 Γ( α ) θ α x α1 e x/θ ,0x<. f( x )= 1 Γ( α ) θ α x α1 e x/θ ,0x<.
(1)

Hence, w, the waiting time until the α α th change in a Poisson process, has a gamma distribution with parameters α α and θ=1/λ θ=1/λ .

Function f( x ) f( x ) actually has the properties of a p.d.f., because f( x )0 f( x )0 and

f( x )dx= 0 x α1 e x/θ Γ( α ) θ α dx, f( x )dx= 0 x α1 e x/θ Γ( α ) θ α dx, which, by the change of variables y=x/θ y=x/θ equals

0 ( θy ) α1 e y Γ( α ) θ α θdy= 1 Γ( α ) 0 y α1 e y dy = Γ( α ) Γ( α ) =1. 0 ( θy ) α1 e y Γ( α ) θ α θdy= 1 Γ( α ) 0 y α1 e y dy = Γ( α ) Γ( α ) =1.

The mean and variance are: μ=αθ μ=αθ and σ 2 =α θ 2 σ 2 =α θ 2 .

Example 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 λ=1/2 λ=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 α=2,θ=1/λ=2. α=2,θ=1/λ=2. Hence,

p( X>5 )= 5 x 21 e x/2 Γ( 2 ) 2 2 dx= 5 x e x/2 4 dx= 1 4 [ ( 2 )x e x/2 4 e x/2 ] 5 = 7 2 e 5/2 =0.287. p( X>5 )= 5 x 21 e x/2 Γ( 2 ) 2 2 dx= 5 x e x/2 4 dx= 1 4 [ ( 2 )x e x/2 4 e x/2 ] 5 = 7 2 e 5/2 =0.287.

We could also have used equation with λ=1/θ λ=1/θ , because α α is an integer P( X>x )= k=0 α1 ( x/θ ) k e x/θ k! . P( X>x )= k=0 α1 ( x/θ ) k e x/θ k! . Thus, with x=5, α α =2, and θ=2 θ=2 , this is equal to

P( X>x )= k=0 21 ( 5/2 ) k e 5/2 k! = e 5/2 ( 1+ 5 2 )=( 7 2 ) e 5/2 . P( X>x )= k=0 21 ( 5/2 ) k e 5/2 k! = e 5/2 ( 1+ 5 2 )=( 7 2 ) e 5/2 .

Chi-Square Distribution

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

Definition 2:
Let X have a gamma distribution with θ=2 θ=2 and α=r/2 α=r/2 , where r is a positive integer. If the p.d.f. of X is
f( x )= 1 Γ( r/2 ) 2 r/2 x r/21 e x/2 ,0x<. f( x )= 1 Γ( r/2 ) 2 r/2 x r/21 e x/2 ,0x<.
(2)
We say that X has chi-square distribution with r degrees of freedom, which we abbreviate by saying is χ 2 ( r ) χ 2 ( r ) .

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

μ=αθ=( r 2 )2=r μ=αθ=( r 2 )2=r and σ 2 =α θ 2 =( r 2 ) 2 2 =2r. σ 2 =α θ 2 =( r 2 ) 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.

Note:

the relationship between the mean μ=r μ=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( x )= 0 x 1 Γ( r/2 ) 2 r/2 w r/21 e w/2 dw . F( x )= 0 x 1 Γ( r/2 ) 2 r/2 w r/21 e w/2 dw .
(3)

Example 3

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

P( 1.145X12.83 )=F( 12.83 )F( 1.145 )=0.9750.050=0.925 P( 1.145X12.83 )=F( 12.83 )F( 1.145 )=0.9750.050=0.925

and P( X>15.09 )=1F( 15.09 )=10.99=0.01. P( X>15.09 )=1F( 15.09 )=10.99=0.01.

Example 4

If X is χ 2 ( 7 ) χ 2 ( 7 ) , two constants, a and b, such that P( a<X<b )=0.95 P( a<X<b )=0.95 , 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 α α 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 χ α 2 ( r ) χ α 2 ( r ) is a number such that P[ X χ α 2 ( r ) ]=α P[ X χ α 2 ( r ) ]=α

That is, χ α 2 ( r ) χ α 2 ( r ) is the 100(1- α α ) percentile (or upper 100a percent point) of the chi-square distribution with r degrees of freedom. Then the 100 α α percentile is the number χ 1α 2 ( r ) χ 1α 2 ( r ) such that P[ X χ 1α 2 ( r ) ]=α P[ X χ 1α 2 ( r ) ]=α . This is, the probability to the right of χ 1α 2 ( r ) χ 1α 2 ( r ) is 1- α α . SEE fugure 3.

Example 5

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

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