GAMMA AND CHI-SQUARE DISTRIBUTIONS In the (approximate) Poisson process with mean 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 th change occurs and let find the distribution of W. The distribution function of W ,when w≥0 is given by F( w )=P( W≤w )=1−P( W>w )=1−P( fewer_than_α_changes_occur_in_[ 0,w ] ) =1− ∑ k=0 α−1 ( λw ) k e −λw k! , since the number of changes in the interval [ 0,w ] has a Poisson distribution with mean λw . Because W is a continuous-type random variable, 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 α−1 [ k ( λw ) k−1 λ k! − ( λw ) k λ k! ] =λ e −λw − e −λw [ λ− λ ( λw ) α−1 ( α−1 )! ] = λ ( λw ) α−1 ( α−1 )! e −λw .

Gamma Distribution If w<0 , then F( w )=0 and 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. The gamma function is defined by Γ( t )= ∫ 0 ∞ y t−1 e −y dy ,0<t. This integral is positive for 0<t , 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 Γ( t )= [ − y t−1 e −y ] 0 ∞ + ∫ 0 ∞ ( t−1 ) y t−2 e −y dy =( t−1 ) ∫ 0 ∞ y t−2 e −y dy= ( t−1 )Γ( t−1 ). Let Γ( 6 )=5Γ( 5 ) and Γ( 3 )=2Γ( 2 )=( 2 )( 1 )Γ( 1 ) . Whenever t=n , a positive integer, we have, be repeated application of Γ( t )=( t−1 )Γ( t−1 ) , that Γ( n )=( n−1 )Γ( n−1 )=( n−1 )( n−2 )...( 2 )( 1 )Γ( 1 ). However, Γ( 1 )= ∫ 0 ∞ e −y dy=1 . Thus when n is a positive integer, we have that Γ( n )=( n−1 )! ; and, for this reason, the gamma is called the generalized factorial. Incidentally, Γ( 1 ) corresponds to 0!, and we have noted that 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 α−1 e −x/θ ,0≤x<∞. Hence, w, the waiting time until the α th change in a Poisson process, has a gamma distribution with parameters α and θ=1/λ . Function f( x ) actually has the properties of a p.d.f., because f( x )≥0 and ∫ −∞ ∞ f( x )dx= ∫ 0 ∞ x α−1 e −x/θ Γ( α ) θ α dx, which, by the change of variables y=x/θ equals ∫ 0 ∞ ( θy ) α−1 e −y Γ( α ) θ α θdy= 1 Γ( α ) ∫ 0 ∞ y α−1 e −y dy = Γ( α ) Γ( α ) =1. The mean and variance are: μ=αθ and σ 2 =α θ 2 .

Gamma Distribution The c.d.f. graph. The p.d.f. graph. The p.d.f. and c.d.f. graphs of the Gamma Distribution.

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 . 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. Hence, p( X>5 )= ∫ 5 ∞ x 2−1 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 , because α is an integer P( X>x )= ∑ k=0 α−1 ( x/θ ) k e −x/θ k! . Thus, with x=5, α =2, and θ=2 , this is equal to P( X>x )= ∑ k=0 2−1 ( 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. Let X have a gamma distribution with θ=2 and 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/2−1 e −x/2 ,0≤x<∞. We say that X has chi-square distribution with r degrees of freedom, which we abbreviate by saying is χ 2 ( r ) . The mean and the variance of this chi-square distributions are μ=αθ=( r 2 )2=r and 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.

The p.d.f. of chi-square distribution for degrees of freedom r=2,3,5,8. the relationship between the mean 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/2−1 e −w/2 dw . Let X have a chi-square distribution with r =5 degrees of freedom. Then, using tabularized values, P( 1.145≤X≤12.83 )=F( 12.83 )−F( 1.145 )=0.975−0.050=0.925 and P( X>15.09 )=1−F( 15.09 )=1−0.99=0.01. If X is χ 2 ( 7 ) , two constants, a and b, such that 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 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 ) is a number such that P[ X≥ χ α 2 ( r ) ]=α That is, χ α 2 ( r ) is the 100(1- α ) percentile (or upper 100a percent point) of the chi-square distribution with r degrees of freedom. Then the 100 α percentile is the number χ 1−α 2 ( r ) such that P[ X≤ χ 1−α 2 ( r ) ]=α . This is, the probability to the right of χ 1−α 2 ( r ) is 1- α . SEE fugure 3. Let X have a chi-square distribution with seven degrees of freedom. Then, using tabularized values, χ 0.05 2 ( 7 )=14.07 and χ 0.95 2 ( 7 )=2.167. These are the points that are indicated on Figure 3.

χ 0.05 2 ( 7 )=14.07 and χ 0.95 2 ( 7 )=2.167.