The geometric distribution - Math

1 The geometric distributionSo far, we have seen only examples of random variables that have a finitenumber of possible values. However, our rules of probability allow us toalso study random variables that have a countable [but possibly infinite]number of possible values. The word countable means that you can labelthe possible values as 1,2,.. A theorem of Cantor states that the numberof elements of the real line is uncountable. And so is the number ofelements of any nonempty closed/open interval. Therefore, the conditionthat a random variableXhas a countable number of possible values is say thatXhas thegeometric distributionwith parameter!:= 1 "ifP{X=#}="# 1!(#=1$2$%%%)%We have seen this distribution before. Here is one way it arises naturally:Suppose we toss a!-coin [ ,P(heads)=!]

2 ,P(tails)="=1 !] untilthe first heads arrives. IfXdenotes the number of tosses, thenXhas theGeometric(!) the geometric (!) distribution . ThenP{X 3}=P{X=1}+P{X=2}+P{X=3}=!+!"+!"2%Here is an alternative expression forP{X 3}:P{X 3}=1 P{X 4}=1 #=4"# 1!=1 ! &=3"&%Simple facts about geometric series then tell us thatP{X 3}=1 !("31 ")=1 "3$555612since!=1 ". More generally, if'is a positive integer 1, thenP{X '}=' #=1"# 1!=!' #=1"# 1=!' 1 &=0"&=!(1 "'1 ")=1 "'$since!=1 ".Example isE(X)whenXhas the geometric (!) distribution ? Well,clearly,E(X)= #=1#"# 1!=! #=1#"# 1%How can we calculate this? Note that#"# 1=((""#. Therefore a standardfact from calculus [term-by-term differentiation of an infinite series] tellsus thatE(X)=!((" #=1"#=!((" #=0"#=!(("(11 ")=!(1(1 ")2)=1!%Note that we benefited greatly from studying the problem for a generalvariable!))))))))

3 [as opposed to!=1/2, say, or some such choice]. In this way,we were able to use the rules of calculus in order to obtain the is Var(X)whenXhas the geometric (!) distribution ? Weknow thatE(X)=1/!. Therefore, it suffices to computeE(X2). ButE(X2)= #=1#"# 1!=! #=1#2"# 1%In order to compute this we can write the sum as #=1#2"# 1= #=1#(# 1)"# 1+ #=1#"# 1%We saw earlier that #=1#"# 1=((" #=0"#=1!2%Similarly, #=1#(# 1)"# 1=" #=1#(# 1)"# 2="(2("2 #=0"#=2"!3%Therefore, #=1#2"# 1=2"! 3+! 2, and henceE(X2)=2"!2+1!=2 !!2%Var(X)=2 !!2 1!2="!2%Equivalently, SD(X)= "/!.The Poisson distribution57 The negative binomial distributionThe negative binomial distribution is a generalization of the geometric [andnotthe binomial, as the name might suggest]. Let us fix an integer) 1; then we toss a!)))

4 -coin until the)th heads occur. LetX)denote thetotal number of 4(The negative binomial distribution ).What is the distributionofX)? First of all, the possible values ofX)are)$)+1$)+2$%%%. Nowfor all integers* ),P{X)=*}=P{The first* 1tosses have) 1heads, and the*th toss is heads}%Therefore,P{X)=*}=(* 1) 1)!) 1"(* 1) () 1) !=(* 1) 1)!)"* )(* ))%This looks a little bit like a binomial probability, except the variable ofthe mass function is*. This distribution is called thenegative binomialdistributionwith parameters()$!).Example (X))and Var(X)). LetG1denote the number oftosses required to obtain the first heads. Then defineG2to be the num-ber of additional tosses required to obtain the next heads ..G)to bethe number of additional tosses required to obtain the next [and)th]head. Then,G1$G2$%%%$G*are independent random variables each withthe Geomteric(!

5 distribution . Moreover,X)=G1+ +G)%Therefore,E(X))=E(G1)+ +E(G))=)!$andVar(X))=Var(G1)+ +Var(G))=)"!2%SD(X))= )"!%The Poisson distributionWe say that a random variableXhas thePoisson distribution with pa-rameter >0whenP{X='}=, ''!for'=0$1$2$%%%%Two questions arise: First, is the preceding a well-defined probability dis-tribution? And second, why this particular form?5812In order to answer the first question we have to prove that '=0, '/'!=1. But the Taylor s expansion of-(.)=,.tells us that, = '=0 ''!%Divide by, in order to find that '=0P{X='}=1, as one would next proposition [the law of rare events] shows how the Poissondistribution can be viewed as an approximation to the Binomial distributionwhen! 1/*.Proposition 1(Poisson).Suppose is a positive constant that does notdepend on*.

6 IfXhas the Binomial distribution with parameters*and /*, thenP{X='} , ''!for all'=0$%%%$*$provided that*is exact expression we want follows: For all'=0$%%%$*,P{X='}=(*')( *)'(1 *)* '=*(* 1) (* '+ 1)'!( *)'(1 *)* '%Because'is held fixed, we have*(* 1) (* '+ 1) *'as* .Therefore,P{X='} 1'! '(1 *)* ' 1'! '(1 *)*$since(1 *)' 1as* [again because'is fixed]. It suffices to provethat(1 *)* , as* %Equivalently,*ln(1 *) as* %But this is clear becauseln(1 .) 1 .if. 0,thanks to the Taylor expansion ofln(1 .).!Now because Poisson( )is approximately Binomial(*$ /*), one mightimagine that the expectation of a Poisson( )is approximately the expec-tation of Binomial(*$ /*)which is* /*= . And that the variance ofThe Poisson distribution59a Poisson( )is approximately the variance of Binomial(*$ /*)which is*( /*)(1 *) %Although the preceding argument is logically flawed,it does produced the correct the Poisson( ) distribution .

7 In order to computeE(X)weproceed as follows:EX= '=0', ''!= '=1', ''!= '=1, '(' 1)!= &=0, &+1&!= , &=0 &&!= %Similarly,E(X2)= '=0'2, ''!= '=1'2, ''!= '=1'(' 1), ''!+ '=1', ''!= '=1'(' 1), ''!+ [from the computation forEX].Now '=1'(' 1), ''!=, 2 '=1'(' 1) ' 2'!=, 2(2( 2 '=1 ''!=, 2(2( 2, = 2%Consequently,E(X2)= 2+ %Var(X)= SD(X)= %Theorem 1(A central limit theorem).LetX have a Poisson distributionwith parameter . Then the standardization ofX has approximately astandard normal distribution . That is, for all / 0 ,P{/ X 0} (0) (/)when is large%))))