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NOTEBOOK FOR SPATIAL DATA ANALYSIS Part I. Spatial Point Pattern Analysis ______________________________________________________________________________________ ________________________________________________________________________ ESE 502 A1-1 Tony E. Smith APPENDIX TO PART I In this Appendix, designated as A1 (appendices A2 and A3 are for Parts II and III, respectively), we shall again refer to equations in the text by section and equation number, so that (2.4.3) refers to expression (3) in section 2.4 of Part I. Also, references to previous expressions in this Appendix (A1), will be written the same way, so that (A1.1.3) refers to expression (3) of section 1 in Appendix A1. A1.1. Poisson Approximation of the Binomial This standard result appears in many elementary probability texts [such as Larsen and Marx (2001, p.247)]. Here one starts with the fundamental limit identity (A1.1.1) lim 1nxnxne that defines the exponential function. Given this relation, observe that since (A1.1.2) !(1)(1)()!(1)(1)!( )! !( )! !nnnnknknnnkknk knk k  it follows that expression (2.2.3) can be written as (A1.1.3) !() ()1!( )! () ()knknaC aCknk aR aR    ( 1) ( 1) () ()1!()()/() ()11 ()()11!()()knkkkknknnn nk aC aCnkaRaRnaR aCnn n k aC aCnn n k aR aR      But if we now evaluate expression (A1.1.3) at the sequence in (2.3.2) and recall that /( ) 0mmnaR , then in the limit we can replace / ( )mmnaR by  in the second factor. Moreover, since ( )/ 1mmnhn for all 0,1,.., 1hk , it also follows that the first factor in (A1.1.3) goes to one. In addition, the last factor also goes to one since () ()/() 0mmaR aC aR  . Hence by taking limits we see that (A1.1.4) !() ()lim 1!( )!() ()mknkmmmm mnaC aCkn k aR aR   NOTEBOOK FOR SPATIAL DATA ANALYSIS Part I. Spatial Point Pattern Analysis ______________________________________________________________________________________ ________________________________________________________________________ ESE 502 A1-2 Tony E. Smith [()] ()(1) lim 1 (1)!()mnkmmaC aCkaR ()[()] ()[/()]lim 1![()] ()lim 1![()]!mmnkmmmmnkmmkaCaC aC n aRknaC aCknaCek   A1.2. Distributional Properties of Nearest-Neighbor Distances under CSR Given that the nn-distance, D , for a randomly selected point has cdf (A1.2.1) 2() 1 Pr( ) 1dDFd D d e   By differentiating (A1.2.1) we obtain the probability density Df of D as (A1.2.2) 2() () 2dDDfd Fd de This distribution is thus seen to be an instance of the Rayleigh distribution (as for example in Johnson and Kotz, 1970, p.197). This distribution is closely related to the normal distribution, which can be used to calculate its moments. To do so, recall first that since ()0EX  for any normal random variable, 2~(0,)XN, it follows that the variance of X is simply its second moment, i.e., (A1.2.3) 2222var( ) ( ) ( ) ( )XEX EX EX  But since this normal density 22( ) exp( ) / 2xx is symmetric about zero, we then see that (A1.2.4) 22 2222 22/2 2/2 22001()222 22xxEXxe dx xe dx    Hence by setting 21/(2 )  so that 21/(2 )  , we obtain the identity (A1.2.5) 22012 11 1 142 4 22xxe dx      NOTEBOOK FOR SPATIAL DATA ANALYSIS Part I. Spatial Point Pattern Analysis ______________________________________________________________________________________ ________________________________________________________________________ ESE 502 A1-3 Tony E. Smith 2201(2 )2xxe dx  So to obtain the mean, ()ED , of D observe from (A1.2.2) and (A1.2.5) that (A1.2.6) 22200 01() () (2 ) 22xxDE D xf x dx x xe dx x e dx     To obtain the variance, var( )D , of D we first calculate the second moment, 2()ED. To do so, observe first from the integration-by-parts identity (as for example in Bartle, 1975, Section 22) that for any differentiable functions, ()fx and ()gx on [0, ) , (A1.2.7) 00() () ()() (0)(0) lim ()()xfxg xdx f xgxdx f g f xgx whenever these integrals and limits exist. Hence letting 2()fxx and 2()xgx e , it follows that (A1.2.8) 22 22 200(2 ) (2)( ) (0) lim 0xx xxxxedx xedx xe       But by (A1.2.2) we have, (A1.2.9) 220001() 1 2 1 2xxDf x dx xe dx xe dx    which together with (A1.2.8) now shows that (A1.2.10) 2222 200 01() () (2 ) 2xxDE D x f x dx x xe dx xe dx    Finally, by combining (A1.2.6) and (A1.2.10) we obtain1 (A1.2.11) 22211 114var() ( )[()]442DED ED   1 I am indebted to Christopher Jodice for pointing out several errors in my original posted derivations of these moments.NOTEBOOK FOR SPATIAL DATA ANALYSIS Part I. Spatial Point Pattern Analysis ______________________________________________________________________________________ ________________________________________________________________________ ESE 502 A1-4 Tony E. Smith A1.3. Distribution of Skellam’s Statistic under


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