An Introduction to Integrtion nd Probbility Density Functions Kevin Mitchell Deprtment of Mthemtics nd Computer Science Hobrt nd Willim Smith Colleges Genev, NY USA 56 Mrch 999 c999 by Kevin Mitchell All rights reserved Do not reproduce without permission of the uthor This nd other relted mterils re vilble on-line t: http://scienceholycrossedu/deprtments/biology/kprestwi/behvior/ess/ess index frmsethtml This is supplement to Kenneth Prestwich's mteril \The Mixed ESS Solution to the Wr of Attrition"
FINDING EXPECTED LIFETIME NET BENEFITS Finding Expected Lifetime Net Benets Recll the sitution under discussion t \The Mixed ESS Solution to the Wr of Attrition" website Benets re only obtined by the focl strtegist when she wins ie, when the focl strtegist is willing to py higher cost thn her opponent from the mix (x <m, where m is the cost the focl strtegist will py) Net Benet to x(x = m) in win = V, x; () where V is the resource vlue nd x is the cost the opponent from mix is willing to py Unfortuntely, () is not sucient for our needs The complexity of the wr of ttrition intervenes! Recll tht the mix is composed of n innite number of component strtegies x(x = m) only fces one of these supporting strtegies in ny given contest Thus, () only describes the net gin in one specic contest You should relize tht this prticulr contest will probbly be quite rre given the mny dierent strtegists tht x(x = m) could fce from the mix Thus, one prticulr contest nd its benets will hve little if ny importnt lifetime eect on x(x = m)'s tness Single contests cnnot describe the net benet tht the focl supporting strtegy expects to gin from lrge number ( lifetime) of contests To get n ccurte mesurement of lifetime net gins, we need to tke into ccount ll types (costs) of contests tht x(x = m) will win (ie, those where the opposing strtegy x is t most m) nd the probbility of ech Net Benet = X xm [(V, x) (Probbility of fcing x)]: () Note tht this is n innite sum becuse there re n innite number of dierent costs (strtegies) x between nd m To hndle this type of sum, we will need to use clculus, which wenow briey consider We will return the question of lifetime net benets once we hve introduced the pproprite clculus techniques An Introduction to Integrtion The probbility of fcing prticulr strtegy x is determined by probbility density function or pdf We will illustrte the ide by exmining number of dierent situtions Sitution : Suppose tht there re n innite number of strtegies x tht plyers dopt with < x Assume further tht ech strtegy x with <x :5 is just s likely to occur s ny other in this intervl Finlly, ssume tht ech strtegy x with :5 <x istwice s likely to occur s ny strtegy x with <x :5 If we ssign positive probbility p to ny strtegy x in (; :5], then since there re n innite number of other eqully likely strtegies, they, too, would hve probbility p But then summing these probbilities would produce n innitely lrge vlue, not For this reson, the probbility tht ny prticulr strtegy x is encountered must be A better question to sk is wht the probbility is of fcing strtegy x tht flls within certin intervl For exmple, let p denote the probbility of encountering strtegy x tht lies in the intervl (; :5] I clim tht p ==3 Here's why: The strtegies in (:5; ] re twice s likely to be encountered s those in (; :5], so they hve totl probbility p, ie, twice the probbility of those in (; :5]
AN INTRODUCTION TO INTEGRATION But since the only strtegies re those between nd, we must hve =p +p =3p, or p ==3 Suppose, next, tht we wnted to know the probblity of encountering strtegy x with < x :5? Well, since ll the strtegies in (; :5] re eqully like, nd since their totl probbility is p = =3, nd since (; :5] is exctly hlf the intervl (; :5], we conclude tht the probblity tht <x :5 must be hlf of p, tht is p===6 In the sme wy you should be ble to show tht the probblity tht :75 <x is=3 Figure : A geometric reliztion or probbility density function p(x) for sitution =3 =3 5 The function p(x) in Figure gives geometricl representtion of sitution It hs the following properties p(x )=p(x ) for ny two strtegies x nd x in (; :5] becuse they re eqully likely p(x )=p(x ) for ny two strtegies x nd x in (:5; ] becuse these, too, re eqully likely But p(x )=p(x ) for ny strtegy x in (:5; ] nd x in (; :5] becuse strtegy x is twice s likely s x The region under the grph of p(x) on the intervl from to 5 is rectngle whose re is = Which is the probblity tht x is in (; :5] 3 3 The region under the grph of p(x) on the intervl from 5 to is rectngle whose re is = which is the probblity tht x is in (:5; ] 3 3 The region under the grph of p(x) over the entire intervl from to is + = which is 3 3 the probblity tht x is in (; ] More generlly, the re under p(x) on the intervl [; b] represents the probbility of strtegy x where x b The function p(x) is n exmple of probbility density function or pdf Such functions must stisfy two conditions: p(x) for ll x, the totl re under p(x) must be For given x the function p(x) mesures the compritive or reltive likelihood of strtegy x This is why the grph of p is twice s high on (:5; ] s it is on (; :5]
AN INTRODUCTION TO INTEGRATION 3 Siuttion : Assume tht the only strtegies x re those such tht x nd tht ll of these strtegies re eqully likely Since they re ll eqully likely, p(x) must be constnt on[; ], so tht the region under the grph is rectngle Since the re under the pdf over [; ] must be, the the constnt height (for the rectngle) must be becuse the bse is (see Figure ()) Figure : () The probbility density function p(x) when ll strtegies between nd re eqully likely The re under the grph from to b is (b, ) 5 5 b Since ll strtegies re eqully likely, the probbility tht prticulr strtegy x lies in [; ] is just or hlf the totl probblity Of course this corresponds exctly to the re under p(x) in Figure () over the intervl [; ]: it is rectngle In fct, if b is ny numberin[; ], then the probbility of encountering strtegy x in the intervl [;b] is the re under the curve from to b, tht is, b = b The probblity tht one encounters strtegy with witing time less thn or equl to x is clled the the cumultive probbility distribution of quitting times, nd is denoted by P (x) Note the uppercse P for the cumultive function nd the lowercse p for the density function In this exmple, P (x) is just the re of the rectngle from to x with height, so P (x) =x=; x : If we wnt to nd the probbility of encountering strtegy x between nd b, we could nd the re of the rectngle from to b Its bse would be b, nd the height b,, so its re is But this probblity cn be expressed more generlly using P (x) The probbility wewnt is just the re from to b minus the re from to s in Figure (b) But this is just Sitution 3: P (b), P () = b, = b, : Suppose gin tht the only possible strtegies x re those such tht x nd tht the pdf is p(x) =x= Notice tht this is legitimte pdf since p(x) nd the region under the grph of p is tringle whose re = (see Figure 3 ()) Wht is the cumultive distribution P (x) for this sitution? Well, P (x) is just the re under p from to x which is just tringle with bse x nd height p(x) (see Figure 3 (b)) Thus, P (x) = x p(x) = x x = x : For exmple, the probbility tht strtegy x lies in the intervl [; :5] is P (:5) = (:5) = :5 =:65;
AN INTRODUCTION TO INTEGRATION Figure 3: () The probbility density function p(x) =x= for x (b) P (x) =x = is the re under p from to x 5 5 5 5 x p(x) while the probbility tht x is in the intervl [:5; :5] is P (:5), P (:5) = (:5), (:5) = :5, :5 =:5: Sitution : Suppose now tht the possible strtegies re restricted to x nd tht p(x) =3x (see Figure ()) Clerly p(x), so to show tht p(x) ispdf,we need to show tht the re under this curve is But how dowe nd the re of curved region? Figure : () The probbility density function p(x) =3x for x (b) P (x) is the re under p from to x 3 3 3 We cn't directly use the re formul of rectngle to determine the re, but we cn use it indirectly to pproximte the re under the curve Suppose we divide the intervl [; ] into ve subintervls of equl width x =, =:: 5 Approximte the re under the curve onech subintervl by using rectngle whose height is determined by evluting p t the right-hnd endpoint of the subintervl (see Figure (b)) These ve right-hnd endpoints will be x =:, x =:, x 3 =:6, x =:8, nd x 5 = So in this
AN INTRODUCTION TO INTEGRATION 5 cse, the ve heights would be p(x )=p(:), p(x )=p(:), p(x 3 )=p(:6), p(x )=p(:8), nd p(x 5 )=p() Thus, pproximte re is 5X k= p(x k )x = p(:) :+p(:) :+p(:6) :+p(:8) :+p() : =:[p(:) + p(:) + p(:6) + p(:8) + p()] =:[3(:) + 3(:) + 3(:6) + 3(:8) + 3() ] =:[: + :8 + :8 + :9 + 3] =:(6:6) =:3: If we used ten rectngles insted ech with width x =: (see Figure (c)), the pproximtion is even better This time, the re is X k= p(x k )x = p(:) :+p(:) :++ p(:9) :+p() : =:[p(:) + p(:) + + p(:9) + p()] =:[3(:) + 3(:) + + 3(:9) + 3() ] =:(:55) =:55: The sme process using one-hundred rectngles (no drwing for this!) yields n pproximtion of 55 nd with one-thousnd rectngles the pproximtion is 55 It ppers tht these pproximtions re getting close to s the number of rectngles gets lrge Mthemticins dene the exct re under the curve to be the limit of this rectngulr pproximtion proccess s the number of rectngles n becomes innitely lrge, lim n! nx k= p(x k )x: We denote the limit of this summtion process more compctly s Z p(x) dx: This is red s, \The integrl of p(x) from to " The integrl sign, R, is n elongted S, reminder to us tht n integrl is relly sum The lower nd upper limits of integrtion (here nd, respectively) re the beginning nd ending points of the intervl where the sum is tking plce The expression p(x)dx is ment to remind us of p(x)x, the re of rectngle of height p(x) nd width x Think of p(x)dx s being the re of innitesimlly thin rectngle of height p(x) In our prticulr cse, with p(x) =3x, it ppers tht Z 3x dx =;
AN INTRODUCTION TO INTEGRATION 6 nd this is, in fct, correct The Fundmentl Theorem of Clculus tells us tht under certin circumstnces such integrls cn be esily evluted using functions clled ntiderivtives In fct, the cumultive distribution function is just n ntiderivtive ofp(x) In this prticulr situttion using clculus, P (b), tht is, the re under p(x) =3x over the intervl from to b, is given by the formul P (b) = 3x dx = b 3 : Since P (b) =b 3, then the cumultive distribution function is P (x) =x 3, which clculus students will recognize s n ntiderivtive ofp(x) =3x Using methods developed in integrl clculus, such ntiderivtives cn be found for wide vriety of functions Using the nottion of integrls, we cn express the probbility of encountering R strtegy x in b the intervl [; b] This is just the re under the curve p(x) on this intervl, or p(x) dx But we sw tht erlier tht this is just the re from to b minus the re from to (s in Figure (b)) so, p(x) dx = It is customry to use the symbol P (x) b p(x) dx, Z p(x) dx = P (b), P (): (3) to denote the dierence P (b), P () So we would write p(x) dx = P (x) b : In the cse of sitution, with p(x) =3x nd P (x) =x 3, the probbility tht x lies in the intervl [; b] is 3x dx = x 3 b = b3, 3 : sub For exmple, the probbility tht x lies in the intervl [:; :8] is Z :8 3x dx = : A more relistic sitution: 3x dx = x 3 :8 : =(:8) 3, (:) 3 =:5, :8 = :5: Generlly speking, strtegies re not restricted to nite intervls such s we hve used in the previous exmples Tht is, x cn tke tke on ny nonnegtive rel vlue, x Tht mens tht pdf p(x) must be dened on the innite intervl [; +), not just some nite intervl such s [; ] or [; ] On the other hnd, the re under this curve must still be since it represents the probbility of encountering some strtegy, tht is, Z p(x) dx =: R b Such expressions re evluted using limits We rst evlute the expression p(x) dx Then we see wht hppens to this expression s b gets innitely lrge The nottion for this is lim b! p(x) dx:
AN INTRODUCTION TO INTEGRATION 7 Here's simple exmple: Suppose tht the pdf ws p(x) ==(x +) for x> To show tht this is pdf, we need to show tht the re under the curve (see Figure 5) is Tht is we need to show tht Z dx =: (x +) Figure 5: The probbility density function p(x) ==(x +) for x The totl re under this innitely long curve is 3 It turns out tht the cumultive distribution function (or in clculus terms, n ntiderivtive for p(x) isp (x) =, Using this nd limits to evlute the integrl bove x+ Z dx = lim (x +) b! Now sb gets lrge pproches So b+ Z dx = lim (x +) b! dx = lim (x +), b + b!, = lim b! b x +, b +, (), : (5) +,, =(, ), (, )= + s expected Our excursion to clculus is now complete nd you my return return to the discussion of lifetime benets t Prestwich's website Clculus students note tht the generl ntiderivtive for p(x) = (x+) =(x +), is P (x) =,(x +), + c We must choose c so tht P () =, becuse the probbility tht n opponent quiting t cost less thn is