MASSACHUSETTS INSTITUTE OF TECHNOLOGY 6.265/15.070J Fall 2013 Lecture 4 9/16/2013. Applications of the large deviation technique

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1 MASSACHUSETTS ISTITUTE OF TECHOLOGY 6.265/5.070J Fall 203 Lecure 4 9/6/203 Applicaios of he large deviaio echique Coe.. Isurace problem 2. Queueig problem 3. Buffer overflow probabiliy Safey capial for a isurace compay Cosider some isurace compay which eeds o decide o he amou of capial S 0 i eeds o hold o avoid he cash flow issue. Suppose he isurace premium per moh is a fixed (o-radom quaiy C > 0. Suppose he claims are i.i.d. radom variable A 0 for he ime =,2,. The he capial L = a ime is S = S 0 + (C A. The compay wishes o avoid he siuaio where he cash flow S is egaive. Thus i eeds o decide o he capial S 0 so ha P(,S 0 is small. Obviously his ivolves a radeoff bewee he smalless ad he amou S 0. Le us assume ha upper boud δ = 0.00, amely 0.% is accepable (i fac his is prey close o he bakig regulaio sadards. We have P(,S 0 = P(mi S 0 + (C A 0 = = P(max (A C S 0 = L = If E[A ] C, we have P(max (A C S 0 =. Thus, he ieresig case is E[A ] < C (egaive drif, ad he goal is o deermie he sarig capial S 0 such ha P(max (A C S 0 δ. =

2 2 Buffer overflow i a queueig sysem The followig model is a varia of a classical so called GI/GI/ queueig sysem. I applicaio o commuicaio sysems his queueig sysem cosiss of a sigle server, which processes some C > 0 umber of commuicaio packes per ui of ime. Here C is a fixed deermiisic cosa. Le A be he radom umber packes arrivig a ime, ad Q be he queue legh a ime (asume Q 0 =0. By recursio, we have ha Q = max(q + A C,0 = max(q 2 + A + A 2C,A C,0 = max ( (A k C,0 k= oice, ha i disribuioal sese we have Q = max max (A k C,0 k= I seady sae, i.e. =, we have Q = max max k= (A k C,0 Our goal is o desig he size of he queue legh sorage (buffer B, so ha he likelihood ha he umber of packes i he queue exceeds B is small. I commuicaio applicaio his is impora sice every packe o fiig io he buffer is dropped. Thus he goal is o fid buffer size B > 0 such ha P(Q B δ P(max (A k C B δ k= If E[A ] C, we have P(Q B =. So he ieresig case is E[A ] < C (egaive drif. 3 Buffer overflow probabiliy We see ha i boh siuaios we eed o esimae P(max k= We will do his asympoiclly as B. (A k C B. 2

3 Theorem. Give a i.i.d. sequece A 0 for ad C > E[A ]. Suppose M(θ = E[exp(θA] <, for some θ [0,θ 0. The lim log P(max (A k C B = sup{θ > 0 : M(θ < exp(θc} B B k= Observe ha sice A is o-egaive, he MGF E[exp(θA ] is fiie for θ < 0. Thus i is fiie i a ierval coaiig θ = 0, ad applyig he resul of Lecure 2 we ca ake he derivaive of MGF. The d d M(θ = E[A], exp(θc = C dθ dθ θ=0 θ=0 Sice E[A ] < C, he here exiss small eough θ so ha M(θ < exp(θc, M ( θ exp( θc θ * θ Figure : Illusraio for he exisace of θ such ha M(θ < exp(θc. ad hus he se of θ > 0 for which his is he case is o-empy. (see Figure. The heorem says ha roughly speakig P(max k= (A k C B exp( θ whe B is large. Thus give δ selec B such ha exp( θ B δ, ad we ca se B = log. θ δ B, 3

4 Example. Le A be a radom variable uiformly disribued i [0,a] ad C = 2. The, he mome geeraig fucio of A is M(θ = a 0 exp(θa exp(θa d = θa The exp(θa sup{θ > 0 : M(θ exp(θc} = sup{θ > 0 : exp(2θ} θa Case : a = 3, we have θ = sup{θ > 0 : exp(3θ 3θexp(2θ}, i.e. θ = Case 2: a = 4, we have ha {θ > 0 : exp(3θ 3θexp(2θ} = Ø sice E[A] = 2 = C. Case 3: a = 2, we have ha {θ > 0 : exp(3θ L 3θexp(2θ} = R + ad hus θ =, which implies ha P(max k= (A k C B = 0 by heorem. Proof of Theorem. We will firs prove a upper boud ad he a lower boud. Combiig hem yields he resul. For he upper boud, we have ha P(max (A k C B P( (A k C B k= = k= B = P( A k C + = k= B exp( (θ(c + log M(θ (θ > 0 = = exp( θb exp( (θc log M(θ Fix ay θ such ha θc log M(θ, he iequaliy above gives exp( θb exp( (θc log M(θ 0 = exp( θb[ exp( (θc log M(θ] Simplificaio of he iequaliy above gives log P(max (A log([ exp( (θc log M(θ] k C B θ+ B B k= lim sup log P(max B B k= (A k C B θ for θ : M(θ < exp(θc 4

5 ex, we will derive he lower boud. P(max Fix a > 0, he The, we have (A k C B P( (A k C B, k= k= B P(max (A k C B P( (A k C B k= k= lim if log P(max (A k C B B B k= B B lim if log P (A k C B B k= B B = lim if log P (A k C B B k= = lim if log P( (A k C k= = lim if log P( A k C + k= if I(x (by Cramer s heorem. x>c+ B B P (A k C k= if if I(x (sice we ca choose a arbirary posiive. ( >0 x>c+ We claim ha if if I(x = if I(C + >0 x>c+ >0 Ideed, le x = if x : I(x = (possibly x =. If x C, he I(C + =. Suppose C < x. If is such ha C+ x, he if I(x = x>c+. Therefore i does o make sese o cosider such. ow for c + < x, 5

6 we have I is covex o-decreasig ad fiie o [E[A ],x. Therefore i is coiuous o [E[A ],x, which gives ha ad he claim follows. Thus, we obai if I(x = I(C + x>c+ lim if log P(max (A k C B if I(C + B B >0 k= Exercise i HW 2 shows ha sup{θ > 0 : M(θ < exp(cθ} = if >0 I(C +. 6

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