Probability for mathematicians INDEPENDENCE TAU

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1 Probbility for mthemticis INDEPENDENCE TAU Cotets 2 Cetrl limit theorem 21 2 Itroductio b Covolutio c The iitil distributio does ot mtter d From smooth fuctios to idictors Cetrl limit theorem 2 Itroductio Discrete probbility spces re eough here s log s ll rdom vribles re discrete; to this ed use trigle rrys. Let X 1,X 2,... be idepedet ideticlly distributed rdom vribles, d S = X 1 + +X. 21 Theorem. 1 Let EX 1 = 0 d EX 2 1 = 1. The P < S < b 1 2π b wheever b. e x2 /2 dx s Clerly, the De Moivre-Lplce Theorem 120 is specil cse. More th 10 proofs re well-kow. Some use Stirlig formul. Some use Browi motio. Some prove covergece to the orml distributio. Some prove first covergece to some distributio, d the idetify it. Momet method: first, fid lim E S k ssumig ll momets fiite otherwise, tructe; the pproximte the idictor of itervl by polyomils. Fourier trsform chrcteristic fuctio : first, lim E exp iλ S = exp λ2 ; the pproximte the idictor of itervl by trigoometric sums. 2 Smooth test fuctios Lideberg: first, Ef S Ef S 0 s for f C 3 ; the pproximte the idictor of itervl by such smooth fuctios. This will be doe here. 1 [KS, Sect. 10.1, Th. 10.5]; [D, Sect. 2.4, Theorem 4.1].

2 Probbility for mthemticis INDEPENDENCE TAU b Covolutio The covolutio ν f of probbility distributio ν o R d bouded cotiuous fuctio f : R R is fuctio R R defied by 1 ν fx = fx+yνdy. For discrete ν the covolutio is lier combitio of shifts. I geerl it my be thought of s itegrl combitio of shifts. Probbilisticlly, P X f = Ef+X. 2b1 Lemm. Iff isboudeddcotiuous 1 thelsoµ f is, d µ f f. Here d below the orm is supreml rther th L 2 : f = sup fx. x R Proof. Boudedess: Ef+X sup f. Cotiuity: if the f + x f + x poitwise, thus Ef + X Ef + X by the bouded covergece theorem. For idepedet X,Y we hve P X+Y f = P Y P X f it mes, P Y P X f, sice P X+Y f = Ef+X +Y = f+x+yp X dxp Y dy = f+x+yp X dx P Y dy = P X f+yp Y dy = P Y P X f. We defie the covolutio of two probbility distributios µ, ν byµ νb = µ ν {x,y : x+y B}, the P X+Y = P X P Y for idepedet X,Y, d we my iterpret P Y P X f s P Y P X f eqully well. Covolutio for discrete: P X f = p X xf+x; x p X+Y = X xp Y y = x,y:x+y=p p X xp Y x. x 1 ThedefiitiogeerlizesesilytofiitesigedmesuresdboudedBorelfuctios, but we do ot eed it. 1 Well, it is required by the defiitio bove...

3 Probbility for mthemticis INDEPENDENCE TAU Covolutio for bsolutely cotiuous: P X f = p X xf+xdx; p X+Y = p X xp Y xdx. Some exmples: Biomm,p Biom,p = Biomm+,p, N 1,σ 2 1 N 2,σ 2 2 = N 1 + 2,σ 2 1 +σ 2 2. biomil orml The ltter equlity c be checked by itegrtio, or obtied from the former by limitig procedure, but better ote tht the stdrd two-dimesiol orml distributio N0,1 N0,1 hs the desity 2 1 e x2 /2 1 e y2 /2 = 1 2π 2π 2π e x2 +y2 /2 ivrit uder rottios; thus, X cosα+y siα N0,1 for ll α. 2b2 Lemm. If f hs bouded d cotiuous derivtive, the lso µ f hs, d µ f = µ f. Proof. Wehveboudedcotiuousg stisfyigf+b = f+ b gxdx. Thus, µ f+b = = = µ f+ b b+y f+b+yµdy = f+y+ +y b f+yµdy+ gx+ydx µdy = gx+yµdy dx = µ f+ b gx dx µdy = µ gxdx. The sme holds for f d f. 3 2 I dditio, itegrtigit i polrcoorditesweget 1 2π 0 e r2 /2 rdr 2π 0 dϕ = 1, which shows tht 1/ 2π is the right coefficiet for the desity of N0,1. See lso Proof of Ad so o, of course, but we eed oly three derivtives.

4 Probbility for mthemticis INDEPENDENCE TAU c The iitil distributio does ot mtter Let µ, ν be two probbility distributios o R stisfyig xµdx = xνdx = 0, x 2 µdx = x 2 νdx = 1. We cosider idepedet rdom vribles X 1,...,X distributed µ, d idepedet rdom vribles Y 1,...,Y distributed ν. Note tht EX 1 = EY 1 = 0 d EX1 2 = EY1 2 = 1. 2c1 Propositio. If f,f,f,f re cotiuous d bouded o R the Y1 + +Y Ef Ef 0 s. The proof will be give fter corollry. 2c2 Corollry. Ef 1 + 2π fxe x2 /2 dx. Proof of the corollry. Let Y 1 be orml N0,1, the Y 1 + +Y is lso orml, thus Y1 + +Y Ef = 1 + 2π fxe x2 /2 dx for ll. We strt provig the propositio. We hve +bx+cx 2 µdx = +bx+cx 2 νdx for ll,b,c R. Similrly, +bx+cx 2 µ dx = +bx+cx 2 ν dx; here d below µ is the distributio of X 1 /, d ν of Y 1 / ; tht is, f x µdx = f dµ d the sme for ν. These µ,ν re useful, sice 2c3 Ef = µ µ f0 = µ f0, d the sme for Y d ν.

5 Probbility for mthemticis INDEPENDENCE TAU c4 Lemm. There exist ε 0 such tht for every f s i 2c1 d every, ε f dµ f dν f + f. 2c5 Remrk. These ε deped o µ,ν but ot f. If µ,ν hve third momets the moreover 1 f dµ f dν E X f 1 3 +E Y 1 3. Proof of the lemm. We defie g by fx = f0+f 0x+ 1 2 f 0x 2 +gx; g is cotiuous but ot bouded; gx f 1 6 x 3. We hve f gdµ = f gdν, therefore f dµ f dν g dµ + g dν, which leds immeditely to 2c5, but we eed rgumet tht does ot require the third momets. We ote tht 1 2 f 0x 2 + gx 1 2 f x 2, therefore gx f x 2, d split the itegrl: 1 g x µdx g dµ = f 1 x 3 µdx x 6 } 1/12 {{} O 1/12 1/2 3 =O 5/4 ε the sme holds for g dν. + f x 2 µdx x > } 1/12 {{} o1/ f + f where 1 ε = mx 6 1/4, x 2 µdx ; x > 1/12 1 The expoet 1/12 my be replced with y other umber betwee 0 d 1/6.

6 Probbility for mthemticis INDEPENDENCE TAU Proof of Propositio 2c1. By 2c3 it is sufficiet to prove tht µ f0 ν f0 0. Applyig Lemm 2c4 to shifted fuctio x f+x we get µ f ν f ε f + f. We tur µ µ f ν f = ito ν grdully: 1 k=0 µ k ν k f µ k 1 ν k+1 f = 1 = k=0 µ k 1 µ f k ν f k, where f k = ν k f. Now, f k f, f k f, d µ k ; thus, µ s. 1 f ν f k=0 ε f + f = ε f + f 0 2d From smooth fuctios to idictors 2d1 Lemm. There exists fuctio ϕ : R R hvig three bouded derivtives d such tht ϕx = 0 for ll x 1, ϕx = 1 for ll x 0. Proof. The fuctio ψx = 1 x 2 4 for x 1, otherwise 0, hs two i fct, three cotiuous derivtives. We tke ϕx = 1 2x+1 ψtdt where c c = ψtdt. Let X 1,...,X be s i 2c1. By 2c2, for every R d ε > 0, 1 Eϕ 1 1 ϕ ε 2π ε x e x2 /2 dx s. Tkig ito ccout tht P we get lim supp Eϕ 1 ε 1 1 ϕ 2π ε x e x2 /2 dx.

7 Probbility for mthemticis INDEPENDENCE TAU Theright-hdsidecovergesto 1 2π e x2 /2 dxsε 0. Thus, limsupp X 1 + +X P ξ whereξ N0,1; orequivletly, limifp X X < P ξ <. Similrly, limsupp X 1 X P ξ, thtis, limsupp X 1 + +X P ξ, or equivletly, limsupp X 1 + +X P ξ. We hve P ξ < limifp X 1 + +X < limsupp X 1 + +X P ξ = P ξ <, therefore lim P X 1 + +X < = P ξ <.