Probability Theory. Exercise Sheet 4. ETH Zurich HS 2017

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1 ETH Zurich HS 2017 D-MATH, D-PHYS Prof. A.-S. Szita Coordiator Yili Wag Probability Theory Exercise Sheet 4 Exercise 4.1 Let X ) N be a sequece of i.i.d. rado variables i a probability space Ω, A, P ). Defie the two sequeces of rado variables Y ) N ad M ) N by Y := i 1 i X i ad M := ax 1 i X i a) Let X 1 be uiforly distributed o the iterval [0, 1]. Show that Y coverges i distributio to a expoetial rado variable Z with paraeter 1, i.e., the desity of Z is e x 1 [0, ) x), x R. b) Let X 1 be expoetially distributed with paraeter 1. Show that M log coverges i distributio to a rado variable Z with Gubel distributio, i.e. the desity of Z is e x exp e x ), x R. Solutio 4.1 a) Take y R. For y < 0, because the expoetial distributio is cocetrated o [0, ), we have that P [Y y] = 0 = P [Z y] for all N. Hece without loss of geerality we assue y 0. It follows that for all y: P [Y y] = P = 1 [ Y y ] = P ] P [ X i > y { } X i y = 1 P = 1 1 y ). Now let, we hece obtai that y li P [Y y] = 1 e y = e x 1 [0, ) x) dx. { X i > y } 1 / 7

2 b) Take y R. The for all e y we have that: P [M log y] = P [M y + log ] = P {X i y + log } = P [X i y + log ] = }{{} =1 e y log =1 e y 1 e y ). Now let, we hece obtai that y li P [M log y] = exp e y ) = e x exp e x ) dx. 2 / 7

3 Exercise 4.2 a) Let X ) N be a sequece of real rado variables covergig i probability to a rado variable X. Show that X ) N coverges to X i distributio. b) The coverse does ot hold i geeral. Istead, show that if the sequece X ) N coverges i distributio to a costat rado variable X = c, the X ) N coverges i probability to c. Solutio 4.2 a) We deote the distributio fuctios of X ad X by F ad F respectively. We have to show that li F y) = F y) for cotiuity poits y of F. So, let y R be a cotiuity poit of F ad let ε > 0. By the cotiuity of F i y there is a δ > 0 such that F y) ε F x) F y) + ε, x [y δ, y + δ]. 1) Sice the X coverge to X i probability, there is a N N such that Now, for N, ad so that Thus, P [ X X > δ ] ε, N. 2) F y) = P [X y] P [{X y + δ} { X X > δ}] P [X y + δ] + P [ X X > δ] 2) F y + δ) + ε 1) F y) + 2ε F y) = P [X y] P [{X y δ} \ { X X > δ}] P [X y δ] P [ X X > δ] 2) F y δ) ε 1) F y) 2ε, F y) 2ε F y) F y) + 2ε. F y) 2ε li if F y) li sup F y) F y) + 2ε. But this holds for all ε > 0, so we are doe. b) We assue that for a c R The costat c has distributio fuctio X c i distributio. 3) F x) = 1 [c, ), 3 / 7

4 which is cotiuous except i c. So we kow fro 3) 0 if z < c, F z) = P [X z] 1 if z > c. 4) We wat to show that for all ε > 0 li P [ X c ε] = 0. Now, so that P [ X c ε] = P [X c ε] + P [X c + ε], li P [ X c ε] li P [X c ε] + li P [X c + ε]. By 4), we have li P [X c ε] = 0 for all ε > 0. Furtherore P [X c + ε] = 1 P [X < c + ε] 1 P [ X c + ε ]. 2 Thus, li P [X c + ε] 1 li [X P c + ε ] 4) = / 7

5 Exercise 4.3 a) Let f be a ot ecessarily Borel-easurable) fuctio fro R to R. Show that the set of discotiuities of f, defied as is Borel-easurable. U f := { x R f is discotiuous i x }, b) Assue that X X i distributio. Let f be easurable ad bouded, such that P [X U f ] = 0. Use ) ) fro the lecture otes to show that we have E[fX )] E[fX)]. c) Let f be easurable ad bouded o [0, 1], with U f of Lebesgue easure 0. Show that the correspodig Riea sus coverge to the itegral of f, i.e. Solutio 4.3 a) Let f : R R ad 1 f k=1 ) k 1 fx)dx. 0 V ɛ,δ := { x R y, z x δ, x + δ) s.t. fy) fz) ɛ }. i) Clai: V ɛ,δ is ope. Let x V ɛ,δ. The there are y, z x δ, x + δ) such that fy) fz) ɛ. We set r := δ ax { y x, z x } > 0. x x r, x+r) it holds that y x y x + x x < y x +r δ, ad siilarly for z. Fro this it follows that y, z x δ, x+δ) ad fy) fz) ɛ, which gives x V ɛ,δ. So x r, x + r) V ɛ,δ, ad the clai follows. ii) Clai: U f = V 1, 1. Let x U f. The there is a N, such that N y x 1, x + 1 ) s.t. fx) fy) 1. We assue that for soe N, x V 1, 1,. The there are y, z x 1, x + 1 ) so that fy) fz) 1. Fro this it follows that either fy) fx) or fz) fx) 2 ust hold. I other words N, such that N y x 1, x + 1 ) : fy) fx) 1 2, which iplies that f is discotiuous i x. 5 / 7

6 Sice the V 1, 1 are ope, they are Borel easurable. Ad sice ay σ algebra is closed uder coutable uios ad itersectios, U f = V 1, 1 ust also be Borel easurable. b) By ) ) of the lecture otes, there exist Y d = X, ad Y d = X, such that Y Y, P -alost surely o a probability space Ω, F, P ). Of course, we also have fy ) = d fx ), ad fy ) = d fx), so that we have E[fX )] = E [fy )], ad E[fX)] = E [fy )], where we deote by E the expectatio w.r.t. P. Thus, it suffices to show that E [fy )] E [fy )]. 5) Now, sice Y Y, P -alost surely, we have a set N, with P N) = 0, such that O the other had, we have { } ω Y ω) Y ω) N = Ω. 6) {ω Y ω) Y ω)} {ω f cot. i Y ω), Y ω) Y ω)} {ω f discot. i Y ω)} {ω fy ω)) fy ω))} {ω Y ω) U f }. Cosequetly, it follows fro equatio 6) that we have {ω fy ω)) fy ω))} {ω Y ω) U f } N = Ω. But, by assuptio P Y U f ) = P X U f ) = 0 recall that Y ad X have the sae distributio). Therefore, we get fy ) fy ), P -alost surely. Fially f is a bouded fuctio, so by the Doiated Covergece Theore equatio 5) holds. c) Let λ be the Lebesgue easure o [0, 1], ad, for all a [0, 1], let δ a deote the Dirac delta easure o [0, 1]. Let X, 1, be rado variables with distributio 1 k=1 δ k. Note that E[fX )] = 1 ) k f. Let X be a uifor rado variable o [0, 1], hece it has distributio λ, ad we ote that E[fX)] = Thus, it suffices to show that we have E[fX )] 1 0 k=1 fx)λdx). E[fX)]. 7) Sice by assuptio P [X U f ] = λu f ) = 0, part b) iplies that equatio 7) is a cosequece of the followig: X d X. 8) 6 / 7

7 To show equatio 8), ote that for all N, 0, a < 0, [a] P [X a] =, 0 a 1, 1, 1 < a. Sice we have a 1 [a] a, we get [a] obtai, for 0 a 1, P [X a] a, for all 0 a 1. Thus, we λ[0, a]) = P [X a], which iplies equatio 8), by defiitio. Cases for a < 0 ad a > 1 are trivially verified.) Subissio deadlie: 13:15, Oct. 24. Locatio: Durig exercise class or i the tray outside of HG E 65. Office hours Präsez): Mo. ad Thu., 12:00-13:00 i HG G Class assiget: Studets Tie & Date Roo Assistat A-Gr Tue HG F 26.5 Yili Wag He-Lag Tue ML H 41.1 Agelo Abächerli Laz-Sa Tue HG F 26.5 Vicezo Igazio Sch-Zh Tue ML H 41.1 Lukas Goo 7 / 7