Stat 200C HW 1 Solution (typeset and submitted by Hao Ho)

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1 Stat 00C HW Solutio (tyeset ad submitted by Hao Ho) HW solutio Problem. (..5) For fixed x as, the robability (..) satisfies P (x) x! x x q x = x! ( q )x q. To verify this asymtotic equivalet relatio, we ca directly show that the it of the ratio is equal to. We have P (x) x x! x x q! x!( x)! x q x x! x x q x j= x + j = [ ( x j )] = = j= j= x j where the order of roduct ad it ca be exchaged ad Thus, we have P (x) x! x x q x = ( x! q )x q. = 0, sice x is fixed. Alterative roof. Oe might use the Stirlig s Formula to fid a simlified sequece which is asymtotic equivalet to the robability P (x). We have P (x) =! x!( x)! x q x x! x q x π( e ) π( x)( x e )( x) = π x! x q x [ + x π( x) x ] x ( e )x x! x q x e x ( e )x = x! x x q x = x! ( q )x q. Problem. (..) Let X have the egative biomial distributio Nb(, m) show i Table.6.. For = m satisfyig m( m ) λ > 0 as m, show that the egative biomial distributio teds to the Poisso distributio P (λ). For the coveiece, we let q m = m ad rewrite the it by mq m λ > 0 as m. From the it above we kow that q m = O (m ), the q m = o (). That is q m 0 as m.

2 The we have P m(x) ( ) m + x m m mqm x (m + x )! m (m )!x! mqm x x! m m [(m + j )q m ] = x! [ ( q m) m ] j= HW solutio [ (m + j )q m] where the order of roduct ad it ca be exchaged, sice x is fixed. We recogize that last it terms i the roduct are (m + j )q m mq m + (j ) q m = λ + (j )0 = λ. Now we evaluate the rest term ( q m) m ex{m log( q m)} = ex{[ mq m] log[ ( ) qm ]} q m = e λ log e = e λ where the order of it, ex ad log ca be exchaged sice both are cotiuous fuctios. Fially, we ca lug these results ito the origial equatio ad get P m(x) = x! [ ( q m) m ] [ (m + j )q m] = x! eλ j= j= j= λ = λx x! eλ which is the robability mass fuctio of the Poisso distributio. Thus, we roved that the egative biomial distributio teds to the Poisso distributio. Problem 3. (.4.6) Relace each of the followig quatities by a simler oe which cosists of oly oe of the terms ad is asymtotically equivalet (i) log + ; (ii) log + log(log ); (iii) + e. Solutio. (i) log +. log + log + + =. Here we alied the L Hosital s Rule i the equatio, sice log =, = ad = 0.

3 (ii) log + log(log ) log. HW solutio log + log(log ) log = + log(log ) log = + log = + log =. Here we alied the L Hosital s Rule i the equatio, sice log =, log(log ) = log ad = 0. (iii) + e e. + e e e + e + e + =. Here we alied the L Hosital s Rule i the equatio, sice =, =, e = ad e = 0. Problem 4. (.4.8) Show that [ + c + o( )] e c as First we claim that for ay δ > 0, there exists a iteger N 0 (δ) such that The we have [ + c δ o( ) < δ for all > N 0(δ). ] < [ + c + o( )] < [ + c + δ ] for all > N 0 (δ). Sice we kow that [+ c δ ] = e c δ ad [+ c+δ ] = e c+δ, the for ay give ɛ there exists N (ɛ) ad N (ɛ) such [ + c δ ] e c δ < ɛ for all > N (ɛ) ad [ + c + δ ] e c+δ < ɛ for all > N (ɛ). Now we wat to cotrol e c e c δ ad e c+δ e c. We have e c e c δ < ɛ δ < log( ɛ e ) c e c+δ e c < ɛ δ < log( + ɛ e ) c 3

4 HW solutio Thus, for ay give ɛ > 0, there exists M(ɛ) = max{n 0 [ log( ɛ ) log( + ɛ )], N e c e c (ɛ), N (ɛ)} such that e ɛ [ + c δ ] < [ + c + o( )] < [ + c + δ ] e + ɛ for all > M(ɛ). We have roved that [ + c + o( )] = e. Proof of claim. Sice o( ) = 0, for ay δ > 0 there exists N 0 (δ) such that o( ) < δ for all > N 0 (δ). Problem 5. Suose {X } = is a sequece of radom variables where X follows Ber( ) ideedetly. Cosider the cases (i) = ad (ii) =. Are they covergig to 0 i robability? What are the tail robability P (X = 0 = X + = )? Solutio. (i) For =, we have Thus, X 0. The tail robability is P (X = 0) ( ) =. P (X = 0 = X + = ) P (X = X + = = X +m = 0) m ( j ) ( )( + ) ( + m + m ) Alterative roof. j= ( + m ) 0 = 0. P (X = 0 = X + = ) m ( j ) j= e P (X = 0 = X + = = X +m ) m j= j 0 = 0. Here we used the fact that e j j ad j= is divergig so that j j= j = for ay. (ii) For =, we have P (X = 0) ( ) =. 4

5 Thus, X 0. The tail robability is HW solutio P (X = 0 = X + = ) P (X = X + = = X +m = 0) m ( j ) j= = m j j= j= j = 0 =. Here we used the Boferroi iequality that P ( m j=a j ) = P ( m j=a c j) m j= P (Ac j), where A j = {ω X j (ω) = 0} ad the fact that j= is covergig to π, so that j 6 j= 0. Problem 6. (..) I Examle..4, show that (i) Y if 0; (ii) E(Y ) if =. (i) From examle..4 we have P (Y = ) =. Thus, if 0 we get P (Y = ) = =. (ii) For =, we have E(Y ) [0( )+( ) ( )] ( ) =. Problem 7. (..8) If there exists M such that ad if Y c the (i) E(Y ) c; (ii) E(Y c) 0. P ( Y c < M) = for all, (i) For coveiece we let Z = Y c, the we have Z 0. Furthermore, there exists M = j 5

6 HW solutio such that P ( Z < M) = for all. It is equivalet to show that E(Z ) = E(Y c) 0. For ay give ɛ > 0, we have Sice Z we have E(Z ) E( Z ) = E{ Z [ [ ɛ, ɛ ](Z ) + [ ɛ, ɛ ]c(z )]} ɛ + M P ( Z > ɛ ). 0, there exist N 0 (ɛ) such that P ( Z > ɛ ) < ɛ M for all > N 0(ɛ). Therefore, E(Z ) ɛ + M P ( Z > ɛ ) < ɛ + M ɛ M = ɛ for all > N 0(ɛ). Thus, we have roved that E(Y ) c. (i) For coveiece we also let Z = Y c, the we have Z 0. Furthermore, there exists M such that P ( Z < M) = for all. It is equivalet to show that E(Z) = E(Y c) 0. For ay give ɛ > 0, we have E(Z) = E{Z[ [ ɛ, ɛ ](Z ) + [ ɛ, ɛ (Z )]} ]c ɛ ɛ + M P ( Z > ). Sice Z 0, there exist N (ɛ) such that P ( Z > ɛ ) < ɛ Therefore, we have M for all > N (ɛ). E(Z ) ɛ + M P ( Z > Thus, we have roved that E(Y c) 0. ɛ ) < ɛ + M ɛ M = ɛ for all > N (ɛ). Problem 8. Suose {X j } j= is a sequece of radom variables, where X j s follow N(µ, ) ideedet idetically ad { if x g(x) = x 0 x =. Let X = j= X j. Show that g( X ) µ for µ. First we claim that, if Y c ad h(y) is cotiuous at c, the h(y ) h(c). Sice X j s follow N(µ, ) ideedet idetically, we have X = j= X j µ by WLLN ad g(x) is a ratioal fuctio which is cotiuous everywhere excet at x =. Thus we have g( X ) g(µ) = for µ. µ 6

7 HW solutio Proof of claim. Sice h(y) is cotiuous at c, for ay give ɛ > 0 there exists δ(ɛ) > 0 such that, Furthermore, sice Y The we have h(y) h(c) < ɛ if y c < δ(ɛ). c, for ay give δ > 0 ad ξ > 0 there exists N 0 (δ) such that, P ( Y c < δ) > ξ for all > N 0 (δ). P ( h(y ) h(c) < ɛ) > P ( Y c < δ) > ξ for all > N 0 (δ(ɛ)). Thus, we roved our claim. Problem 9. Suose {X j } j= is a sequece of radom variables, where X j s follow N(µ, σ j ) ideedetly. Let X = j= X j. (i) Show that if σ j = j the X = µ + o ( α ) for α <. (ii) Excet the trivial case σ = σ = = 0, ca we fid {σ j } j= so that X = µ + o ( )? (i) Sice X j s follow N(µ, σ j ) ideedetly, we have V ar( X ) = j= V ar(x j) = j= j = π 6. For ay ɛ > 0, we aly the Chebyshev iequality ad get 0 P ( X µ α > ɛ) α V ar( X ) ɛ Thus, we have roved that X = µ + o ( α ) for α <. = 0, for α <. 6ɛ α π (i) Sice we excluded the trivial case σ = σ = = 0, there must be at least oe σj > 0. Without loss of geerality, we assume that σ > 0. The we have j= V ar(x j) = j= σ j σ. Sice X j= µ follows N(0, σ j ), we have P ( X µ > σ ) = P ( X µ j= σ j > σ j= σ j ) = Φ( Thus, for ay ɛ < Φ( ) we ca ot fid such N 0 (ɛ) such P ( X µ σ > σ ) < ɛ for all > N 0 (ɛ). j= σ j ) Φ( ) 0.3 for all. We have fiished the roof that excet the trivial case σ = σ = = 0, we caot fid {σ j } j= so that X = µ + o ( ). 7