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26, Jan 2007 Solution Key to Homework Assignment 1 1 Stat 643 Solution Key to Homework Assignment 1 Jan 26, 2007 This is only the sketch of the proof, you will need to ll in the details and show/convince me there is sound reasoning from step to step. Good luck. Q1. Draw the Fi (x) and F (x) as the limit when i (See Fig 1 at the end of solution). 1 Let measure be that (( , x]) = 1 (( , x]) + 1 I(x > 2 ) + 1 I(x > 1.5) 6 3 3 Argue that x {0.5, 1.5}, Fi (x) = i (( , x]) (( , x]) = F (x) Follow the construction of Skorohod s Thm, let us de ne the inverse of c.d.f as hi ( ) = sup{t : i ( , t] < } h( ) = sup{t : ( , t] < } where hi and h are r.v on ((0, 1), B((0, 1)), m). The hardest part is hi , where for i 4 6 1 2+1 i hi ( ) = 6 2 3 1.5 + 1.5 i 6 3 1 1 (0, 12 + 6i ) 1 15 [ 12 + 6i , 12 + 5 1 [ 12 + 6i , 1 ] 2 15 (2, 6] ( 5 , 1) 6 1 6i ) Note: the point of 5 12 + 1 6i could be open or closed. Q2. Rosenthal 10.1.2 Enough to show x C(F ), Fi (x) F (x) f is continuous on a bounded(bdd) interval, by property of Riemann integral we have 1 limi 1 i f ( j ) = 0 f (t)dt = 1 j=1 i i x limi 1 j x f ( j ) = 0 f (t)dt i i i i lim Fi (x) = = = i lim i (( , x]) j x i i lim f(j ) i x 0 f (t)dt 1 0 f (t)dt j i j=1 f ( i ) = (( , x]) = F (x) 2 Solution Key to Homework Assignment 1 Jan 26, 2007 Let Y ( ) = F 1 ( ) = sup{t : (( , t]) < } The question ask us to explicitly construct Yi , so it goes: Yi ( ) = Fi 1 ( ) = sup{t : i (( , t]) < } = Then by Skorohod s Thm Yi d Y J i if J 1 j=1 f (xj ) i j=1 f (xj ) < J j=1 f (xj ) i j=1 f (xj ) Q3. Rosenthal 10.1.4 The measure in Q1 should one give some hint on how to solve this one. To be a.c. w.r.t Lebesque measure on R, all we need is to eliminate jump in the c.d.f. One construction is like this: let ij = Lebesgue measure on [j 1 , j] i i = ai i ij j=1 See Fig 2 at the end for some illumination. Q4. Rosenthal 10.1.6 An application of BCT Q6. Convergence of Type Let Yi = t C(FX ) Xi ai bi , Zi = Xi i i , then Yi d X i i ai t+ ) bi bi i i ai FX ( t + ) bi bi FX (t) FYi ( FZi (t) = i i (1) (2) (3) (1) is simple arithmetic (2) is because C(FX ) is dense in R so t C(FX ), you can nd i large enough so that i i ai is in near neighborhood of t and is a continuous point bi t + bi (3) is because i bi t + i ai bi t and FX is continuous at t Jan 26, 2007 Solution Key to Homework Assignment 1 3 Many of you used Slutsky Thm, which is also ne. Pretend/think a sequence of a constant( ii or ibi i ) as a sequence of r.v will make life much easier in this case. b Q7. A&L 9.7 > 0 and > 0 {Xn } is stochastically bdd > 0, M s.t. supn>1 P (|Xn | > M ) < {Yn } p 0 n0 s.t. n > n0 P (|Yn | > M) 2 < 2 P (|Xn Yn | > ) = P (|Xn Yn | > , |Xn | > M ) + P (|Xn Yn |) > , |Xn | <= M ) = P (|Xn Yn | > , |Xn | > M ) + P (|Xn Yn |) > , |Xn | <= M, |Yn | > P (|Xn | > M ) + P (|Yn | > < ) M ) M so Xn Yn p 0 Q9. Assume the slow varying part will become a constant near the zero, absorb it into c. P (n1/ ( Mn ) > x) = = = = P (Mn < x/n1/ ) P n (X1 < x/n1/ ) (1 c(x/n1/ ) )n (1 cx /n)n exp( cx ) The guys from the third oor will tell you this is a Weibull( , c) distribution. Q10. Rosenthal 11.5.3 Yes, { i } is tight. {xi } is compact, i.e. {xij } converge, say to x, then { ij } d x , the point mass at x.

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