Now we claim that the functions a and 3 satisfy t 0 s

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Now we claim that the functions a(·) and /3( ·) satisfy (t > 0, s > 0) a(ts) = a(t)a(s), f3(ts) = a(t)f3(s) + f3(t) = a(s)f3(t) + f3(s), the last line following by symmetry. To verify these assertions we use G 1 (x) = G(a(t)x + ,8(t)) to conclude that G(a(ts)x + ,8(ts)) = G 15 (x) = (G 5 (x)) 1 = (G(a(s)x + ,8(s))) 1 = G(a(t)[a(s)x + ,8(s)] + f3(t)) = G(a(t)a(s)x + a(t)f3(s) + f3(t)) . Now apply Step (iv). (8.27) (8.28) (8 . 29) Step (vi). Now we prove that there exists() e lR such that a(t) = t 9 . If() = 0, then f3(t) = c logt, for some c e JR. If() "# 0, then {3(t) = c(1- t 9 ), for some c E lit Proof of (vi): Since a(·) satisfies the Hamel equation, a(t) = t 9 for some () e JR. If() = 0, then a(t) = 1 and f3(t) satisfies ,8(ts) = {3(s) + f3(t). So exp{,B( · )} satisfies the Hamel equation which implies that exp{,8(t)} = tc, for some c e lR and thus {3(t) = clogt . If () "# 0, then f3(ts) = a(t),B(s) + ,8(t) = a(s)f3(t) + f3(s). Fix so "# 1 and we get a(t)f3(so) + ,8(t) = a(so)f3(t) + ,8(so), and solving for ,8(t) we get f3(t)(1 - a(so)) = ,8(so)(l - a(t)). Note that 1 - a(so) "# 0. Thus we conclude ( ,8(so) ) 9 f3(t) = (1 - a(t)) = : c(1- t ). 1 - a(so)
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282 8. Convergence in Distribution Step (vii). We conclude that either (a) G 1 (x)=G(x+clogt), (0=0), or Now we show that 0 = 0 corresponds to a limit distribution of type A (x ), that the case 0 > 0 corresponds to a limit distribution of type <l>a and that 0 < 0 corresponds to \f1 a. Consider the case 0 = 0. Examine the equation in (a): For fixed x, the function G 1 (x) is non-increasing in t. Soc < 0, since otherwise the right side of (a) would not be decreasing. If xo e lR such that G (xo) = 1, then 1 = G 1 (xo) = G(xo + clogt), Vt > 0, which implies G(y) = 1, Vy E IR, and this contradicts G non-degenerate. If xo e lR such that G(xo) = 0, then 0 = G 1 (xo) = G(xo + clogt), Vt > 0, which implies G(x) = 0, Vx E IR, again giving a contradiction. We conclude 0 < G(y) < 1, for ally e JR. In (a), set x = 0 and set G (0) = e-K. Then e-tK = G(clogt). Set y = c logt, and we get G(y) = exp{-Keyfc} = exp{-e-<fcr-logK)} which is the type of A (x). The other cases 0 > 0 and 0 < 0 are handled similarly. 0 8.8 Exercises 1. Let Sn have a binomial distribution with parameters nand 0 E [0, 1) . What CLT does Sn satisfy? In statistical terms, 0 := Sn In is an estimator of 0 and Sn- E(Sn) JVar(Sn)
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is an approximate pivot for the parameter 9. If g(9) = log (- 9 -) 1-9 8.8 Exercises 283 is the log-odds ratio, we would use g(O) to estimate g(O). What CLT does g(B) satisfy? Use the delta method . 2. Suppose {X n, n ;::: 1} is a sequence of random variables satisfying 1 P[Xn=n]=-, n 1 P[Xn = 0] = 1- - . n (a) Does {Xn} converge in probability? If so, to what? Why? (b) Does {Xn} converge in distribution? If so, to what? Why? (c) Suppose in addition that {Xn} is an independent sequence. Does {Xn} converge almost surely? What is lim sup X n and lim inf X n n ..... oo n ..... oo almost surely? Explain your answer. 3. Suppose fUn, n 2: 1} are iid U(O, 1) random variables so that P[Uj ::;x]=x, O::;x::; 1. (a) Show nj= 1 uyn converges almost surely . What is the limit? (b) Center and scale the sequence tnj= 1 uJin, n ;::: 1} and show the resulting sequence converges in distribution to a non-degenerate limit .
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