第八讲 动态最优问题求解：(I）....pdf

1lù Ä •`¯K¦)µ £I¤.‚KF•{ •% E ŒÆ²LÆ •% (E ŒÆ²LÆ ) êþ²LÆ£1lù¤ 1 / 14 ù̇SN 1. (½5Ä •`¯K £ã!©a†¦)•{ 2. ¦))·±Ï;O¯K 3. lÑžm!ÕÏ. •`O•¯K 3.1 .‚KF•{¦) 3.2 ÕÏ.¯K 3.3 - NC - Ú=£Ä ‚5Cq)Úƒã 4. ~K†SK 4.1 è’•`Ý] 4.2 ž¤ö•` ‚5 4.3 ˜„z •% g. . •`O• . (E ŒÆ²LÆ ) êþ²LÆ£1lù¤ 2 / 14 1. (½5Ä •`¯K £ã!©a†¦)•{ (½5Ä •`¯K £ã!©a†¦)•{ (½5£deterministic¤Ä CþÚûüCþ|¤ •`¯K¥vk ) •þ x(t)§8I¼ê•Ø^ M ax ‘ÅÀ z(t)§•kS) G Ï""˜„z£ã•µ U (x (t) ; θ) x(t)∈XT s.t. G (x (t) ; θ) = 0 ©a†¦)•{µ k•Ï.£T < ∞¤ )·±Ï;O¯K lÑžmµ.‚KF•{ T P β t ln ct M ax t=1 s.t. kt = w + (1 + r)kt−1 − ct k0 = kT = 0 ∞ P β t ln ct M ax ÕÏ.£T → ∞¤ •`O•¯K t=1 α s.t. kt = zkt−1 − ct k0 = 1 lim h(ct , kt ) = 0 t→∞ •% (E ŒÆ²LÆ ) êþ²LÆ£1lù¤ ëYžmµ•`››•{ R M ax 0T e−ρt ln (c) dt s.t. dk ≡ k̇ = w + rk − c dt k (0) = 0, k (T ) = 0 M ax R∞ 0 e−ρt ln (c) dt s.t. dk ≡ k̇ = zkα − δk − c dt k (0) = 1 lim e−ρt u0 (c) k = 0 t→∞ 3 / 14 2. ¦))·±Ï;O¯K )·±Ï;O¯K M ax T P β t ln ct t=1 s.t. kt = w + (1 + r)kt−1 − ct k0 = kT = 0 t 1 2 ... T T +1 ÏЕþ k0 = 0 k1 ... kT −1 kT = 0 Â\ w w+rk1 ... w + rkT −1 0 ž¤ c1 c2 ... cT 0 Ï6þ •% (E ŒÆ²LÆ ) êþ²LÆ£1lù¤ 4 / 14 2. ¦))·±Ï;O¯K .‚KF•{¦) Ñ.‚KF•§£P R = 1 + r¤µ L= T X t=1 d˜ ^‡ β t ln ct − T X λt (kt − w − Rkt−1 + ct ) t=1 î.•§£Euler equation¤µ βt ∂L ∂ct = 0 ⇒ ct = λt ∂L ∂kt = 0 ⇒ λt = λt+1 R ⇒ ct+1 = βR ct †ýŽ å^‡éá ¤‚5 ˜ ©•§|µ " # " #" # " # ct+1 βR 0 ct 0 = + kt+1 −βR R kt w •% (E ŒÆ²LÆ ) êþ²LÆ£1lù¤ 5 / 14 3. lÑžm!ÕÏ. •`O•¯K 3.1 .‚KF•{¦) .‚KF•{¦)•`O•¯K éuÕÏ. •`O•¯K§^î 5^‡“O>.^‡µ ∞ P M ax β t ln ct t=1 α s.t. kt = zkt−1 − ct k0 = 1, lim h(ct , kt ) = 0 t→∞ ƒA .‚KF•§•µ ∞ ∞ X X α β t ln ct − λt (kt − zkt−1 L= + ct ) t=0 d˜ ^‡ î.•§µ ∂L ∂ct ∂L ∂kt †ýŽ t=0 t = 0 ⇒ βct = λt = 0 ⇒ λt = λt+1 αzktα−1 å^‡éá š‚5 ˜ ⇒ ct+1 = βαzktα−1 ct ©•§|µ ct+1 = βαzktα−1 ct kt+1 = zktα − ct+1 •% (E ŒÆ²LÆ ) êþ²LÆ£1lù¤ 6 / 14 3. lÑžm!ÕÏ. •`O•¯K 3.2 ÕÏ.¯K - Ú=£Ä - Ú=£Ä òš‚5 ˜ ©•§| • xt+1 = F (xt ) /ªµ ct+1 = βαzktα−1 ct ≡ F 1 (ct , kt ) kt+1 = zktα − ct+1 = zktα − βαzktα−1 ct ≡ F 2 (ct , kt ) Äk½Â- £steady state¤ lim xT = x∗"ØJ •þÚž¤©O•µ T →∞ 1 k∗ = (αβz) 1−α , c∗ = Šâ˜ |•x  y§•`O•ˆ - ž ] ˜ •§  1 − 1 k∗ αβ VÐmúª xt+1 = x∗ + DF (x∗ ) (xt − x∗ )§e¡ ‚5 .3- NC =£Ä £transitional dynamics¤ µ " # " #" # ct+1 − c∗ 1 βc∗ f 00 (k∗ ) ct − c∗ = kt+1 − k∗ −1 1/β − βc∗ f 00 (k∗ ) kt − k ∗ Ù¥f 00 (k∗ ) = α (α − 1) z(k∗ )α−2" •% (E ŒÆ²LÆ ) êþ²LÆ£1lù¤ 7 / 14 3. lÑžm!ÕÏ. •`O•¯K 3.3 - NC ‚5Cq)Úƒã - NC ‚5Cq) - äˆ'Ý µ " ∗ ∗ ∗ ∗ J(c , k ) = DF (c , k ) = J(c∗ , k∗ ) kü‡¢A λ1 = ©•§ ϕ+ 1 βc∗ f 00 (k∗ ) −1 1/β − βc∗ f 00 (k∗ ) Šµ £Ù¥ ϕ = 1 + β1 − βc∗ f 00 (k∗ ) > 1 + β1 ¤ p p ϕ2 − 4/β ϕ − ϕ2 − 4/β > 1, λ2 = ∈ (0, 1) 2 2 Ï)• xt = C1 λt1 + C2 λt2 §C1 Ú C2 •ü‡ –½Xê"du λ1 > 1§ t → ∞ ž§²L‡ˆ - U- C1 = 0¶2|^Щ^‡§t = 0 žk C2 = x0" 3- # NCQ:´»£saddle path¤ Ò• . ‚5Cq)•µ ct = c∗ + (c0 − c∗ ) λt2 kt = k∗ + (k0 − k∗ ) λt2 ã¡5 µWikiz‰" •% (E ŒÆ²LÆ ) êþ²LÆ£1lù¤ 8 / 14 3. lÑžm!ÕÏ. •`O•¯K 3.3 - NC ‚5Cq)ڃ㠕`O•¯K ƒã £‡¤©•§´·‚©ÛÄ /þïG •`¯K Ä ó䧧Ø=û½X²L•ªUˆ 0§•Œ±•x²LlЩY²•- LÞ /=£Ä ±^ c − k ²¡þ ƒã£phase diagram¤†*/Ы²LlЩG 0 "·‚„Œ ••ª - ÅÚÂñ Ä L§§ù‚=•Q:´»" ct+1 = βαzktα−1 ct kt+1 = zktα − ct+1 1 ⇒ ⇒ ktss = (αβz) 1−α ss α ss css t = z(kt ) − kt kt > ktss , ct+1 < ct ct > css t , kt+1 < kt Source: Heer and Maubner(2009), Figure 1.2. •% (E ŒÆ²LÆ ) êþ²LÆ£1lù¤ 9 / 14 4. ~K†SK 4.1 è’•`Ý] è’•`Ý]¯K ‰½ k0 = 1!ëê a, b, c > 0§b½è’¡ e¡ •`Ý]¯Kµ M ax Π0 = ∞ P t=0 i h 2 β t akt − 2b kt2 − 2c (kt+1 − kt ) s.t. k0 = 1, lim h(It , kt ) = 0 t→∞ ˜ ^‡•µ ∂Π0 = 0 ⇒ c (kt+1 − kt ) = β (a − bkt+1 + c (kt+2 − kt+1 )) ∂kt+1 P It = kt+1 − kt §˜ ^‡C• cIt = β (a − bkt+1 + cIt+1 )" de¡ ˜ . Ä ©•§|û½µ kt+1 = kt + It It+1 = cb kt+1 + β1 It − ac •% (E ŒÆ²LÆ ) êþ²LÆ£1lù¤ 10 / 14 4. ~K†SK 4.1 è’•`Ý] - !=£Ä †Cq) - ©•§|µ kt+1 = kt + It ≡ G1 (kt , It )   It+1 = cb kt + β1 + cb It − ac ≡ G2 (kt , It ) - • I ∗ = 0, k∗ = ab "3- NC‰˜ " # " # " kt+1 k∗ 1 = + ∗ b It+1 I c äˆ'Ý - VÐm =£Ä •§µ #" # 1 kt − k ∗ 1 + cb β It − I ∗ ü‡¢A Š•£P ϕ = 1 + 1/β + b/c¤ µ p p ϕ + ϕ2 − 4/β ϕ − ϕ2 − 4/β λ1 = > 1, λ2 = ∈ (0, 1) 2 2 NC ‚5Cq)µ kt = k∗ + (k0 − k∗ ) λt2 It = I ∗ + (I0 − I ∗ ) λt2 •% (E ŒÆ²LÆ ) êþ²LÆ£1lù¤ 11 / 14 4. ~K†SK è’•`Ý]¯K 4.1 è’•`Ý] ƒã kt+1 = kt + It It+1 = cb kt+1 + β1 It − ac ⇒ •% Itss = 0
⇒ a bβ ktss − b Itss = c(β−1) (E ŒÆ²LÆ ) It > Itss , kt+1 > kt kt > ktss , It+1 > It êþ²LÆ£1lù¤ 12 / 14 4. ~K†SK ž¤ö•` ‚5 4.2 ž¤ö•` ‚5 g. . g. . b½ž¤ö¡ e¡ ‚5 g.£linear quadratic model¤Ä •`¯ K§Ù¥ ^•ž¤ g¼ê§ Ñ•] ‚5¼ê§ëê β ∈ (0, 1)! a, b > 0!z > 1/βµ M ax ∞ P t=1   bc2 β t act − 2t s.t.kt = zkt−1 − ct k0 = 1, lim h(ct , kt ) = 0 t→∞ 1. 2. ¦ Ñ.‚KF•§Úî.•§¶ . - !- NC =£Ä 3. ‰Ñ . •% •§ÚQ:´» Cq)¶ ƒã" (E ŒÆ²LÆ ) êþ²LÆ£1lù¤ 13 / 14 4. ~K†SK ˜„z 3c© •`O• •`O• δ = 1§Ïd 4.3 ˜„z •`O• . . .¥§·‚b½ ^¼ê•éê/ª u(ct ) = lnct §] òÎÇ e¡ Ä •§|µ ct+1 = βαzktα−1 ct kt+1 = zktα − ct+1 1. ž y ct = (1 − αβ)zktα , kt = αβzktα ´þ㕧| £L«• k0 Úžm t 2. ˜„œ¹e§b½ )§ Ñ ct , k t )Û) ¼ê¤ ¶ ^¼ê•CRRA/ª§òÎÇ δ ∈ (0, 1)§¯KC•µ M ax ∞ P t=1 1−γ c t β t 1−γ α s.t. kt = zkt−1 + (1 − δ) kt−1 − ct k0 = 1, lim h(ct , kt ) = 0 t→∞ ž-#¦)£•)î.•§!- !=£Ä !‚5Cq)Úƒã©Û¤ " •% (E ŒÆ²LÆ ) êþ²LÆ£1lù¤ 14 / 14