problem-set-2-solutions_sherrard

Ry an Sherrard Econ 204A Octob er 14, 2018 Problem Set 2 Romer Problem 1.3 Describ e ho w, if at all, each of the follo wing dev elopmen ts a\ufb00ects the break- ev en and actual in v estmen t lines in our basic diagram for the Solo w mo del: (a) The rate of depreciation falls. If the depreciation rate drops, the slop e of the break ev en in v estmen t curv e falls. As a result, the break ev en in v estmen t line rotates clo c kwise. The actual in v estmen t curv e, how ev er, remains unc hanged. Using our Solo w graph, w e kno w that this c hange should result in higher steady state capital in e\ufb00eciency units. No w, lets pro v e it: \u02d9 k = sf ( k ) \u2212 ( \u03b4 + n + g ) k \u21d2 sf ( k \u2217 ) = ( \u03b4 + n + g ) k \u2217 (Imp ose steady state) \u21d2 sf 0 ( k \u2217 ) dk \u2217 d\u03b4 = k \u2217 + ( \u03b4 + n + g ) dk \u2217 d\u03b4 (Di\ufb00eren tiate) \u21d4 dk \u2217 d\u03b4 = k \u2217 sf 0 ( k \u2217 ) \u2212 ( \u03b4 + n + g ) 0 ( sf 0 ( k \u2217 ) ( \u03b4 + n + g )) (b) The rate of tec hnological progress rises. Graphically , the break even in v estmen t line rotates coun terclockwise, while the actual in v estmen t line sta ys is unc hanged. Hence, w e kno w that k \u2217 decreases as capital per e\ufb03ciency unit is b ecoming \u201cdiluted\u201d b y faster tec hnological gro wth. No w, let\u2019s pro v e it: \u02d9 k = sf ( k ) \u2212 ( \u03b4 + n + g ) k \u21d2 sf ( k \u2217 )=( \u03b4 + n + g ) k \u2217 (Imp ose steady state) \u21d2 sf 0 ( k \u2217 ) dk \u2217 dg = k \u2217 + ( \u03b4 + n + g ) dk \u2217 dg (Di\ufb00eren tiate) \u21d4 dk \u2217 dg = k \u2217 sf 0 ( k \u2217 ) \u2212 ( \u03b4 + n + g ) 0 ( sf 0 ( k \u2217 ) ( \u03b4 + n + g )) 1 (c) The pro duction function is Cobb-Douglas, f ( k ) = k \u03b1 , and capital\u2019s share, \u03b1 , rises. This question is sligh tly tric kier. Because c hanging capital share c hanges the curv ature of the actual in v estmen t line, there is no simple rotation or shift in either curv e. Instead, the new actual in v estmen t line will b e ab o v e the old curv e for lo w capital, but b elow the old curv e for high capital. The break ev en in v estmen t line do es not c hange, ho w ev er. Clearly , our graphical analysis do es not giv e as an unam biguous answ er in this case. Let\u2019s \ufb01nd out what happ ens: \u02d9 k = sf ( k ) \u2212 ( \u03b4 + n + g ) k \u21d2 sk \u2217 \u03b1 = ( \u03b4 + n + g ) k \u2217 (Imp ose steady state) \u21d2 s\u03b1k \u2217 \u03b1 \u2212 1 dk \u2217 + sk \u2217 \u03b1 ln( k \u2217 ) d\u03b1 = ( \u03b4 + n + g ) dk \u2217 (T otal deriv ativ e) \u21d4 dk \u2217 d\u03b1 = sk \u2217 \u03b1 ln( k \u2217 ) ( \u03b4 + n + g ) \u2212 s\u03b1k \u2217 \u03b1 \u2212 1 (Rearrange) Note that in \ufb01nding the total deriv ativ e the rule d dx a x = a x ln ( a ) might be useful. Also recognize that the denominator is alw a ys p ositiv e due to the conca vit y of f ( · ). The n umerator on the other hand has an am biguous sign. When k \u2217 > 1, increasing capital share increases k \u2217 . When k \u2217 1, increasing capital share decreases k \u2217 . The fact that the deriv ative has a di\ufb00eren t sign dep ending on where the steady state is re\ufb02ects the fact that the in vestmen t line isn\u2019t simply \u201cshifting\u201d or \u201crotating.\u201d (d) W ork ers exert more e\ufb00ort, so that output p er unit of e\ufb00ectiv e lab or for a giv en v alue of capital p er unit of e\ufb00ectiv e lab or is higher than b efore. This c hange simply implies that the new actual in v estmen t curv e is everywhere abov e the old actual in v estmen t curv e (ignoring k = 0). F or an y giv en lev el of k , after this added e\ufb00ort is tak en in to accoun t, the level of y that is produced is higher. Th us, unlike in the last part, the new output (and sa vings) curv e, will unambiguously lie abov e its previous sp ot. Also, the break-even- in v estmen t line will remain unmo v ed. Th us, w e will see k \u2217 rise. Romer Problem 1.4 Consider an econom y with technological progress but without population gro wth that is on its balanced gro wth path. No w supp ose there is a one-time jump in the n um b er of w ork ers. (a) A t the time of the jump, do es output p er unit of e\ufb00ective labor rise, fall, or stay the same? Wh y? W e kno w that a one-time c hange in L will not c hange the steady state v alue of k as no funda- men tal parameters (structure) in the econom y ha v e changed. Recall that k = K/ AL . When L jumps up from L 0 to L 1 , then k will jump down from k 0 to k 1 . Next, because y = f ( k ) where f 0 ( · ) > 0, w e kno w that y will also jump do wn w ards. 2 (b) After the initial c hange (if an y) in output p er unit of e\ufb00ectiv e lab or when the new w ork ers app ear, is there an y further c hange in output p er unit of e\ufb00ectiv e lab or? If so, do es it rise or fall? Wh y? After the immediate drop in k and y , the economy will not be in the steady state. W e kno w, ho w ev er, that the economy will con verge to its steady state o ver time. During this transition, the dynamics of the econom y are go v erned b y \u02d9 k = sf ( k ) \u2212 ( \u03b4 + n + g ) k Because k drops immediately after the c hange, k 1 k \u2217 . Hence, w e kno w that sf ( k ) > ( \u03b4 + n + g ) during the transition. Th us, capital p er e\ufb03cien y unit will b e increasing. Moreo v er, b ecause y = f 0 ( k ) > 0, this implies that output p er e\ufb03ciency unit will also b e gro wing o v er time. (c) Once the econom y has again reac hed a balanced gro wth path, is output p er unit of e\ufb00ective lab or higher, lo w er, or the same as it w as b efore the new w ork ers app eared? Wh y? As men tioned in part a, the fundamental structure of the econom y has not changed due to this rise in w ork ers. In other w ords, w e kno w that no fundamen tal parameters ha v e c hange. Thus, the steady state v alues of k , y , and c will all b e the same as b efore. F or extra practice, see if you can dra w the time ev olution of K/L and Y /L as a result of this c hange. Romer Problem 1.6 Consider a Solow econom y that is on its balanced gro wth path. Assume for simplicit y that there is no tec hnological progress. No w supp ose that the rate of p opulation growth falls. (a) What happ ens to the balanced-gro wth-path v alues of capital p er w orker, output p er work er, and consumption p er w ork er? Sk etch the paths of these v ariables as the economy mo ves to its new balanced gro wth path. Note here that b ecause there is no tec hnologial progress, the time paths of capital and output p er e\ufb03ciency unit is qualitativ ely the same as capital and output p er capita. First, let\u2019s \ufb01nd out ho w our steady state v alues c hange. sf ( k \u2217 )=( \u03b4 + n ) k \u2217 (Imp ose steady state) \u21d2 sf 0 ( k \u2217 ) dk \u2217 = ( \u03b4 + n ) dk \u2217 + k \u2217 dn (T otal deriv ativ e) \u21d4 dk \u2217 dn = k \u2217 sf 0 ( k \u2217 ) \u2212 ( \u03b4 + n ) 0 ( sf 0 ( k \u2217 ) + n ) Th us, as p opulation growth increases, the steady state v alue of capital p er work er decreases. Next, for output p er w ork er, w e kno w that y \u2217 = f ( k \u2217 ) at the steady state. Th us 3 dy \u2217 dn = f 0 ( k \u2217 ) dk \u2217 dn 0 , Last, w e kno w that c \u2217 = (1 \u2212 s ) f ( k \u2217 ). The deriv ative w.r.t. n will b e prop ortional to y \u2217 with the same sign. Th us, consumption p er work er will also decrease. Here, n is decreasing. As a result, k \u2217 , y \u2217 , and c \u2217 all increase. The time path for capital p er w ork er is giv en b elo w. The time paths for the other v ariables are qualitativ ely iden tical. time k (b) Describ e the e\ufb00ect of the fall in p opulation gro wth on the path of output (that is, total output, not output p er w ork er). T otal output can b e written as Y = f ( k ) L = y L . Here, I\u2019m simply normalizing A = 1 since it\u2019s not gro wing and is th us a simple scale factor. Using the usual tric k of taking logs and time deriv ativ es, w e can sk etc h the time path of Y . ln ( Y ) = l n ( y ) + l n ( L ) (Def.) \u21d2 \u02d9 Y Y = \u02d9 y y + \u02d9 L L (Di\ufb00eren tiate) W e kno w what the gro wth rate of lab or is, but what ab out \u02d9 y /y ? Consider the follo wing, y = f ( k ) = \u21d2 ln ( y ) = l n ( f ( k )) = \u21d2 \u02d9 y y = f 0 ( k ) f ( k ) \u02d9 k \u21d4 \u02d9 y y = \u03b1 ( k ) \u02d9 k k (Mulitply b y k k ) 4 No w, let\u2019s c hec k what happ ens immediately b efore and immediately following this c hange. Know- ing ho w capital p er e\ufb03ciency unit c hanges will allo w us to w ork bac kw ards to Y . \u02d9 k k \ue00c \ue00c \ue00c \ue00c \ue00c t 0 \u2212 \u03b5 = sf ( k ) k \u2212 ( \u03b4 + n 0 ) and \u02d9 k k \ue00c \ue00c \ue00c \ue00c \ue00c t 0 + \u03b5 = sf ( k ) k \u2212 ( \u03b4 + n 1 ) \u02d9 k k \ue00c \ue00c \ue00c \ue00c \ue00c t 0 + \u03b5 \u2212 \u02d9 k k \ue00c \ue00c \ue00c \ue00c \ue00c t 0 \u2212 \u03b5 = n 0 \u2212 n 1 > 0 That is, w e are lo oking at the instan t b efore and the instan t after the c hange in n . Next, consider the gro wth in y just after the c hange. \u2206 \u02d9 y y = \u03b1 ( k )\u2206 \u02d9 k k = \u03b1 ( k )( n 0 \u2212 n 1 ) > 0 No w, use this relationship to \ufb01nd the c hange in the gro wth rate of Y . \u2206 \u02d9 Y Y = \u03b1 ( k )\u2206 \u02d9 y y + \u2206 \u02d9 L L = \u03b1 ( k )( n 0 \u2212 n 1 )+( n 1 \u2212 n 0 ) \u21d4 \u2206 \u02d9 Y Y = (1 \u2212 \u03b1 ( k ))( n 1 \u2212 n 0 ) 0 Hence, we kno w that at the time of the change, the growth rate of Y immediately c hanges (i.e. there\u2019s a kink). Moreov er, b ecause \u02d9 y y = 0 in the steady state, w e kno w that \u02d9 Y Y \u2192 \u02d9 L L along the balanced gro wth path. Th us, the time path of ln( Y ) is the follo wing: time ln ( Y ) 5

problem-set-2-solutions_sherrard

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