Suppose VS's stock price is currently $20. A six-month call option on VS's stock with an exercise price of $15 has a value of $7
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Answer:
Check the explanation
Explanation:
Yt = Kt + Nt
Taking output per worker, we divide by Nt
Yt/Nt = Kt/Nt + 1
yt = kt + 1
where yt is output per worker and kt is capital per worker.
a) With population being constant, savings rate s and depreciation rate δ.
ΔKt = It - δKt
dividing by Nt, we get
ΔKt/Nt = It/Nt - δKt/Nt ..... [1]
for kt = Kt/Nt, taking derivative
d(kt)/dt = d(Kt/Nt)/dt ... since Nt is a constant, we have
d(kt)/dt = d(Kt/Nt)/dt = (dKt/dt)/Nt = ΔKt/Nt = It/Nt - δKt/Nt = it - δkt
thus, Capital accumulation Δkt = i – δkt
In steady state, Δkt = 0
That is I – δkt = 0
S = I means that I = s.yt
Thus, s.yt – δkt = 0
Then kt* = s/δ(yt) = s(kt+1)/(δ )
kt*= skt/(δ) + s/(δ)
kt* - skt*/(δ) = s/(δ)
kt*(1- s/(δ) = s/(δ)
kt*((δ - s)/(δ) = s/(δ)
kt*(δ-s)) = s
kt* = s/(δ -s)
capital per worker is given by kt*
b) with population growth rate of n,
d(kt)/dt = d(Kt/Nt)/dt =
=
=
= ΔKt/Nt - n.kt
because (dNt/dt)/Nt = growth rate of population = n and Kt/Nt = kt (capital per worker)
so, d(kt)/dt = ΔKt/Nt - n.kt
Δkt = ΔKt/Nt - n.kt = It/Nt - δKt/Nt - n.kt ......(from [1])
Δkt = it - δkt - n.kt
at steady state Δkt = it - δkt - n.kt = 0
s.yt - (δ + n)kt = 0........... since it = s.yt
kt* = s.yt/(δ + n) =s(kt+1)/(δ + n)
kt*= skt/(δ + n) + s/(δ + n)
kt* - skt*/(δ + n) = s/(δ + n)
kt*(1- s/(δ + n)) = s/(δ + n)
kt*((δ + n - s)/(δ + n)) = s/(δ + n)
kt*(δ + n -s)) = s
kt* = s/(δ + n -s)
.... is the steady state level of capital per worker with population growth rate of n.
3. a) capital per worker. in steady state Δkt = 0 therefore, growth rate of kt is zero
b) output per worker, yt = kt + 1
g(yt) = g(kt) = 0
since capital per worker is not growing, output per worker also does not grow.
c)capital.
kt* = s/(δ + n -s)
Kt*/Nt = s/(δ + n -s)
Kt* = sNt/(δ + n -s)
taking derivative with respect to t.
d(Kt*)/dt = s/(δ + n -s). dNt/dt
(dNt/dt)/N =n (population growth rate)
so dNt/dt = n.Nt
d(Kt*)/dt = s/(δ + n -s).n.Nt
dividing by Kt*
(d(Kt*)/dt)/Kt* = s/(δ + n -s).n.Nt/Kt* = sn/(δ + n -s). (Nt/Kt)
using K/N = k
plugging the value of kt*
n
thus, Capital K grows at rate n
d) Yt = Kt + Nt
dYt/dt = dKt/dt + dNt/dt = s/(δ + n -s).n.Nt + n.Nt
using d(Kt*)/dt = s/(δ + n -s).n.Nt from previous part and that (dNt/dt)/N =n
dYt/dt = n.Nt(s/(δ + n -s) + 1) = n.Nt(s+ δ + n -s)/(δ + n -s) = n.Nt((δ + n)/(δ + n -s)
dYt/dt = n.Nt((δ + n)/(δ + n -s)
dividing by Yt
g(Yt) = n.(δ + n)/(δ + n -s).Nt/Yt
since Yt/Nt = yt
g(Yt) = n.(δ + n)/(δ + n -s) (1/yt)
at kt* = s/(δ + n -s), yt* = kt* + 1
so yt* = s/(δ + n -s) + 1 = (s + δ + n -s)/(δ + n -s) = (δ + n)/(δ + n -s)
thus, g(Yt) = n.(δ + n)/(δ + n -s) (1/yt) = n.(δ + n)/(δ + n -s) ((δ + n -s)/(δ + n)) = n
therefore, in steady state Yt grows at rate n.