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Week 1: Overview and Math Review
Monetary Economics
ECOS3010
Overview
Assessment
Midterm exam (30%): Monday, Sep 18, 11 am.
Final exam (70%): during exam period.
Monetary Economics (ECOS3010) Week 1: Overview and Math Review 4 / 18
Overview
Lecture topics
Money
Ination
Price surprises and the Phillips curve
International monetary system
Money and capital
Liquidity and
nancial intermediation
Bank risks, liquidity risks and bank panics
Crisis (if time allows)
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Chain rule, log and exponential functions I
Important Mathematical Concepts
Di¤erentiation and chain rule:
Consider three functions F (x , y), x(a), and y(a).
Q: whats the derivative of F w.r.t to a
A: using chain rule,
∂F (x , y)
∂a
=
∂F (x , y)
∂x
∂x(a)
∂a
+
∂F (x , y)
∂y
∂y(a)
∂a
Monetary Economics (ECOS3010) Week 1: Overview and Math Review 6 / 18
Chain rule, log and exponential functions II
Example:
problem: suppose
rms pro
t function is Π(y) = y4 + 6y2 5,
where Π is the
rms pro
t and y is the amount of output. Assume
rms production function is y = 5L2/3, where L is the amount of
labour input. Apply the Chain Rule to compute the derivative of Π
with respect to L.
solution:
d
dL
(Π(y(L))) = Π0(y) y 0(L)
= [4(y)3 + 12y ] (10
3
L1/3)
= [4(5L2/3)3 + 12(5L2/3)] (10
3
L1/3)
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Chain rule, log and exponential functions III
which after simplifying equals
(4 125L2 + 60L2/3) (10
3
L1/3) = 5000
3
L5/3 + 200L1/3.
Properties of log functions:
ln(ab) = ln(a) + ln(b)
ln(
a
b
) = ln(a) ln(b)
ln(ab) = b ln(a)
d ln(x)
dx
=
1
x
ln(1+ x) x , if x is small
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Chain rule, log and exponential functions IV
Exponential function:
eab =
ea
eb
e ln(a) = a
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Unconstrained optimization I
consider a function y = f (x). Necessary (
rst-order) condition for
optimization of this function:
dy
dx
=
df (x)
dx
= 0
its a maximum if d
2f (x )
dx 2 0 (a concave function), a minimum if
d 2f (x )
dx 2 0 (a convex function).
example 1 (concave):
y = f (x)
= xα x + 1 , where 0 < α < 1
gure: ....
Monetary Economics (ECOS3010) Week 1: Overview and Math Review 10 / 18
Unconstrained optimization II
example 2 (convex):
y = f (x)
= xβ x + 1 , where β > 1
gure: ....
the value of x that maximizes y is called
x = argmax
x
f (x)
y = f (x) is the maximum value of the function.
Monetary Economics (ECOS3010) Week 1: Overview and Math Review 11 / 18
Constrained optimization (Substitution method) I
Example 1:
problem:
Max
x ,y
f (x , y) = xy
s.t. 3 x + 4y = 16
solution:
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Constrained optimization (Substitution method) II
Example 2:
consider the following utility maximization problem of a
consumer/household
max
x ,y
U = U(x , y)
s.t. px x + py y M
solution:
Monetary Economics (ECOS3010) Week 1: Overview and Math Review 13 / 18
Constrained optimization (Lagrangian method) I
It is not always possible to express x as a function of y explicitly. In
that case, we can apply the Lagrangian method.
Example 1:
problem:
Max
x ,y
f (x , y) = xy
s.t. 3 x + 4y = 16
solution:
rst, form a Lagrangian with Lagrange multiplier λ
L = xy + λ [16 3x 4y ]
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Constrained optimization (Lagrangian method) II
next, take the derivatives w.r.t. x , y , and λ yielding
0 = y 3λ
0 = x 4λ
0 = 16 3x 4y
last, solve for three values using the system of equations above, yielding
x =
8
3
y = 2
λ =
2
3
Monetary Economics (ECOS3010) Week 1: Overview and Math Review 15 / 18
Constrained optimization (Lagrangian method) III
Example 2:
consider the following utility maximization problem of a
consumer/household
max
x ,y
U = U(x , y)
s.t. px x + py y M
we can solve this problem using the Lagrangian method
solving the problem:
form the Lagrangian function (note the way the constraint is
written!)
L = U(x , y ) + λ[M px x py y ]
Monetary Economics (ECOS3010) Week 1: Overview and Math Review 16 / 18
Constrained optimization (Lagrangian method) IV
rst-order conditions (FOC)
∂L
∂x
= Ux (x , y ) λpx = 0
∂L
∂y
= Uy (x , y ) λpy = 0
∂L
∂λ
= M px x py y = 0
the solution: combining the
rst two FOCs to eliminate λ, we have
(the maximum utility, subject to the budget constraint, must satisfy
these conditions)
Ux
Uy
=
px
py
and
px x + py y = M
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Exercise
max f (x , y , z) = (1+ x)yz
s.t. x + y + z 1
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