CS代考 ECOS3010 - PowCoder代写

Week 1: Overview and Math Review
Monetary Economics ECOS3010
Monetary Economics (ECOS3010) Week 1: Overview and Math Review 1 / 18

Instructor: Associate Professor
O¢ ce: 552, Social Sciences Building
Phone: 9036 9311
Consultation Hour: Mondays 1:45-2:45 pm via Zoom.
Monetary Economics (ECOS3010) Week 1: Overview and Math Review 2 / 18

Textbook: , and , Modeling Monetary Economics, 5th edition, Cambridge University Press.
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Assessment
Assignment 1 (10%): Available Friday August 26, 5 pm. Due Friday September 16, 11:59 pm.
Assignment 2 (10%): Available Friday October 7, 5 pm. Due Friday October 28, 11:59 pm.
Midterm exam (30%): Monday, Sep 19, 11 am. Final exam (50%): during exam period.
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Lecture topics
Price surprises and the Phillips curve International monetary system
Money and capital
Liquidity and Önancial intermediation Bank risks, liquidity risks and bank panics Crisis
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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: whatís the derivative of F w.r.t to a
A: using chain rule,
∂F(x,y) = ∂F(x,y) ∂x(a) + ∂F(x,y) ∂y(a) ∂a ∂x ∂a ∂y ∂a
Monetary Economics (ECOS3010) Week 1: Overview and Math Review

Chain rule, log and exponential functions II Example:
problem: suppose Örmís proÖt function is Π(y ) = y 4 + 6y 2 5, where Π is the Örmís proÖt and y is the amount of output. Assume Örmís 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.
d (Π(y(L))) =
= [4(5L2/3)3 +12(5L2/3)](10L1/3) 3
Π0(y) y0(L)
= [4(y)3 + 12y] (10L1/3)
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Chain rule, log and exponential functions III which after simplifying equals
(4 125L2 + 60L2/3) (10L1/3) = 5000L5/3 + 200L1/3. 33
Properties of log functions: ln(ab)
= ln(a) + ln(b) = ln(a)ln(b)
ln(ab ) = b ln(a)
ln(1+x) x ,ifxissmall
Monetary Economics (ECOS3010)
Week 1: Overview and Math Review

Chain rule, log and exponential functions IV
Exponential function:
eab = eln(a) = a
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Unconstrained optimization I
consider a function y = f (x ). Necessary (Örst-order) condition for optimization of this function:
dy = df (x) = 0 dx dx
itís a maximum if d 2 f (x ) 0 (a concave function), a minimum if dx2
d2f (x) 0 (a convex function). dx2
example 1 (concave):
= xαx+1 ,where0<α<1 Ögure: .... Monetary Economics (ECOS3010) Week 1: Overview and Math Review 10 / 18 Unconstrained optimization II example 2 (convex): = xβx+1 ,whereβ>1
Ögure: ….
the value of x that maximizes y is called
x =argmaxf(x) x
y = f (x) is the maximum value of the function.
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Constrained optimization (Substitution method) I Example 1:
Max f(x,y) = xy x,y
s.t. 3x+4y = 16
Monetary Economics (ECOS3010)
Week 1: Overview and Math Review

Constrained optimization (Substitution method) II
Example 2:
consider the following utility maximization problem of a consumer/household
maxU = U(x,y) x,y
s.t. pxx+pyy M
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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 f(x,y) = xy x,y
3x+4y = 16
Örst, 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 = 163x4y
last, solve for three values using the system of equations above, yielding
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
maxU = U(x,y) x,y
s.t. pxx+pyy 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 ]
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Constrained optimization (Lagrangian method) IV Örst-order conditions (FOC)
∂L ∂x ∂L ∂y ∂L ∂λ
the solution: combining the Örst two FOCs to eliminate λ, we have (the maximum utility, subject to the budget constraint, must satisfy these conditions)
Ux = px Uy py
px x + py y = M
= Ux(x,y)λpx =0
= Uy(x,y)λpy =0
= Mpxxpyy=0
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max f(x,y,z) = (1+x)yz s.t. x+y+z 1
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