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W3213_-_Math_notes4

# W3213_-_Math_notes4 - W3213 Intermediate Macroeconomics...

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W3213 - Intermediate Macroeconomics, section 002 Fall 2008 Mathematics notes °12/02/08 FINAL version Ricardo Reis Columbia University 1. A few basic principles There are a few mathematical tools that we will use repeatedly: 1.1 Changes/derivatives In a model that relates three variables, Y , T , I linearly: Y = a + b ( Y ° T ) + I; we will use the operator ° Y to mean the change in Y , which you can think of as Y after ° Y before . Evaluating this operator in the equation above: ° Y = b Y ° ° T ) + ° I + ° G: Note that for the constant a , ° a = 0 , and that for the product of a constant and a variable °( bY ) = b ° Y . Say we were studying the e/ect of ° T on ° Y . Then, ° I = ° G = 0 , and the expression above becomes (1 ° b Y = ° b ° T: 1

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If I therefore asked: what happens if T increases by \$100 million? The answer would be that Y falls by \$[ b= (1 ° b )] ± 100 million. Note that, in the limit, when the change between before and after is in±nitesimal ° Y becomes dY , your familiar derivative operator. So, in this limit, the expression above would say: dY dT = ° b 1 ° b : In all cases in the class this equivalence between ° Y and dY will be assumed to be approx- imately true. 1 So, we will refer to applying the ° operator as taking ²total derivatives³and you can use all of the properties of derivatives that you know. In particular, think of the non-linear relation: Y = CI In this case: ° Y = I ° C + C ° I; so if I am studying the e/ect of a change in I on C , keeping Y ±xed then: ° C C = ° ° I I : This takes us to the next topic. 1 For those of you appreciative of mathematical rigor, the equalities involving dY are all exact, whereas those involving ° Y are linear approximations. 2
1.2 Proportional change / growth rates If the level of a variables is Y , and its change is ° Y , then its proportional change is ° Y=Y . The relation above said that C fell by the same proportion as I rose so that their product remained constant. Returning to that problem note that as ° Y = I ° C + C ° I; we can divide both sides by Y using the fact that Y = CI to get: ° Y Y = ° C C + ° I I : So the proportional change in Y is equal to the sum of the proportional changes in C and I . This will be a recurring case in the class, relating the proportional changes in variables. Making the link to derivatives, note the following steps: Y = CI , ln( Y ) = ln( C ) + ln( I ) ) dY Y = dC C + dI I ; where the last step uses the fact that the total derivative of ln( x ) is dx=x . A slightly harder case is Y = X ° , but it is solved by the steps: dY=dX = °X ° ° 1 ) dY = °X ° ° 1 dX ) dY Y = °X ° ° 1 dX X ° ) dY Y = °dX X or ° Y Y = ° ° ° X X ± : We will sometimes talk of growth rates . The growth rate of a variable Y is its proportional change over a unit of time, and we will sometimes denote it by g Y . If time is discrete (1,2,3,...) then a unit of time is 1, and g Y = ° Y=Y . Nothing is new here, so in the Y = CI case, g Y = g C + g I . If time is continuous, then we really want to talk of ( dY=dt ) =Y . But dividing 3

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the expression above by dt , you get immediately: dY=dt Y = dC=dt C + dI=dt I , g Y = g C + g I ; and you get the same thing.
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