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Integral interpretation

  1. Mar 20, 2016 #1
    • Member warned to take more care in starting threads
    1. The problem statement, all variables and given/known data

    Ct= (∫0 1 (Ct(i) ̂(1-1/ε)) di)^ (ε/ε-1)

    So Ct (i) denotes number of good i consumed by the household.

    But what is Ct ? What does it mean to take the intgral from zero to 1 for all the Ct (i) ?


    2. Relevant equations


    3. The attempt at a solution
     
    Last edited: Mar 20, 2016
  2. jcsd
  3. Mar 20, 2016 #2

    Ray Vickson

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    Here is what is meant by what you wrote:
    [tex] Ct = \frac{1}{\epsilon} \left( \int_0^1 \left( Ct(i)^1 - \frac{1}{\epsilon} \right) \ di \right)^{\epsilon} -1 [/tex]
    It that really what you meant?
     
    Last edited: Mar 20, 2016
  4. Mar 20, 2016 #3
    Sorry, I added some paranthesis so it should be right now. How do write math the way you do?
     
  5. Mar 20, 2016 #4

    Ray Vickson

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    I use LaTeX.

    What you wrote now has exponent ##\epsilon/\epsilon - 1##. Did you mean ##\epsilon/(\epsilon-1)##? If so, use parentheses around the terms in the denominator, like this: ε/(ε-1).

    Remember, when you are writing in plain text, a+b/c-d means ##a+\frac{b}{c}-d##, while (a+b)/(c-d) means ##\frac{a+b}{c-d}##.

    Anyway, you now (probably) have:
    [tex]
    Ct = \left( \int_0^1 \left( Ct(i)^{(\epsilon-1)/\epsilon} \right) \ di \right)^{\epsilon/(\epsilon-1)} [/tex]
    What is ##\epsilon##?
     
    Last edited by a moderator: Mar 20, 2016
  6. Mar 20, 2016 #5
    Good question. From what I understand it says something about consumption preferences. And it is presented in the complicated way to give a more meaningful expression later.

    What I now know is that this is simmilar to a sum and you could use sum notation to present the same and ( can it be correct?) that the integral from 0 to 1 is another way of saying the integral from 0 to infinity.

    So gbasically my interpretation now boils down to Ct being some sort of sum of all the small c`s, in other words an index.

    And yes, your new expression is the one I meant.
     
  7. Mar 20, 2016 #6

    Ray Vickson

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    I don't know the context of your problem, but I can give a possible interpretation (and which is possibly wrong). You have some kind of "consumption profile" ##Ct(i), 0 \leq i \leq 1##, and a "utility function" for consumption of the form ##u(c) = c^{\alpha}##; it happens that ##\alpha = (\epsilon-1)/\epsilon## in your case, but never mind that for now.

    Anyway, the total utility for the consumption profile is ##U(Ct) = \int_0^1 u(Ct(i)) \, di = \int_0^1 Ct(i)^{\alpha}\, di##. The quantity ##U(Ct)^{1/\alpha}## is the "utility-equivalent" consumption, which is the single consumption level ##C_{\alpha}## that has the same utility as the profile ##\{ Ct \}##. That is the quantity computed in your post.

    I don't know if that really means anything, but it is similar to the type of thing that people look at in Finance, when they talk about "certainty equivalents" for risky investments, for example.
     
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