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I don't know how to compute the average income and the average utility for the second case. It is confused because I have to consider two densities $[0,0.5]$ and $[0.5,1.2]$.Can anybody tell me how to compute this?

Thank you in advance!

Edit: $\bar {z} =$ the average income, and $z_i$ = the income of an individual $i$. $U(\cdot) =$ utility function for all.

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  • $\begingroup$ First, instead of pasting a picture of the question, consider typing it out so that it becomes more searchable, which may benefit future visitors of this site. Second, explain the notations. For instance, what is $\bar z$? $\endgroup$ – Herr K. Apr 10 '18 at 18:05
  • $\begingroup$ Sorry. I edited it. $\endgroup$ – shk910 Apr 11 '18 at 2:23
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These are not two densities, but one density defined on the interval $[0,1.2]$, taking two different values. The average income is then the integral of the density function, i.e.

$$ \int_0^{0.5} 1 x \, dx + \int_{0.5}^{1.25} \frac{2}{3} x \, dx = \frac{9}{16} = 0.5625 $$

In order to obtain the average income, you do the same with the Utility and $\bar{z} = \frac{9}{16}$.

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  • $\begingroup$ Could you explain what $1$ on $[0,0.5]$ and $2/3$ on $[0.5,1.2]$ really mean? I understand your formula, but not clearly understand what the author want to say here. $\endgroup$ – shk910 Apr 10 '18 at 10:20
  • $\begingroup$ It demonstrates how income is distributed, normalized such that the area below the density sums up to 1. Initially, you had a uniform distribution between $0$ and $1$, i.e. the density had also the value $1$ on this interval. Then, the distribution becomes more stretched for the richer half of the population, i.e. the income range extends beyond the former maximum of $1$. In order to still represent a density, the relative mass above $0.5$ needs to decrease. Try to draw both income distributions to get an idea what's going on. $\endgroup$ – E. Sommer Apr 10 '18 at 11:48

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