We let $g(z)$ be a strictly monotonous function so: $$\frac{dg(z)}{dz}>0$$ Consumer 1 has preferences given by the utility function $u(x_1,x_2)=ln(x_1)+2ln(x_2)$, while consumer 2 has preferences given by n $v(x_1,x_2)=g(x_1x_2^2)$. Then I have to show that consumer 2 got same preferences as consumer 1. I think I have to use MSR on $x_1x_2^2$ and on $v(x_1,x_2)$. For MSR on $x_1x_2^2$ I get: $$MRS=-\frac{\frac{\partial }{\partial x_1}}{\frac{\partial }{\partial x_2}}=-\frac{x_2^2}{2x_1x_2}$$ But How can I find MSR on $v(x_1,x_2)$ (I think I got the same MRS if I use the chain rule, but I'm not sure?), and how can I use this to conclude that the ranking of the two indifference curves is the same when $g(z)$ is monotonous? I hope that someone can help me?

  • $\begingroup$ Note that ln(x1)+2*ln(x2) = ln(x1)+ln(x2^2) = ln (x1 * x2^2) and then verify that MRS is invariant to any positive monotone transforation. $\endgroup$ – Bayesian Nov 26 '20 at 16:41

Note that $$ln(x_1)+2*ln(x_2) = ln(x_1)+ln(x_2^2) = ln (x_1 * x_2^2),$$ and note that $$MRS_v = \frac{g'(x_1 * x_2^2) x_1 * 2 x_2}{g'(x_1 * x_2^2) x_2^2}$$ such that the derivative of $g$ cancels out.

For more intuition, see here.

  • $\begingroup$ Nice that makes sense. Thank you $\endgroup$ – Lifeni Nov 26 '20 at 16:53
  • $\begingroup$ You should also see that if you prefer bundle x1,x2 to y1,y2 then it must be that u(x1,x2) > u(y1,y2). If you now apply a positive monotone transformation g, you can see that u(x1,x2) > u(y1,y2) if and only if g(u(x1,x2)) > g(u(y1,y2)). $\endgroup$ – Bayesian Nov 26 '20 at 16:55
  • $\begingroup$ @Lifeni if you think this answer answered your question consider accepting it. $\endgroup$ – 1muflon1 Nov 26 '20 at 16:56
  • $\begingroup$ Hi Bayesian that makes sense and the intuition in the link makes sense but have can I formally show that the ranking of the two indifference curves is the same when 𝑔(𝑧) is monotonous? $\endgroup$ – Lifeni Nov 26 '20 at 17:06
  • $\begingroup$ See my comment if you compare any two bundles then you have u(x1,x2) > u(y1,y2) if and only if g(u(x1,x2)) > g(u(y1,y2)) by definition of g being a positive monotone transformation. Therefore the ranking is not changed. $\endgroup$ – Bayesian Nov 26 '20 at 17:17

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