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?
1 Answer
$\begingroup$
$\endgroup$
5
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.
-
-
$\begingroup$ 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)). $\endgroup$– BayesianCommented Nov 26, 2020 at 16:55
-
$\begingroup$ @Lifeni if you think this answer answered your question consider accepting it. $\endgroup$– 1muflon1 ♦Commented Nov 26, 2020 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? $\endgroup$– LifeniCommented Nov 26, 2020 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. $\endgroup$– BayesianCommented Nov 26, 2020 at 17:17