# Proof that two utility functions $u(x_1,x_2) \text{ and } v(x_1,x_2)$ with the same MRS represent the same preference

I am having some difficulty proving that two utility functions with the same marginal rate of substitution represent the same preference. I think I need two use some argument related to strictly increasing monotonic transformation, but I am really lost. If someone could help me, it would be great.

Thank you very much!

• (1) Use the formula for the MRS of your utility function. (2) Apply a strictly increasing monotonic transformation to your utility function. (3) Use the formula for the MRS again, but for the transformed utility function. (4) Compare the results of (1) and (3). – VARulle Apr 19 at 14:19
• Thank you very much! – Jackaba Apr 19 at 14:52
• @VARulle That seems to go in the other direction. You show that two utility functions representing the same preferences have the same MRS, but the question is about the other direction. – Michael Greinecker Apr 19 at 16:08
• @MichaelGreinecker Oops, you're right, that was too quick... – VARulle Apr 20 at 5:51
• Next try: You know that (1) the MRS is the slope of the indifference curve, (2) the slope is the derivative, and (3) given the derivative and some initial value you can reconstruct the curve by integrating. Use (1)-(3) to show that the indifference curves for $u$ and $v$ are identical. Additionally assume that preferences are monotonically increasing to conclude that the preferences are the same. – VARulle Apr 20 at 7:12

I think that you need to also assume that the utilities are non-decreasing in the goods. That is, if $$x'_1>x_1$$, then $$u(x'_1,x_2)>u(x_1,x_2)$$.

The utility functions having the "same preferences" means

$$\forall (x_1,x_2), (x'_1,x'_2): u(x_1,x_2) > u (x'_1,x'_2) \text{ iff } v(x_1,x_2) > v(x'_1,x'_2)$$

So let's assume $$u(x_1,x_2) > u (x'_1,x'_2)$$. Take a curve going from $$(x_1,x_2)$$ to a point $$(x'_1, x''_2)$$ by changing $$x_1$$ to $$x'_1$$ and having the second coordinate change according the MRS so that this curve has constant value of $$u$$ [1]. Then since $$u$$ is constant along this curve, $$u(x'_1,x''_2) = u(x_1,x_2)$$. Since $$u(x_1,x_2) > u(x'_1,x'_2)$$, $$u(x'_1,x''_2) > u(x'_1,x'_2)$$, and from the non-decreasing utility property, we have $$x''_2 >x'_2$$.

Since this curve was defined by the MRS of $$u$$, and the MRS of $$u$$ is equal to that of $$v$$, we also have that $$v(x'_1,x''_2) = v(x_1,x_2)$$, and so $$v(x_1,x_2) > v(x'_1,x'_2)$$

[1] This can be done, for example, by taking $$C = (x_1+ t, x_2+\int_0^tMRS_{x_1x_2}dt)$$. Then taking $$t=x'_1-x_1$$, we get the point $$\left(x'_1, \int_0^{x'_1-x_1}MRS_{x_1x_2}dt \right)$$

• This is more an intuitive explanation than a proof. The MRS is a local concept, while your argument uses a discrete jump from one bundle to another and a bit of handwaving ("this curve was defined by the MRS of $u$..."). – VARulle Apr 21 at 8:33
• @VARulle It's not a discrete jump. It's a continuous traversal of a contour (i.e. curve with constant utility). I thought the trace of C was reasonable inferable, but I've made it more explicit. – Acccumulation Apr 21 at 16:57