How do I prove that it is convex? How to prove it with the typical t in [0,1] definition? I am having a bad time typing equations here, sorry. please help me out
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$\begingroup$ Do you insist on proving it with the [0,1] definition, or would you consider using derivatives? $\endgroup$– GiskardNov 19, 2022 at 13:03
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$\begingroup$ @Giskard using derivative? If i have to use the Hessian definition, it should be positive semi-definite? But I don't know what is positive semi-definite though? Also, here is there any difference in proving the preference is convex vs the function is convex? $\endgroup$– reindeerNov 19, 2022 at 13:24
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$\begingroup$ Is it $\ln[(x_1)^2]$ or $[\ln(x_1)]^2$? $\endgroup$– Richard HardyNov 20, 2022 at 8:35
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$\begingroup$ I think it is enough clear that it is $ln [(x_1)^2]$ The formula in the question is correct, maybe there are superfluous brackets, but with brackets maybe is better: without brackets, a square, as there is a subscript near the $x_i$, could be confused with a superscript. $\endgroup$– BakerStreetNov 22, 2022 at 12:35
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$\begingroup$ @reindeer Just to clarify, you are asking whether this utility function is convex? (As opposed to whether it represents convex preferences, or generates convex indifferences curves.) $\endgroup$– afreelunchNov 22, 2022 at 14:51
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3 Answers
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We can make life easier taking a monotonic transformation of $$U=log (x_1)^2+log (x_2)^2.$$
This is possible as monotonic transformations of a utility function represent the same preferences.
We can see that the utility function above can be reconducted to a Cobb-Douglas utility function $U(x_1,x_2)= x_1^ax_2^b$, $a,b>0$.
Take
$U_1(x_1,x_2)=e^{log (x_1)^2+log (x_2)^2}= x_1^2 x_2^2, $
which can be furthermore transformed (as the square is a monotonic transformation for positive values) as:
$$U_2(x_1,x_2)=x_1 x_2.$$
The indifference curves are given by:
$$U_2(x_1,x_2)=x_1 x_2=c.$$ $c$ constant,
that is:
$$x_1=\frac{c}{x_2}.$$
The indifference curves are hyperbolas, that are convex, the preferences are convex.
answered Nov 20, 2022 at 10:56
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$\begingroup$ Isn't the question whether the utility function is convex? (As opposed to whether it generates convex indifference curves, or rationalises convex preferences.) $\endgroup$ Nov 22, 2022 at 14:50
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$\begingroup$ They are connected questions. A preference can defined as ' convex if the upper contour set is convex". This definition is equivalent to the definition with $\lambda$ and $[0,1]$ and so on. To see these definitions you can refer to Mas Colell, Microeconomic Theory. $\endgroup$ Nov 22, 2022 at 15:06
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$\begingroup$ I don't think the connection is clear at all. For example, Cobb Douglas preferences can be represented using a concave utility function, but generate convex indifference curves. In contrast, there is a connection between convexity of indifference curves and quasi-concavity of the utility function. $\endgroup$ Nov 22, 2022 at 15:08
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$\begingroup$ Yes, you are right, but here we are talking about the convexity of preferences and indifference curves, the question of the OP is about convexity of preferences, that can be defined the two way I mentioned above. We are not talking about concavity of utility functions and convexity of indifference curves, this is another question.. $\endgroup$ Nov 22, 2022 at 15:14
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$\begingroup$ @BakerStreet If you read OP's question, it seems to be about the convexity of the utility function. Like you, I suspected that this was not the intended question. $\endgroup$ Nov 22, 2022 at 15:21
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A proof that used a calculator:
Let $y = (\frac{3}{4},\frac{1}{4}), z = (\frac{1}{4},\frac{3}{4}), x = (e,1)$
We have that
$U(x) = (\log(e))^2 + (\log(1))^2 = 1^2 + 0^2 = 1 + 0 = 1$,
$U(y) = (\log{\frac{3}{4}})^2 + (\log{\frac{1}{4}})^2 = (\log(3) - \log(4))^2 + (-\log(4))^2 = (\log(4) - \log(3))^2 + (\log(4))^2 = (\log(4))^2 - 2 \log(3)\log(4) + (\log(4))^2 = 2 (\log(4))^2 - 2 \log(3) \log(4)$
$= 2 (2 \log(2))^2 -4 \log(3) \log(2) = 8 \log(2) - 4 \log(3) \log(2) = (8-4\log(3)) \log(2) \approx 2$
Note that $U(x_1,x_2)$ is symmetric in $x_1, x_2$.
From symmetry, $U(z) = (8-4\log(3)) \log(2) \approx 2$.
This implies that $y \geq x$ and $z \geq x$.
But for $\lambda = \frac{1}{2} \in [0,1]$,
$\lambda y + (1-\lambda)z = \frac{1}{2} y + \frac{1}{2} z = (\frac{1}{2},\frac{1}{2})$,
and
$U(\frac{1}{2},\frac{1}{2}) = (\log(\frac{1}{2}))^2 + (\log(\frac{1}{2}))^2 = 2 (\log(\frac{1}{2}))^2 = 2 (-\log(2))^2 = 2 (\log(2))^2 \approx 0.96 < 1 = U(x)$
Therefore, $\lambda y + (1-\lambda)z < x$.
From this, we can conclude that the preference relation represented by the utility function is not convex.
answered Nov 19, 2022 at 21:47
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$\begingroup$ the other answer from @BakerStreet contradicts this?? $\endgroup$– reindeerNov 22, 2022 at 15:07
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2$\begingroup$ My answer if for (ln(x_i))^2 and @BakerStreet ‘s is for (ln(x_i^2)). They’re not the same function. I thought you meant the former, hence I answered like that. $\endgroup$ Nov 22, 2022 at 16:19
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In one line: the utility function you posted is monotonic and continous. Any monotonic function is quasi-concave. Quasi-concavity implies convex preferences.
Quickest possible way to see that this function is strictly concave is to simply notice (no algebra) that the Hessian is negative semi-definite, as the cross derivatives are zero.
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$\begingroup$ Maybe be you refer to functions from $\mathbb{R}$ to $\mathbb{R}$. A function defined on an interval of $\mathbb{R}$ is quasi-concave if it is monotonic, that’ s true. But the function in the question is a function of two variables. You can’t extend this statement to function from $\mathbb{R^2}$ to $\mathbb{R}$, because in $\mathbb{R^2}$ , in mathematical analysis, we don’t speak of ‘monotonic function’. A definition of monotonicity requires an order. In $\mathbb{R}$ there is an order, in $\mathbb{R^2}$ an order isn’t defined $\endgroup$ Nov 22, 2022 at 22:13