# Can i check monotonicity with the marginal utility?

For example I have a Cobb-Douglas utility function $$U(x,y)$$ and I want to check the monotonicity property. Can I use the marginal utility functions to see that they are always positive to conclude that $$U(x,y)$$ has the monotonicity property?

• What exactly do you mean by the monotonicity property? Can you provide a definition? (Doing so may help you answer the question yourself.) – Giskard Sep 21 '19 at 16:29
• Hi Giskard, Monotone is that for any $x,y\in X$, if $y\gg x$, then $y\succ x$. That means that the agent prefers all consumption bundles that have more of all goods – neto333 Sep 21 '19 at 17:06
• I suppose you could, but why not just look at that function and determine whether it is increasing with each argument? Or look at the marginal utility of $y$ and determine if it had a larger effect on the utility function than $x$ does such that it satisfies the definition of monotone you provided in the above comment – Brennan Sep 21 '19 at 23:34

Yes. But you should review the definition of monotonicity (and perhaps of marginal utility) so that you understand why.

• For strict monotonicity, the marginal utility of both goods should be strictly positive everywhere. However, marginal utility of x will be 0 at y=0 (and similarly for marginal utility of y). Is the function still considered strictly monotone? – PGupta Jul 9 at 8:04

Consider $$f:\mathbb{R}\to\mathbb{R}$$ which has a positive derivative. Then by the fundamental theorem of calc we have

$$f(a)=\int_{0}^a\frac{df}{dx}dx-f(0)$$

Therefore, $$f$$ is monotonic (you are integrating a positive function over a larger area when you increase $$x$$). For a function $$U$$ of two variables with positive partial derivatives we have similar results: If $$a,b>0$$ then $$U(x,y) and $$U(x,y). This follows because we can always fix $$y_0$$ and view $$U(x,y_0)$$ as a function of one variable call it $$f_{y_0}(x)$$. Then $$\frac{\partial U}{\partial x}(x,y_0)=\frac{df_{y_0}}{dx}(x)$$ and apply the above results.

Hence we have,

$$U(x,y)