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I'm working on an economics paper, and in the model I've made I've basically gotten myself a little bit stuck. I need to show that there exists a nondecreasing concave function $u$ and numbers $P$ and $\theta$ with $P>\theta>0$, and $\gamma\in[0,1]$ such that:$$u(P-\theta)-u(-\theta)>\frac{1}{1-\gamma}(u(P)-u(0))$$

And in fact what I would like to show is that for any monotonically increasing $u$ which is strictly concave, we can find $P$, $\theta$, and $\gamma$ which satisfy the equality. I don't have much experience proving that kind of a statement though and I'm having some trouble getting started. Does anyone have any idea what would be a good way to start proving that statement (if it's even true—when I draw pictures of what I want it looks like it should be true but maybe it's not)?

EDIT

I've been puzzling over this and I realize that it will follow directly from a lemma: for $a>b$ and $c>0$, $u(a)-u(b)>u(a+c)-u(b+c)$, since then I can say that $u(P-\theta)-u(-\theta)>u(P)-u(0)$ just by adding $\theta$ to the arguments of $u$ on the left hand side. So I just need to prove that lemma. I can see geometrically why it must be true but I still can't quite make it follow form the concavity of $u$. I'm close though, so I may end up just answering my own question. Writing it out like this is helping.

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does $\gamma$ belong to $[0,1]$ or we need $\gamma$ to be strictly between zero and one? Based on your expression it seems to me that $\gamma \in (0,1)$ is more appropriate (take $\gamma=1$ for instance). –  Cristian Sep 17 '12 at 19:09
    
well $\gamma$ is meant to be a probability so $\gamma=1$ would be kind of a degenerate case –  crf Sep 17 '12 at 23:11
    
but then your inequality cannot hold. –  Cristian Sep 17 '12 at 23:21
    
@Cristian Good point. This inequality came from some manipulations on another inequality in which there is no $1/(1-\gamma)$ term, so I'll need to specify that this is only in the case where $\gamma<1$. Thanks! EDIT actually I realize now that I'm not making the claim for every $\gamma\in[0,1]$ —I'm just saying that some such $\gamma$ exists. So it's perfectly fine to let $\gamma\in (0,1)$. It just happens to never hold for $\gamma =1$ which makes quite nice sense in the model. –  crf Sep 17 '12 at 23:26
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up vote 1 down vote accepted

Consider the function $h(x)=U(x+c)-U(x)$. Its derivative is $U\prime(c+x)-U\prime(x)$. Because $U(x)$ in concave then $U\prime\prime<0$, which means that $U\prime$ is decreasing and hence $h\prime$ is negative. Hence $h(x)$ is decreasing.

Your inequality follows directly: $h(a)<h(b)$ for $a>b$

We can also prove a more general result without resorting to derivatives. Let $F$ be a concave function in an interval and let $x<y\le z<w$ be points in this interval. Then the following holds:$$\frac{F(y)-F(x)}{y-x} \ge \frac{F(w)-F(z)}{w-z}$$ To prove this write $y$ as a linear combination of $x$ and $w$ thus:$$y=\frac{w-y}{w-x}x+\frac{y-x}{w-x}w$$ This implies: $$ \begin{align*} F(y) &\ge \frac{w-y}{w-x}F(x)+\frac{y-x}{w-x}F(w) \\ &= F(x) + \frac{y-x}{w-x}(F(w)-F(x))\\ &= F(x) + \frac{y-x}{w-x}(F(w)-F(y)+F(y)+F(x))\\ \end{align*}$$

This is equivalent to:$$\frac{F(y)-F(x)}{y-x}\ge\frac{F(w)-F(y)}{w-y}$$ In the same way we can prove that (note this goes in the other direction):$$\frac{F(w)-F(y)}{w-y}\le\frac{F(w)-F(x)}{w-x}$$

Combining the last two inequalities leads to our result:$$ \frac{F(y)-F(x)}{y-x}\ge\frac{F(z)-F(y)}{z-y}\ge\frac{F(w)-F(z)}{w-z} $$

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Ahh, perfect. It took me a couple of minutes to see how $(F(y)−F(x))/(y−x)\geq (F(w)−F(z))(w−z)$ would help but I got it. Thanks so much! –  crf Sep 17 '12 at 23:28
    
Just a quick note. I did not make use of monotonicity to prove this. So you can drop that assumption if that makes sense in the model. –  ivan Sep 18 '12 at 5:38
    
Good point, and yeah that makes sense graphically now that I think of it that monotonicity shouldn't make a difference. However, utility functions tend to increase monotonically by assumption, so it's in there anyways. For the record, the concavity means risk-aversion. So basically I'm basically proving that risk-averse agents take riskier risks when they already have some money in the bank. Thanks so much for your help! –  crf Sep 18 '12 at 6:06
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