1
$\begingroup$

I encounter the following problem today. It seems a simple question.

Let $U$ be a real function from $R^+\rightarrow \bar{R}$ satisfying the following conditions:

(1) $U$ is concave, continuous, and strictly increasing,

(2) $\limsup_{x\rightarrow +\infty}\frac{xU^{'}(x)}{U(x)} <1.$

(3) $ U^{'}(0+) = +\infty, \mbox{ and }U^{'}(+\infty) = 0.$

Is the following statement true?

$\bf Claim:$ For any non-negative random variable $\xi,$ if $E[U(\xi)] < +\infty,$ then we have $$E[\xi U^{'}(\xi)] < +\infty.$$

$\bf Remark:$ If $U(0) >-\infty,$ it is trivial if one notices that $$0\leq \xi U^{'}(\xi) \leq U(\xi) - U(0).$$ But in general, the property $U(0) >-\infty$ does NOT hold. For example $\ln(x).$

Any comment and suggestion are welcome. Thanks.

Improve this question
$\endgroup$
2
  • $\begingroup$ As you may assume wlog '$U(\infty)>0$', it should follow directly from the asymptotic elasicity condition (2), no? As this inequality holds for $x$ sufficiently large, you can just separate the domain of integration and use for the second part the inequality... $\endgroup$
    – Stephan Sturm
    Commented Jun 5, 2012 at 20:45
  • $\begingroup$ Yes. If $x$ is large, there is no problem. But problem arises when $x$ goes to zero such as $\ln(x).$ $\endgroup$
    – Jun
    Commented Jun 5, 2012 at 21:20

1 Answer 1

Reset to default
2
$\begingroup$

Take $U(x) = -e^{1/x^2}$ and the probability density function $f(x) = C e^{-1/x^2}$, $x \in [0;1]$, where $C$ is normalization constant.

Then, $U'(x) = \frac{2 e^{1/x^2}}{x^3}$, $E(U(\xi)) = \int_0^1 U(x) f(x) dx = -C$, $E(\xi U'(\xi)) = \int_0^1 x U'(x) f(x) dx = 2C \int_0^1 \frac{dx}{x^2} = +\infty$

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .