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Let $u(x)$ be a continous bounded function such that $u'(x) >0$, $u''(x) < 0$. Define $A(x) = -\frac{u''(x)}{u'(x)}>0$. Let $Y$ be a random variable with $\mathbb{E}|Y|<\infty$. I solve a maximization problem $$ \mathsf Eu\left( (K-x)C+ x Y \right) \to \max_{x} $$ This optimization problem may be interpreted as a problem of choosing a part $x$ of all available capital $K>0$ that we will spent on risky assets with random price $Y$, the remaining part $K-x$ of capital will be spent on non-risky assets with fixed price $C$. A solution of the problem will be denoted by $x(K)$.

How to show that there is a relation between $x'(K)$ and $A(x)$? $$ A(\cdot) \uparrow \implies x'(\cdot)<0 \\ A(\cdot) \equiv \mathrm{const} \implies x'(\cdot) = 0 \\ A(\cdot) \downarrow \implies x'(\cdot)>0 $$ where $(\cdot)$ means that this property holds for any possible arguments.

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What $x'$ means here? – Alecos Papadopoulos Jul 31 '13 at 16:02
@AlecosPapadopoulos, it is the derivative of $x$ with respect to $K$ – Nimza Jul 31 '13 at 20:59
You ask for a proof that there is a negative relationship between the coefficient of absolute risk aversion $A(w)$ and the amount of wealth invested in the risky asset, $x$, as wealth increases. I don't think this has a definite answer. The known result is that there is a negative relation between the coefficient of relative risk aversion, $wA(w)$ and the fraction of wealth invested in the risky asset, $x/K$, as wealth increases. Perhaps this latter relation is the one for which you seek proof? – Alecos Papadopoulos Aug 5 '13 at 0:57

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