Changes in multiplicative risks and optimal portfolio choice: new interpretations and results

Abstract

This paper reconsiders the conditions determining the optimal response of a decision maker in case of stochastic changes in multiplicative risks. In particular, we focus on an optimal portfolio choice where the return of the risky asset exhibits an Nth-degree risk increase. We provide two interpretations of the conditions analyzed. The first interpretation involves a comparison between the elasticities with respect to the investment in the risky asset of the Nth derivative of the utility function and of the distance between the Nth moments of the two risks. The second interpretation refers to the direction of the change in the utility premium when the investment in the risky asset changes. We then study the linkages between the conditions determining optimal responses of risky investment in the case of risk increases of different degrees. We show that, under some assumptions, the optimal behavior of an agent in the case of Nth degree risk increase makes it possible to infer agent’s behavior in case of risk increases of lower degrees.

Proofs of lemmas

Proof of Lemma 1

• (i) It follows from Proposition 3 in Menegatti (2015).

• (ii) We prove this result by induction. Let us start with the case $n=2$. Strict concavity implies that $u'(x)$ is strictly decreasing, and being also strictly positive, $u'(x)$ converges to some limit $0 \le l < + \infty$ as $x \rightarrow +\infty$. Condition $u^{(3)} (x)>0$ for all $x \ge a$, implies that $u''(x)$ is strictly increasing, hence it converges, as $x \rightarrow + \infty$, to some real (non-positive) number. By the asymptote criterion, $\displaystyle \lim \nolimits _{x\rightarrow + \infty } u''(x) = 0$.

  Let $n>2$ and assume that $\lim \nolimits _{x\rightarrow + \infty } u^{(n)}(x) = 0.$ If n is even (resp. odd), the function $u^{(n+1)}$ is nonnegative and decreasing (resp. non-positive and increasing); therefore, it converges to some limit $l \ge 0$ (resp. $l\le 0$). By the asymptote criterion $l=0$.

$\square$

Proof of Lemma 2

Since $\lim _{x\rightarrow + \infty } u''(x) = \lim _{x\rightarrow + \infty } \frac{1}{x} = 0$, the expression $-x u''(x) = \frac{- u''(x)}{\frac{1}{x}}$ takes the indeterminate form $0/0$ as $x \rightarrow +\infty$, as well as $(-1)^{k-1} x^{k-1} u^{(k)}(x) = \frac{(-1)^{k-1} u^{(k)}(x)}{\frac{1}{x^{k-1}}}$, for $k=1, \ldots , n$. Applying L’Hôspital’s rule n times, we get:

$$\begin{aligned} \lim _{x\rightarrow +\infty } \frac{- u''(x)}{x^{-1}}= & {} \lim _{x\rightarrow +\infty } \frac{u^{(3)}(x)}{x^{-2}} = \lim _{x\rightarrow +\infty } \frac{- u^{(4)}(x)}{2x^{-3}} = \cdots \nonumber \\= & {} \lim _{x\rightarrow +\infty } \frac{(-1)^{n-1} u^{(n)}(x)}{\frac{(n-2)!}{x^{n-1}}} = \frac{L'}{(n-2)!} < +\infty \end{aligned}$$

(6)

$\square$

Proof of Lemma 3

It is sufficient to observe that

$$\begin{aligned} r'_n(x,z)= & {} \frac{\left[ -u^{(n+1)}(x+z) - xu^{(n+2)}(x+z) \right] u^{(n)}(x+z) +x\left[ u^{(n+1)}(x+z) \right] ^2 }{\left[ u^{(n)}(x+z)\right] ^2} \\= & {} - \frac{u^{(n+1)}(x+z)}{u^{(n)}(x+z)} \left[ 1-r_{n+1}(x,z) + r_n(x,z) \right] \end{aligned}$$

and that $- \frac{u^{(n+1)}(x+z)}{u^{(n)}(x+z)} > 0$ because of the sign alternation in the derivatives. $\square$

Proof of Lemma 4

Fix $z\ge 0$ and let ${\tilde{r}}_n(x) =r_n(x,z).$ Assume by contradiction that ${\tilde{r}}_{n-1}$ is not weakly decreasing. Then, either it is weakly increasing, and strictly increasing on some interval $(x_a, x_b)$ ($x_b$ may be $+ \infty$), or it admits some maximum or minimum. Let us first suppose that it is monotone and strictly increasing on $(x_a, x_b)$, or equivalently ${\tilde{r}}_{n-1}(x) +1 > {\tilde{r}}_{n}(x)$ on $(x_a, x_b)$. Then we can find $x_2 > x_1 \ge x_a$ such that

$$\begin{aligned} {\tilde{r}}_{n-1}(x_2) +1> {\tilde{r}}_{n-1}(x_1) +1 > {\tilde{r}}_{n}(x_1) \ge {\tilde{r}}_n(x_2). \end{aligned}$$

(7)

Both functions ${\tilde{r}}_{n-1}$ and ${\tilde{r}}_{n}$ are monotone, therefore they admit limit at $+\infty$. In particular, let $\lim _{x \rightarrow +\infty } {\tilde{r}}_{n-1}(x) = l \in [0, + \infty ]$.

We prove that l cannot be $+ \infty$. Indeed, assume $l=+\infty$. Then for some M, ${\tilde{r}}_{n-1}(x) > {\tilde{r}}_n(x)$ for all $x\ge M$. Let $a_n(x) = -\frac{u^{(n+1)}(x+z)}{u^{(n)}(x+z)} = \frac{r_n(x,z)}{x}.$ Then, working as in the proof of Lemma 3, we show that $a_n$ is weakly (strictly) increasing if and only if ${\tilde{r}}_n \ge {\tilde{r}}_{n+1}$ (${\tilde{r}}_n > {\tilde{r}}_{n+1}$). But then, since ${\tilde{r}}_n(x) < 1+ {\tilde{r}}_n(x)\le {\tilde{r}}_{n+1}$ and ${\tilde{r}}_{n-1}(x) > {\tilde{r}}_{n}(x)$ for $x \ge M$ we have that $a_n$ is strictly decreasing and $a_{n-1}$ is strictly decreasing on $[M, +\infty )$. This implies that both $a_{n-1}(x)$ and $a_n(x)$ admit limit at $+\infty$, and $\lim _{x \rightarrow +\infty } a_n(x)$ is finite since the function is decreasing; moreover, there exist $x_4>x_3 > M$ such that

$$\begin{aligned} \lim _{x \rightarrow +\infty } a_n(x)< a_n(x_4)< a_n(x_3)< a_{n-1}(x_3)< a_{n-1}(x_4)< \lim _{x \rightarrow +\infty } a_{n-1}(x) \end{aligned}$$

(8)

(the third inequality follows from the fact that ${\tilde{r}}_{n-1}(x) > {\tilde{r}}_{n}(x)$ is equivalent to $a_{n-1}(x) > a_{n}(x)$). However, since Lemma 2(ii) holds, we can apply L’Hospital’s theorem and get

$$\begin{aligned} \lim _{x \rightarrow +\infty } a_{n-1}(x)= & {} \lim _{x \rightarrow +\infty } -\frac{u^{(n)}(x+z)}{u^{(n-1)}(x+z)}=\lim _{x \rightarrow +\infty } -\frac{u^{(n+1)}(x+z)}{u^{(n)}(x+z)} \\= & {} \lim _{x \rightarrow +\infty } a_n(x)< +\infty \end{aligned}$$

which contradicts (8). As a consequence, l must be finite.

Since $l < +\infty$, by definition of limit, we have that definitely $|{\tilde{r}}_{n-1}(x) - l | < \varepsilon$ for every $\varepsilon > 0$, or equivalently

$$\begin{aligned} (l-\varepsilon ) |u^{(n-1)}(x)| \le (-1)^{n+1}x u^{(n)}(x) < (l+\varepsilon ) |u^{(n-1)}(x)|, \end{aligned}$$

(9)

hence $\lim _{x \rightarrow +\infty } x u^{(n)}(x) = 0$. Then, L’Hôspital’s theorem entails

$$\begin{aligned} l= \lim _{x \rightarrow +\infty } {\tilde{r}}_{n-1}(x) = \lim _{x \rightarrow +\infty } \left( -1 - \frac{xu^{(n+1)}(x)}{u^{(n)}(x)}\right) = -1 + \lim _{x \rightarrow +\infty } {\tilde{r}}_{n}(x) \end{aligned}$$

Thus, we have

$$\begin{aligned} l +1 \ge {\tilde{r}}_{n-1}(x_2) +1> {\tilde{r}}_{n-1}(x_1)+ 1 > {\tilde{r}}_n(x_1) \ge {\tilde{r}}_{n}(x_2) \ge l+ 1 \end{aligned}$$

for $x_1 < x_2$ which is clearly a contradiction. This shows that if ${\tilde{r}}_{n-1}$ is not weakly decreasing, than it cannot be monotone. As a consequence, it must admit either a maximum or a minimum. The argument above shows that it cannot be strictly increasing on an interval of the form $[x_a, + \infty )$. Therefore, the function must admit a strict local maximum point at some point $x_m$, namely ${\tilde{r}}'_{n-1} (x_m) =0$, or equivalently ${\tilde{r}}_{n-1}(x_m) +1 = {\tilde{r}}_n(x_m)$ and there exists some ${\bar{x}}<x_m$ where ${\tilde{r}}_{n-1}({\bar{x}}) < {\tilde{r}}_{n-1}(x_m)$ and ${\tilde{r}}'_{n-1}({\bar{x}}) > 0$. This implies that ${\tilde{r}}_n({\bar{x}})< {\tilde{r}}_{n-1}({\bar{x}}) + 1 < {\tilde{r}}_{n-1}(x_m) + 1= {\tilde{r}}_n(x_m)$ which contradicts the fact that ${\tilde{r}}_n(x)$ is weakly decreasing. So our claim is proved.

If ${\tilde{r}}_n (x)$ is strictly decreasing, then ${\tilde{r}}_n(x) +1 > {\tilde{r}}_{n+1}(x)$. Moreover ${\tilde{r}}_{n-1}$ is weakly decreasing by the previous result. If ${\tilde{r}}_{n-1}$ were constant on some interval $(x_a, x_b)$, then there would exist $x_1 < x_2 \in (x_a, x_b)$ such that ${\tilde{r}}_n(x_1) = {\tilde{r}}_{n-1}(x_1)-1 ={\tilde{r}}_{n-1}(x_2) -1 = {\tilde{r}}_{n-1}(x_2) - 1 = {\tilde{r}}_n(x_2)$ which contradicts the strict monotonicity of ${\tilde{r}}_n(x)$. Therefore ${\tilde{r}}_{n-1}$ must be strictly decreasing as well. $\square$