On temperance and risk spreading

Abstract

This paper shows that temperance is the highest order risk preference condition for which spreading N independent and unfair risks provides the highest level of welfare than any other possible allocations of risks. These results are also interpreted through the concept of N-superadditivity of the utility premium. This paper provides a novel application of temperance, not in terms of two risks as it is common, but in terms of N risks.

Appendix: Proof of Proposition 1

The proof uses the theorem of Eeckhoudt et al. (2009) and its corollary.

Theorem

(Eeckhoudt et al. 2009) Suppose that$Y_{i}$ dominates $\eta$$_{i}$ via $n_{i}$th order stochastic dominance for$i=1,2$ ($\eta _{1}\preceq _{{\text {SD}}_{n_{1}}}Y_{1}$and$\eta _{2}\preceq _{{\text {SD}}_{n_{2}}}Y_{2}$) and suppose that$\eta _{1},\eta _{2},Y_{1},Y_{2}$are mutually independent risks. The 50–50 lottery$[\eta _{1}+Y_{2},Y_{1}+\eta _{2}]$dominates the 50–50 i $[\eta _{1}+\eta _{2},Y_{1}+Y_{2}]$ via $(n_{1}+n_{2})$th order stochastic dominance.

Corollary

(Eeckhoudt et al. 2009) Suppose that$Y_{i}$dominates$\eta _{i}$ via $n_{i}$th order Ekern dominance for$i=1,2$ ($\eta _1 \preceq _{{\text {Ekern}}_{n_1}} Y_1$and$\eta _2 \preceq _{{\text {Ekern}}_{n_2}} Y_2$) and suppose that$\eta _{1},\eta _{2},Y_{1},Y_{2}$are mutually independent risks. The 50–50 lottery$[\eta _{1}+Y_{2},Y_{1}+ \eta _{2}]$dominates the 50–50 lottery$[\eta _{1}+ \eta _{2},Y_{1}+Y_{2}]$ via $(n_{1}+n_{2})$th order Ekern dominance.

When $Y_{1}=Y_{2}=0$ and using the Expected Utility Theory model, the EST (2009) theorem and its corollary, respectively, can be rewritten as follows:

EST (2009) Theorem: the inequality $E[u(x+\eta _{1})]+E[u(x+\eta _{2})]\ge u(x)+E[u(x+\eta _{1}+\eta _{2})]$ holds for all utility function u such that $(-1)^{(s+1)}u^{(s)}>0$$\forall s=1, \ldots , n_{1}+n_{2}$.

EST (2009) Corollary: the inequality $E[u(x+\eta _{1})]+E[u(x+\eta _{2})]\ge u(x)+E[u(x+\eta _{1}+\eta _{2})]$ holds for all utility function u such that $(-1)^{(n_1+n_2+1)}u^{(n_1+n_2)}>0$.

In the case $N=2$, Proposition 1 and Corollary 1, respectively, coincide with the EST theorem and its corollary. To prove the result in the case of N risks with $N \ge 3$, we use the property of convolution stability of $SD_{k}$ (see Propositions 3–11 in Denuit et al. (1998) for more details):

Convolution stability propertySuppose two random independent variables $X_1$ and $X_2$such that $X_{1}\preceq _{{\text {SD}}_{k}}0$ and $X_{2} \preceq _{{\text {SD}}_{k}}0$ for all$k=1,2$, then$X_{1}+X_{2}\preceq _{{\text {SD}}_{k}} 0$ for all$k=1,2$.

We make the proof of Eq. (6) by steps. Let us rewrite Eq. (6) as follows:

$$\begin{aligned}&\Bigl (\{X_{i}\}_{N}^{N}\Bigr )\succeq _1 \Bigl (\{X_{i}\}_{N}^{N-1}\Bigr ) \succeq _2 \Bigl (\{X_{i}\}_{N}^{N-2}\Bigr ) \succeq _3 \Bigl (\{X_{i}\}_{N}^{N-3} \Bigr ) \nonumber \\&\quad \succeq _4 \cdots \succeq _{N-2} \Bigl (\{X_{i}\}_{N}^{2}\Bigr ) \succeq _{N-1} \Bigl (\{X_{i}\}_{N}^{1} \Bigr ), \end{aligned}$$

(11)

where indexes $1,2,\ldots , N-1$ indicate the different steps. In step 1, we show the first part of the preference relation, $\Bigl (\{X_{i}\}_{N}^{N} \Bigr )\succeq _1 \Bigl (\{X_{i}\}_{N}^{N-1}\Bigr )$, in step 2 we show the second part of the preference relation, $\Bigl (\{X_{i}\}_{N}^{N-1}\Bigr ) \succeq _2 \Bigl (\{X_{i}\}_{N}^{N-2}\Bigr )$, in step 3 we show the third part of the preference relation, $\Bigl (\{X_{i}\}_{N}^{N-2}\Bigr ) \succeq _2 \Bigl (\{X_{i}\}_{N}^{N-3}\Bigr )$, etc.

Step 1$\Bigl (\{X_{i}\}_{N}^{N}\Bigr )\succeq _{1}\Bigl ( \{X_{i}\}_{N}^{N-1}\Bigr )$ (P1)

The left-hand side term of (P1) writes as $\sum _{i=1}^{N}E[u(x+X_{i})]$. In the set of N risks $X_{i}$, we choose two risks that we name $X_{t}$ and $X_{l}$, and that we regroup into a single risk $X_{t}+X_{l}$ in the right-hand side term of (P1). Using this, (P1) rewrites as

$$\begin{aligned}&\sum _{i=1}^{N}E[u(x+X_{i})]+E[u(x+X_{t})]+E[u(x+X_{l})]\\&\quad \ge \sum _{i=1}^{N}E[u(x+X_{i})]+E[u(x+X_{t}+X_{l})]+u(x) \end{aligned}$$

for all $t\le N$, $l\le N$, $t\ne l$, $i\ne t$ and $i\ne l$ that is equivalent to

$$\begin{aligned} E[u(x+X_{t})]+E[u(x+X_{l})]\ge +E[u(x+X_{t}+X_{l})]+u(x) \end{aligned}$$

which holds following EST (2009) theorem for all u such that $(-1)^{(s+1)}u^{(s)}>0$$\forall s=1,\ldots ,2k$ if $X_{1}\preceq _{{\text {SD}}_{k}}Y_{1}$ and $X_{2}\preceq _{{\text {SD}}_{k}}Y_{2}$, $k=1,2$. This inequality also holds following EST (2009) corollary for all u such that $u^{(4)}<0$ in the particular case where risks are zero-mean risks, i.e. $X_{1}\preceq _{{\text {Ekern}}_{2}}Y_{1}$ and $X_{2}\preceq _{{\text {Ekern}}_{2}}Y_{2}$.

In the expression $\sum _{i=1}^{N}E[u(x+X_{i})]+E[u(x+X_{t}+X_{l})]+u(x)$ (with $t\le N$, $l\le N$, $t\ne l$, $i\ne t$ and $i\ne l$), we can consider that we have $(N-1)$ risks. Indeed, let us write $X_{i}^{1}=X_{i}$$\forall i=1,\ldots ,N-2$, $i\ne t$ and $i\ne l$, and $X_{N-1}^{1}=X_{t}+X_{l}$, then the above expression can be equivalently rewritten as $\sum _{i=1}^{N-1}E[u(x+X_{i}^{1})]+u(x)$ with $X_{i}^{1}\preceq _{{\text {SD}}_{k}}0$$\forall i$, for all $k=1,2$ following the convolution stability property. Using the same technique as the one used at the beginning of step 1 to rewrite $\sum _{i=1}^{N}E[u(x+X_{i})]$, the expression $\sum _{i=1}^{N-1}E[u(x+X_{i}^{1})]+u(x)$ can thus be equivalently rewritten as $\sum _{i=1}^{N-1}E[u(x+X_{i}^{1})]+E[u(x+X_{t}^{1})]+E[u(x+X_{l}^{1})]+u(x)$ for all $t\le N-1$, $l\le N-1$, $i\ne t$ and $i\ne l$.

Step 2$\Bigl (\{X_{i}\}_{N}^{N-1}\Bigr )\succeq _{2}\Bigl ( \{X_{i}\}_{N}^{N-2}\Bigr )$ (P2)

Following the same methodology as the one used in step 1, (P2) rewrites equivalently as

$$\begin{aligned}&\sum _{i=1}^{N-1}E[u(x+X_{i}^{1})]+E[u(x+X_{t}^{1})]+E[u(x+X_{l}^{1})]+u(x) \\&\quad \ge \sum _{i=1}^{N-1}E[u(x+X_{i}^{1})]+E[u(x+X_{t}^{1}+X_{l}^{1})]+2u(x) \end{aligned}$$

for all $t\le N-1$, $l\le N-1$, $t\ne l$, $i\ne t$ and $i\ne l$ that is equivalent to

$$\begin{aligned} E[u(x+X_{t}^{1})+E[u(x+X_{l}^{1})\ge E[u(x+X_{t}^{1}+X_{l}^{1})+u(x) \end{aligned}$$

which holds following EST (2009) theorem for all u such that $(-1)^{(s+1)}u^{(s)}>0$$\forall s=1,\ldots ,2k$ if $X_{1}\preceq _{{\text {SD}}_{k}}Y_{1}$ and $X_{2}\preceq _{{\text {SD}}_{k}}Y_{2}$, $k=1,2$. This inequality also holds following EST (2009) corollary for all u such that $u^{(4)}<0$ in the particular case where risks are zero-mean risks, i.e. $X_{1}\preceq _{{\text {Ekern}}_{2}}Y_{1}$ and $X_{2}\preceq _{{\text {Ekern}}_{2}}Y_{2}$.

In the expression $\sum _{i=1}^{N-1}E[u(x+X_{i}^{1})]+E[u(x+X_{t}^{1}+X_{l}^{1})]+2u(x)$, we can consider that we have $(N-2)$ risks $X_{i}^{2}$ (i.e. $(N-3)$ risks $X_{i}^{1}$ and 1 risk $X_{t}^{1}+X_{l}^{1}$). Thus, it can be equivalently rewritten as $\sum _{i=1}^{N-2}E[u(x+X_{i}^{2})]+2u(x)$ with $X_{i}^{2}\preceq _{{\text {SD}}_{k}}0$ for all $k=1,2$ following the convolution stability property. It can also be rewritten as $\sum _{i=1}^{N-2}E[u(x+X_{i}^{2})]+E[u(x+X_{t}^{2})]+E[u(x+X_{l}^{2})]+2u(x)$ for all $t\le N-2$, $l\le N-2$, $t\ne l$, $i\ne t$ and $i\ne l$.

Step 3$\Bigl (\{X_{i}\}_{N}^{N-2}\Bigr )\succeq _{3}\Bigl ( \{X_{i}\}_{N}^{N-3}\Bigr )$ (P3)

Following the same methodology as the one used in step 2, (P3) rewrites equivalently as

$$\begin{aligned}&\sum _{i=1}^{N-2}E[u(x+X_{i}^{2})]+E[u(x+X_{t}^{2})]+E[u(x+X_{l}^{2})]+2u(x) \\&\quad \ge \sum _{i=1}^{N-2}E[u(x+X_{i}^{2})]+E[u(x+X_{t}^{2}+X_{l}^{2})]+3u(x) \end{aligned}$$

for all $t\le N-2$, $l\le N-2$, $t\ne l$, $i\ne t$ and $i\ne l$ that is equivalent to

$$\begin{aligned} E[u(x+X_{t}^{2})+E[u(x+X_{l}^{2})\ge +E[u(x+X_{t}^{2}+X_{l}^{2})]+u(x) \end{aligned}$$

which holds following EST (2009) theorem for all u such that $(-1)^{(s+1)}u^{(s)}>0$$\forall s=1,\ldots ,2k$ if $X_{1}\preceq _{{\text {SD}}_{k}}Y_{1}$ and $X_{2}\preceq _{{\text {SD}}_{k}}Y_{2}$, $k=1,2$. This inequality also holds following EST (2009) corollary for all u such that $u^{(4)}<0$ in the particular case where risks are zero-mean risks, i.e. $X_{1}\preceq _{{\text {Ekern}}_{2}}Y_{1}$ and $X_{2}\preceq _{{\text {Ekern}}_{2}}Y_{2}$.

In the expression $\sum _{i=1}^{N-2}E[u(x+X_{i}^{2})]+E[u(x+X_{t}^{2}+X_{l}^{2})]+3u(x)$, we can consider that we have $(N-4)$ risks $X_{i}^{3}$ (i.e. $(N-2)$ risks $X_{i}^{2}$ and 1 risk $X_{t}^{2}+X_{l}^{2}$). Thus, it can be equivalently rewritten as $\sum _{i=1}^{N-3}E[u(x+X_{i}^{3})]+3u(x)$ with $X_{i}^{3}\preceq _{{\text {SD}}_{k}}0$ for all $k=1,2$ following the convolution stability property. It can also be rewritten as $\sum _{i=1}^{N-3}E[u(x+X_{i}^{3})]+E[u(x+X_{t}^{3})]+E[u(x+X_{l}^{3})]+3u(x)$ for all $t\le N-3$, $l\le N-3$, $t\ne l$, $i\ne t$ and $i\ne l$.

Step m$\Bigl (\{X_{i}\}_{N}^{N-(m-1)}\Bigr )\succeq _{m}\Bigl ( \{X_{i}\}_{N}^{N-m}\Bigr )$ (Pm)

(Pm) rewrites equivalently as

$$\begin{aligned}&\sum _{i=1}^{N-(m-1)} E[u(x+X_i^{m-1})] + E[u(x+X_t^{m-1})] + E[u(x+X_l^{m-1})] + (m-1)u(x) \\&\quad \ge \sum _{i=1}^{N-(m-1)} E[u(x+X_i^{m-1})] + E[u(x+X_t^{m-1}+X_l^{m-1})] +mu(x) \end{aligned}$$

for all $t\le N-(m-1)$, $l\le N-(m-1)$, $t\ne l$, $i\ne t$ and $i\ne l$ that is equivalent to

$$\begin{aligned} E[u(x+X_{t}^{m-1})+E[u(x+X_{l}^{m-1})\ge +E[u(x+X_{t}^{m-1}+X_{l}^{m-1})]+u(x) \end{aligned}$$

which holds following EST (2009) theorem for all u such that $(-1)^{(s+1)}u^{(s)}>0$$\forall s=1,\ldots ,2k$ if $X_{1}\preceq _{{\text {SD}}_{k}}Y_{1}$ and $X_{2}\preceq _{{\text {SD}}_{k}}Y_{2}$, $k=1,2$. This inequality also holds following EST (2009) corollary for all u such that $u^{(4)}<0$ in the particular case where risks are zero-mean risks, i.e. $X_{1}\preceq _{{\text {Ekern}}_{2}}Y_{1}$ and $X_{2}\preceq _{{\text {Ekern}}_{2}}Y_{2}$.

In the expression $\sum _{i=1}^{N-(m-1)}E[u(x+X_{i}^{m-1})]+E[u(x+X_{t}^{m-1}+X_{l}^{m-1})]+mu(x)$, we can consider that we have $(N-m)$ risks $X_{i}^{m}$ (i.e. $(N-m-1))$ risks $X_{i}^{m-1}$ and 1 risk $X_{t}^{m-1}+X_{l}^{m-1}$). Thus, it can be equivalently rewritten as $\sum _{i=1}^{N-m}E[u(x+X_{i}^{m})]+E[u(x+X_{t}^{m})]+E[u(x+X_{l}^{m})]+mu(x)$ with $X_{i}^{m}\preceq _{{\text {SD}}_{k}}0$ for all $k=1,2$ following the convolution stability property.

$$\begin{aligned} \ldots \end{aligned}$$

Step N-2$\Bigl (\{X_{i}\}_{N}^{3}\Bigr )\succeq _{N-2} \Bigl ( \{X_{i}\}_{N}^{2}\Bigr )$$(P(N-2))$

$(P(N-2))$ rewrites equivalently as

$$\begin{aligned}&\sum _{i=1}^{N-(N-3)} E[u(x+X_i^{N-3})] + E[u(x+X_t^{N-3})] + E[u(x+X_l^{N-3})] + (N-3)u(x) \\&\quad \ge \sum _{i=1}^{N-(N-3)} E[u(x+X_i^{N-3})] + E[u(x+X_t^{N-3}+X_l^{N-3})] +(N-2)u(x) \end{aligned}$$

for all $t\le N-(N-3)$, $l\le N-(N-3)$, $t\ne l$, $i\ne t$ and $i\ne l$ that is equivalent to

$$\begin{aligned} E[u(x+X_{t}^{N-3})+E[u(x+X_{l}^{N-3})\ge +E[u(x+X_{t}^{N-3}+X_{l}^{N-3})]+u(x) \end{aligned}$$

which holds following EST (2009) theorem for all u such that $(-1)^{(s+1)}u^{(s)}>0$$\forall s=1,\ldots ,2k$ if $X_{1}\preceq _{{\text {SD}}_{k}}Y_{1}$ and $X_{2}\preceq _{{\text {SD}}_{k}}Y_{2}$, $k=1,2$. This inequality also holds following EST (2009) corollary for all u such that $u^{(4)}<0$ in the particular case where risks are zero-mean risks, i.e. $X_{1}\preceq _{{\text {Ekern}}_{2}}Y_{1}$ and $X_{2}\preceq _{{\text {Ekern}}_{2}}Y_{2}$.

In the expression $\sum _{i=1}^{N-(N-3)} E[u(x+X_i^{N-3})] + E[u(x+X_t^{N-3}+X_l^{N-3})] +(N-2)u(x)$, we have 2 risks $X_i^{N-2}$ (1 risk $X_i^{N-3}$ and 1 risk $X_t^{N-3}+X_l^{N-3}$). Thus, it can be equivalently rewritten as $\sum _{i=1}^{2} E[u(x+X_i^{N-2})] + (N-2) u(x)$ with $X_{i}^{N-2} \preceq _{{\text {SD}}_{k}} 0$ for all $k=1,2$ following the convolution stability property.

Step N-1 $\Bigl (\{X_{i}\}_{N}^{2}\Bigr )\succeq _{N-1} \Bigl ( \{X_{i}\}_{N}^{1}\Bigr )$ $(P(N-1))$

$(P(N-1))$ rewrites equivalently as

$\sum _{i=1}^{2}E[u(x+X_{i}^{N-2})]+(N-2)u(x)\ge E[u(x+X_{1}^{N-2}+X_{2}^{N-2})]+(N-1)u(x)$ which holds following EST (2009) theorem for all u such that $(-1)^{(s+1)}u^{(s)}>0$$\forall s=1,\ldots ,2k$ if $X_{1}\preceq _{{\text {SD}}_{k}}Y_{1}$ and $X_{2}\preceq _{{\text {SD}}_{k}}Y_{2}$, $k=1,2$. This inequality also holds following EST (2009) corollary for all u such that $u^{(4)}<0$ in the particular case where risks are zero-mean risks, i.e. $X_{1}\preceq _{{\text {Ekern}}_{2}}Y_{1}$ and $X_{2}\preceq _{{\text {Ekern}}_{2}}Y_{2}$.