Does value-at-risk encourage diversification when losses follow tempered stable or more general Lévy processes?

Abstract

In this paper, we address the question of when portfolio selection based on Value-at-risk encourages diversification [in the sense of Ibragimov (Quant Financ 9(5):565–580, 2009)]. Specifically, we give sufficient conditions for the case when losses follow a Lévy process. When the process has finite variation, these conditions are also necessary. We then specialize our results to the case when losses have tempered stable distributions.

Appendix

The proof of Theorem 1 is based on the following result, which is a specialization of Theorems 2.1 and 3.2 in Samorodnitsky and Taqqu (1994) [see also Theorem 2.2 in Samorodnitsky and Taqqu (1993)].

Lemma 1

Let $\{X_t:t\ge 0\}$ and $\{Y_t:t\ge 0\}$ be Lévy processes such that $X_1\sim ID_0(0,M_X,0)$ and $Y_1\sim ID_0(0,M_Y,0)$ with $M_X((-\infty ,0])=M_Y((-\infty ,0])=0$. We have

$$\begin{aligned} M_X((r,\infty )) \le M_Y((r,\infty )) \quad \text{ for } \text{ every } r>0 \end{aligned}$$

if and only if

$$\begin{aligned} P(X_t>r) \le P(Y_t>r) \quad \text{ for } \text{ very } r>0, t>0. \end{aligned}$$

Proof (Proof of Theorem 1)

[Proof of Theorem 1] We only prove results related to Schur-convexity of $g_r(\cdot )$ as proofs of the other results are similar.

We begin with the case when $\mu \in \fancyscript{P}$. By positivity, we have $\mathrm{VaR}_\gamma (X_{\varvec{u},t})\le \mathrm{VaR}_\gamma (X_{\varvec{v},t})$ for all $t>0$ and all $\gamma \in (0,1)$ if and only if $P(X_{\varvec{u},t}>r)\le P(X_{\varvec{v},t}>r)$ for all $t>0$ and all $r>0$. By Lemma 1, this holds if and only if $g_r(\varvec{u})=M_{\varvec{u}}((r,\infty )) \le M_{\varvec{v}}((r,\infty ))=g_r(\varvec{v})$ for all $r>0$. This holds for all $\varvec{u},\varvec{v}\in \mathbb R^n_+$ with $\varvec{u}\prec \varvec{v}$ if and only if $g_r(\cdot )$ is Schur-convex for all $r>0$.

Now assume that $\mu \in \fancyscript{S}$. By symmetry, $\mathrm{VaR}_\gamma (X_{\varvec{u},t})\le \mathrm{VaR}_\gamma (X_{\varvec{v},t})$ for all $t>0$ and all $\gamma \in (.5,1)$ if and only if $P(X_{\varvec{u},t}>r)\le P(X_{\varvec{v},t}>r)$ for all $t>0$ and all $r>0$.

First, consider the case when $\int _{\mathbb R}\left( |x|\wedge 1\right) M(\mathrm{d}x)<\infty$. For any $\varvec{w}\in \mathbb R^n_+$ and any $t>0$, the distribution of $X_{\varvec{w},t}$ is symmetric around zero and thus $X_{\varvec{w},t} \mathop {=}\limits ^{d}X_{\varvec{w},t}^{(1)}-X^{(2)}_{\varvec{w}, t}$ where $\{X_{\varvec{w},t}^{(1)}:t\ge 0\}$ and $\{X_{\varvec{w},t}^{(2)}:t\ge 0\}$ are independent Lévy processes with $X^{(1)}_{\varvec{w},1},X^{(2)}_{\varvec{w},1}\mathop {\sim }\limits ^{\mathrm{iid}}ID_0(0,M'_{\varvec{w}},0)$ and

$$\begin{aligned} M'_{\varvec{w}}(A) = \int \limits _{[x>0]} 1_A(x) M_{\varvec{w}}(\mathrm{d}x), \quad A\in \mathfrak B(\mathbb R), \end{aligned}$$

where $M_{\varvec{w}}$ is as in (7). Note that $|X_{\varvec{w},t}|\mathop {=}\limits ^{d}X^{(1)}_{\varvec{w},t}+X^{(2)}_{\varvec{w},t}\sim ID_0(0,2M'_{\varvec{w}},0)$. If $r>0$ then, by symmetry,

$$\begin{aligned} P(X^{(1)}_{\varvec{w},t}+X^{(2)}_{\varvec{w},t}>r) = P(|X_{\varvec{w},t}|>r)= 2P(X_{\varvec{w},t} >r). \end{aligned}$$

Thus,

$$\begin{aligned} P(X_{\varvec{u},t} >r)\le P(X_{\varvec{v},t} >r), \quad \text{ for } \text{ all } t>0, r>0, \varvec{u}\prec \varvec{v} \end{aligned}$$

is equivalent to

$$\begin{aligned} P(X^{(1)}_{\varvec{u},t}+X^{(2)}_{\varvec{u},t}>r)\le P(X^{(1)}_{\varvec{v},t}+X^{(2)}_{\varvec{v},t}>r), \quad \text{ for } \text{ all } t>0, r>0, \varvec{u}\prec \varvec{v}. \end{aligned}$$

By Lemma 1 this is equivalent to

$$\begin{aligned} 2M'_{\varvec{u}}((r,\infty )) \le 2M'_{\varvec{v}}((r,\infty )), \quad \text{ for } \text{ all } r>0, \varvec{u}\prec \varvec{v}. \end{aligned}$$

Since $M'_{\varvec{u}}((r,\infty )) = g_{r}(\varvec{u})$ and $M'_{\varvec{v}}((r,\infty )) =g_{r}(\varvec{v})$, this is equivalent to

$$\begin{aligned} g_{r}(\varvec{u}) \le g_{r}(\varvec{v}), \quad \text{ for } \text{ all } r>0,\varvec{u}\prec \varvec{v}, \end{aligned}$$

which is equivalent to $g_r(\cdot )$ being Schur-convex for each $r>0$.

Now, consider the case $\int _{\mathbb R}\left( |x|\wedge 1\right) M(\mathrm{d}x)=\infty$. Assume that $g_r(\cdot )$ is Schur-convex for each $r>0$. Here the Lévy measure is not well behaved near zero, so we begin by truncating it. For a vector $\varvec{w}\in \mathbb R_+^n$ define

$$\begin{aligned} M_{\varvec{w}}^{(m)}(A) = \int \limits _{[|x|>1/m]}1_A(x) M_{\varvec{w}}(\mathrm{d}x), \quad A\in \mathfrak B(\mathbb R). \end{aligned}$$

Let $\{X^{(m)}_{\varvec{w},t}:t\ge 0\}$ be a Lévy process where $X^{(m)}_{\varvec{w},1}\sim ID(0,M^{(m)}_{\varvec{w}},0)$. Since $M^{(m)}_{\varvec{w}}$ is symmetric, $\int _{\mathbb R} \left( |x|\wedge 1\right) M^{(m)}_{\varvec{w}}(\mathrm{d}x)<\infty$, and $M_{\varvec{w}}^{(m)}((r,\infty ))=g_{r\vee (1/m)}(\varvec{w})$ is Schur-convex for all $r>0$, we can use arguments as in the previous case to show that

$$\begin{aligned} P(X^{(m)}_{\varvec{u},t} >r)\le P(X^{(m)}_{\varvec{v},t} >r), \quad \text{ for } \text{ all } t>0,r>0,\varvec{u}\prec \varvec{v}. \end{aligned}$$

We now remove the truncation by taking limits. Using standard results for convergence of infinitely divisible distributions (see e.g. Theorem 15.14 in Kallenberg (2002)) it is straightforward to show that $X^{(m)}_{\varvec{u},t}\mathop {\rightarrow }\limits ^{d}X_{\varvec{u},t}$ and $X^{(m)}_{\varvec{v},t}\mathop {\rightarrow }\limits ^{d}X_{\varvec{v},t}$ as $m\rightarrow \infty$. Further, so long as $\varvec{u},\varvec{v}\ne \varvec{0}$, we have $M_{\varvec{u}}(\mathbb R)=M_{\varvec{v}}(\mathbb R)=\infty$ and thus, by Theorem 27.4 in Sato (1999), both $X_{\varvec{u},t}$ and $X_{\varvec{v},t}$ have continuous distributions. Hence, for every $t>0$ and $r>0$,

$$\begin{aligned} P(X_{\varvec{u},t} >r)=\lim _{m\rightarrow \infty }P(X^{(m)}_{\varvec{u},t} >r)\le \lim _{m\rightarrow \infty } P(X^{(m)}_{\varvec{v},t} >r)=P(X_{\varvec{v},t} >r). \end{aligned}$$

If $\varvec{u}=\varvec{0}$ or $\varvec{v}=\varvec{0}$ then $\varvec{u}=\varvec{v}=\varvec{0}$ and the result is immediate.

Before proving Corollary 1, we give the following fact, which follows from the proof of Proposition A.6 in Ibragimov (2009).

Lemma 2

Let $X_1,X_2,\ldots ,X_n$ and $Y_1,Y_2,\ldots ,Y_n$ be two sets of independent random variables all having symmetric unimodal densities. If $\mathrm{VaR}_\gamma (X_i)\le \mathrm{VaR}_\gamma (Y_i)$ for all $=1,2,\ldots ,n$ and all $\gamma \in (.5,1)$ then $\mathrm{VaR}_\gamma (\sum _{i=1}^nX_i)\le \mathrm{VaR}_\gamma (\sum _{i=1}^nY_i)$ for all $\gamma \in (.5,1)$.

Proof (Proof of Corollary 1)

We focus on the case $t=1$ as the other cases are similar. If $a=0$ the result follows by Theorem 1. Now assume that $a\ne 0$. Let $Z_{1},\ldots , Z_{n}\mathop {\sim }\limits ^{\mathrm{iid}}N(0,1)$ and let $Y_{1},\ldots ,Y_{n}\mathop {\sim }\limits ^{\mathrm{iid}}ID(0,M,0)$ be independent sequences of random variables. If $X_{1}, \ldots ,X_{n}\mathop {\sim }\limits ^{\mathrm{iid}}\mu$ then $(X_1,\ldots ,X_n)\mathop {=}\limits ^{d}(\sqrt{a}Z_1+Y_1, \ldots ,\sqrt{a}Z_n+Y_n)$. For any $\varvec{w}\in \mathbb R^n_+$ we have $X_{\varvec{w}} \mathop {=}\limits ^{d}\sqrt{a} Z_{\varvec{w}} + Y_{\varvec{w}}$. If $\varvec{u}\prec \varvec{v}$ then Theorem 1 implies that $\mathrm{VaR}_\gamma (Y_{\varvec{u}})\le \mathrm{VaR}_\gamma (Y_{\varvec{v}})$ and Proposition A.7 in Ibragimov (2009) implies that $\mathrm{VaR}_\gamma (\sqrt{a}Z_{\varvec{u}})\le \mathrm{VaR}_\gamma (\sqrt{a}Z_{\varvec{v}})$. By Proposition A.5 in Ibragimov (2009) both $Z_{\varvec{u}}$ and $Y_{\varvec{u}}$ have symmetric and unimodal densities. Thus the result follows by Lemma 2.

Proof (Proof of Proposition 1.)

[Proof of Proposition 1.] If $f_r(\cdot )$ is convex then $g_r(\cdot )$ is Schur-convex by Proposition C.1 on page 92 in Marshall et al. (2011). Now assume that $M$ has a continuous Lebesgue density $m$. This implies that $f_r(s)$ is a continuous function of $s$ and thus it is convex if and only if $g_r(\cdot )$ is Schur-convex by Proposition C.1 on page 92 and Proposition C.1.c on page 95 of Marshall et al. (2011). By Leibniz’s rule $f_r(s)$ is differentiable in $s$ and

$$\begin{aligned} \frac{\partial }{\partial s}f_r(s) = \frac{\partial }{\partial s}\int \limits _{r/s}^\infty m(x) \mathrm{d}x = rs^{-2}m(r/s) = r^{-1}(r/s)^{2}m(r/s). \end{aligned}$$

From here, the fact that a differentiable function is convex if and only if its derivative is non-decreasing gives the equivalence between the convexity of $f_r(\cdot )$ and the non-increasing of $x^2m(x)$. Conditions for $g_r(\cdot )$ to be Schur-concave can be shown in a similar way using the corresponding properties of concave and Schur-concave functions.

Proof (Proof of Corollary 3)

[Proof of Corollary 3] By Theorem 30.1 in Sato (1999), $W_Y\sim ID(0,M^*,0)$ where $M^*$ has a continuous density $m^*(x)$ given by

$$\begin{aligned} m^*(x)= \int \limits _{0}^\infty \frac{1}{\sqrt{2\pi s}}e^{-x^2/(2s)}m(s)\mathrm{d}s =\frac{|x|}{\sqrt{2\pi }} \int \limits _{0}^\infty e^{-u/2}m(x^2/u)u^{-3/2}\mathrm{d}u. \end{aligned}$$

Thus

$$\begin{aligned} x^2m^*(x)= \frac{1}{\sqrt{2\pi }} \int \limits _{0}^\infty e^{-u/2} (|x|/\sqrt{u})^3 m((|x|/\sqrt{u})^2)\mathrm{d}u. \end{aligned}$$

From here the result follows by Corollary 2.