In his book Risk and Asset Allocation, Meucci writes (last paragraph)

Indeed, since all the indices of satisfaction $\mathcal{S}$ discussed in Chapter 5 are consistent with weak stochastic dominance, for a given level of variance of the objective, higher expected values of the objective are always appreciated, no matter the functional expression of $\tilde{\mathcal{H}}$.

I'm trying to figure out the mathematical content/proof behind this statement. The context:

Let $\mathbf{M} : \Omega \to \mathbb{R}^n$ be a random vector, $\mathbf{\alpha} \in \mathbb{R}^n$.

Define a random variable $\mathbf{\Psi}_\mathbf{\alpha} = \mathbf{\alpha}^T \mathbf{M}$, the objective.

Let $\mathcal{S}(\mathbf{\alpha}) = \tilde{\mathcal{H}}(\mathrm{E}\{\mathbf{\Psi}_\mathbf{\alpha}\}, \mathrm{Var}\{\mathbf{\Psi}_\mathbf{\alpha}\})$, the index of satisfaction.

Moreover assume that $\mathcal{S}$ is consistent with weak stochastic dominance, ie.

$$\mathbf{\Psi}_\mathbf{\alpha} \leq_{\mathrm{weak}} \mathbf{\Psi}_\mathbf{\beta} \Rightarrow \mathcal{S}(\mathbf{a}) \leq \mathcal{S}(\mathbf{b})$$

I feel that Meucci is saying that if $\mathrm{Var}\{\mathbf{\Psi}_\mathbf{\alpha}\} = \mathrm{Var}\{\mathbf{\Psi}_\mathbf{\beta}\}$ and $\mathrm{E}\{\mathbf{\Psi}_\mathbf{\alpha}\} \leq \mathrm{E}\{\mathbf{\Psi}_\mathbf{\beta}\}$ then $\mathbf{\Psi}_\mathbf{\alpha} \leq_{\mathrm{weak}} \mathbf{\Psi}_\mathbf{\beta}$. But this isn't true.

  • $\begingroup$ Can you remind me of the definition of the weak stochastic dominance? Is that one CDF dominates another? $\endgroup$ – Ulysses Jun 8 '15 at 6:50
  • $\begingroup$ @Ulysses: yes, that is correct $\endgroup$ – user16469 Jun 8 '15 at 10:30

This is because the author has assumed the approximation (6.67) $$\mathcal{S}(\alpha)\approx \tilde{\mathcal{H}} (\mathrm{E}\{\Phi_\alpha\},\mathrm{Var}\{\Phi_\alpha\})$$ That is, the index of satisfaction $\mathcal{S}(\alpha)$ depends only on the first two moments of the marginals. As explained by the author in section 6.5.1, this is a good approximation because in a wide range of applications, the market $\mathbf M$ is elliptically distributed (see (6.126)).

One important example is $$ \mathbf{M}\sim \mathcal{N}(\boldsymbol\mu,\,\boldsymbol\Sigma) $$ where $\mathcal{N}$ is the multivariate Gaussian distribution with mean $\boldsymbol \mu$ and covariance matrix $\boldsymbol \Sigma$. In this case, if $\mathrm{Var}(\boldsymbol \Psi_\alpha)=\mathrm{Var}(\boldsymbol\Psi_\beta)$ and $\mathrm{E}[\boldsymbol \Psi_\alpha]\le\mathrm{E}[\boldsymbol\Psi_\beta]$, then indeed we have weak stochastic dominance, i.e. for all $t\in \mathbb{R}$, $$ P\{\boldsymbol\Psi_\alpha\le t\}=\int_{-\infty}^t \frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x - \mu_1)^2}{2 \sigma^2}}dx\ge\int_{-\infty}^t\frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x - \mu_2)^2}{2 \sigma^2}}dx= P\{\boldsymbol\Psi_\beta\le t\} $$ where $\sigma^2=\mathrm{Var}(\boldsymbol \Psi_\alpha)=\mathrm{Var}(\boldsymbol\Psi_\beta)=\boldsymbol\alpha^T\boldsymbol \Sigma\boldsymbol\alpha= \boldsymbol\beta^T\boldsymbol \Sigma \boldsymbol \beta$, $\mu_1=\mathrm{E}[\boldsymbol \Psi_\alpha]=\boldsymbol \alpha^T\mathrm{E}[\boldsymbol\Psi]$, and $\mu_2=\mathrm{E}[\boldsymbol \Psi_\beta]=\boldsymbol \beta^T\mathrm{E}[\boldsymbol\Psi]$. (Recall that we say $\boldsymbol \Psi_\alpha\le_{\mathrm{weak}} \boldsymbol \Psi_\beta$ if $$P\{\boldsymbol\Psi_\alpha\le t\}\ge P\{\boldsymbol\Psi_\beta\le t\}$$ — the inequality is reversed — see (5.35).)

The reasoning for a general elliptic distribution is similar. We will show that the distribution functions of $\boldsymbol \Psi_\alpha$ and $\boldsymbol \Psi_\beta$ are translates of each other. Since $\mathbf{M} $ is elliptically distributed, its characteristic function $\varphi_{\mathbf{M } }$ is of the form $$ \varphi_{\mathbf{M } }( \boldsymbol t) = e^{i \boldsymbol t^T\boldsymbol\mu} \chi(\boldsymbol t^T\boldsymbol\Sigma \boldsymbol t) $$for some $\boldsymbol\mu \in \mathbb{R}^N$, positive-definite matrix $\boldsymbol\Sigma$, and characteristic function $\chi$. Hence, for any $\boldsymbol \alpha \in \mathbb{R}^N$, the characteristic function $\varphi_{\boldsymbol \Psi_\alpha}$ of $\boldsymbol \Psi_\alpha=\boldsymbol\alpha^T \mathbf{M}$ is $$ \varphi_{\boldsymbol \Psi_\alpha}(t)=\varphi_{\mathbf{M } }(t\boldsymbol\alpha)= e^{it\boldsymbol \alpha^T\boldsymbol\mu} \chi(t^2 \boldsymbol\alpha^T\boldsymbol\Sigma \boldsymbol \alpha) $$ After calculating the derivatives of $\varphi_{\boldsymbol \Psi_\alpha}$, we find that $$ E[\boldsymbol\Psi_\alpha ] =\boldsymbol\alpha^T\boldsymbol\mu $$ and $$ \mathrm{Var}(\boldsymbol\Psi_\alpha)=2\boldsymbol\alpha^T \boldsymbol\Sigma \boldsymbol\alpha\chi '(0) $$ Thus, if $\mathrm{Var}(\boldsymbol\Psi_\alpha)=\mathrm{Var}(\boldsymbol\Psi_\beta)$, then $\boldsymbol\alpha^T \boldsymbol\Sigma \boldsymbol\alpha=\boldsymbol\beta^T \boldsymbol\Sigma\boldsymbol\beta$. Therefore, $$ \varphi_{\boldsymbol \Psi_\beta}(t) =e^{it(\boldsymbol \beta^T\boldsymbol\mu-\boldsymbol \alpha^T\boldsymbol\mu)}\varphi_{\boldsymbol \Psi_\alpha}(t) $$which implies that $$ f_{\boldsymbol \Psi_\beta}(x)=f_{\boldsymbol \Psi_\alpha}(x-(\boldsymbol \beta^T\boldsymbol\mu-\boldsymbol \alpha^T\boldsymbol\mu)) $$that is, the density functions $f_{\boldsymbol \Psi_\alpha}$ and $f_{\boldsymbol \Psi_\beta}$ are translates of each other. Since $\boldsymbol \beta^T\boldsymbol\mu-\boldsymbol \alpha^T\boldsymbol\mu=E[\boldsymbol\Psi_\beta ]-E[\boldsymbol\Psi_\alpha ]\ge0$ by assumption, we have shown that $\boldsymbol\Psi_\alpha\le_{\mathrm{weak}}\boldsymbol\Psi_\beta $.


I think I figured it out. In an earlier paragraph Meucci says

Suppose that we can focus on the two first moments only and neglect all the higher moments.

So two random variables are equivalent if their first two moments agree. Therefore any random variable can be replaced by $\mathcal{U}(\mu - \sqrt{3}\sigma, \mu + \sqrt{3}\sigma)$ as far as $\mathcal{S}$ is concerned.

By looking at the cdf below, it is clear that $$\mathcal{U}(\mu_1 - \sqrt{3}\sigma, \mu_1 + \sqrt{3}\sigma) \leq_\mathrm{weak} \mathcal{U}(\mu_2 - \sqrt{3}\sigma, \mu_2 + \sqrt{3}\sigma)$$ when $\mu_1 \leq \mu_2$.

enter image description here


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