I am reading Sebastien Bossu's "A new Approach For Modelling and Pricing Correlation Swaps" (link). I am recalling some of the definitions from the paper and would like to understand how to prove one of the very first claims made. Here we go.

The universe $S = (S_i) \quad i = 1..N$. The weight vector $w = (w_i) \quad i = 1..N$.

  1. Denote $S_i(t)$ the price of stock $S_i$ at the time $t$, with convention $S(0) = 1$, and we define their geometric as: $I(t) = \prod_{i=1}^N S_i(t)^{w_i}$.

Under a probability space $(\Omega, E, P)$ with $P$-filtration $F$, and assuming that the vector $S$ of stock prices is an $F$-adapted, positive Ito process.

Given a time period $T$ and a positive Ito process $X$, we define: $|\tau| = \int_\tau ds$

$\sigma^X(\tau) = \sqrt{ |\tau|^{-1} \int_\tau (d \ln X_s)^2}$

$\overline{\sigma}^S(\tau) = \sqrt{\sum_1^n w_i (\sigma^{S_i} (\tau))^2}$

$\epsilon(\tau) = \sqrt{\sum_1^N w^2_i (\sigma^{S_i}(\tau))^2}$

Now that we have defined the terms, the claim is that, $\overline{\sigma}^S(\tau) >= \sigma^I(\tau)$. I tried to prove this identity in vain. I am not sure what I am missing. Here are my attempts.

$\ln (I) = \ln (\prod_{i=1}^N S_i(t)^{w_i}) = \sum w_i \ln S_i(t)$

$d \ln (I) = \sum w_i d \ln S_i(t)$

$\overline{\sigma}^S(\tau) = \sqrt{\sum_1^N w_i \frac{1}{\tau} \int_\tau (d \ln S_i)^2} = \sqrt{\sum_1^N w_i \sigma_i^2}$

$\sigma^I(\tau) = \sqrt{|\tau|^{-1} \int_\tau (d \ln I)^2}$

$=\sqrt{|\tau|^{-1} \int_\tau (\sum w_i d \ln S_i(t))^2 } = \sum_{i,j =1}^N w_i w_j \rho_{ij} \sigma_i \sigma_j$

$\sigma^I(\tau)^2 = \sum_{i,j} \rho_{ij} w_i w_j \sigma_i \sigma_j \le \sum_{i,j} w_i w_j \sigma_i \sigma_j$

I am thinking that inductive proof should work well, but it goes nowhere; Here it is, for $N = 1$ it is true, now if I assume it is true for $N$ we have to show that,

$w_{N+1} \sigma_{N+1}^2 \ge \sum_{i=1}^{N+1} w_i w_{N+1} \rho_{i{N+1}} \sigma_i \sigma_{N+1}$

Clearly, $\sum_{i=1}^{N+1} w_i w_{N+1} \rho_{i{N+1}} \sigma_i \sigma_{N+1} \le \sum_{i=1}^{N+1} w_i w_{N+1} \sigma_i \sigma_{N+1}$ $\le \sum_{i=1}^{N+1} w_{N+1} \sigma_i \sigma_{N+1}$

Now I am stuck because the Identity I chose clearly $\ge$ the left hand side. I have tried several other ways to go about it but with similar results. I am not sure what I am missing.


1 Answer 1


What you seem to be missing is $$\sum_{i,j} w_i w_j \sigma_i \sigma_j = \left(\sum_i w_i\sigma_i\right)^2$$

Now apply Jensen's inequality to get $$\left(\sum_i w_i\sigma_i\right)^2 \leq \sum_i w_i\sigma_i^2 $$

QED (note that nonnegative weights is a crucial assumption here)

  • 1
    $\begingroup$ Thank you, I figured that out myself as well. But good to get a confirmation. $\endgroup$ Commented Oct 14, 2019 at 21:45

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