3
$\begingroup$

In this paper The Interplay between Stochastic Volatility and Correlations in Equity Autocallables by Alvise De Col, Patrick Kuppinger (2017) https://papers.ssrn.com/sol3/papers.cfm?abstract_id=3228065, it mentioned enter image description here

enter image description here

enter image description here

I am confused by the last equation, how did the paper get $dW_{s1} dW_{v2} = \rho_s \rho_{sv} dt$, given that

$dW_{s1} dW_{v1}=dW_{s2} dW_{v2}=\rho_{sv}dt$

and $dW_{s1} dW_{s2} = \rho_s dt$,

the $dW_{s1} dW_{v2}$ should be either

$[\rho_s \rho_{sv} - \sqrt{1-\rho_s^2}\sqrt{1-\rho_{sv}^2} ] dt$ or

$[\rho_s \rho_{sv} + \sqrt{1-\rho_s^2}\sqrt{1-\rho_{sv}^2} ] dt$

Any help are appreciated.

Btw, thanks @noob2 for editing, it's much easier to read now.

$\endgroup$
1
$\begingroup$

The usual ansatz for these kind of setups is to find those components of a Cholesky decomposition of the correlation matrix of your stochastic drivers $dW_{S_1}, dW_{S_2}, dW_{V_1}, dW_{V_2}$ such that all conditions are fulfilled.

Let us assume a 4x4 correlation matrix $R$ that we decompose using Cholesky to

$$ L(R) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ a & d & 0 & 0 \\ b & e & g & 0 \\ c & f & h & i \end{pmatrix} $$
such that $LL^T=R$, i.e.

$$ LL^T = \begin{pmatrix} 1 & a & b & c \\ . & a^2+b^2 & ab+de & ac+df \\ . &. & b^2+e^2+g^2 & bc+ef+gh \\ .& . & . & c^2+f^2+h^2+i^2 \end{pmatrix} $$

Let us now identify the rows / columns with $S_1$, $S_2$, $V_1$, $V_2$, and put in all assumptions from your text plus the usual assumptions regarding the diagonals of the correlation matrix:

  • $\mathrm{E}(dW_{S_1}dW{S_2})=a=\rho_Sdt$
  • $\mathrm{E}(dW_{S_1}dW{V_1})=b=\rho_Vdt$
  • $\mathrm{E}(dW_{S_1}dW{V_2})=c=\rho_S\rho_Vdt$
  • $\mathrm{E}(dW_{S_2}dW_{V_1})=ab+de=\rho_S\rho_Vdt$
  • $\mathrm{E}(dW_{S_1}dW_{S_1})=1dt$
  • $\mathrm{E}(dW_{S_2}dW_{S_2})=a^2+b^2=1dt$
  • $\mathrm{E}(dW_{V_1}dW_{V_1})=b^2+e^2+g^2=1dt$
  • $\mathrm{E}(dW_{V_2}dW_{V_2})=c^2+f^2+h^2+i^2=1dt$

You may then proceed to solve for all variables. Close inspection shows that there's one additional degree of freedom:

  • $\mathrm{E}(dW_{V_1}dW_{V_2})=bc+ef+gh=Adt$

With these ingredients, you can quite simply and iteratively solve for $a,b,c,d,e,f,g,h,i$ and obtain a correlation matrix fulfilling all conditions, i.e.

$$ R=\mathrm{E} \begin{pmatrix} dW_{S_1}dW_{S_1} & dW_{S_1}dW_{S_2} & dW_{S_1}dW_{V_1} & dW_{S_1}dW_{V_2}\\ dW_{S_1}dW_{S_2} & dW_{S_2}dW_{S_2} & dW_{S_2}dW_{V_1} & dW_{S_2}dW_{V_2}\\ dW_{S_1}dW_{V_1} & dW_{S_2}dW_{V_1} & dW_{V_1}dW_{V_1} & dW_{V_1}dW_{V_2}\\ dW_{S_1}dW_{V_2} & dW_{S_2}dW_{V_2} & dW_{V_1}dW_{V_2} & dW_{V_2}dW_{V_2} \end{pmatrix}= \begin{pmatrix} 1 & \rho_S & \rho_V & \rho_S\rho_V \\ \rho_S & 1 & \rho_S\rho_V & \rho_V \\ \rho_V & \rho_S\rho_V & 1 & A \\ \rho_S\rho_V & \rho_V & A & 1 \end{pmatrix}dt $$

You may think of (the vector of) your correlated stochastic drivers as a linear transformation of uncorrelated stochastic drivers $d\tilde{W}_i$, transformed by the lower Cholesky:

$$ dW=Ld\tilde{W} $$ and thus

\begin{align} \mathrm{E}\left(dW\left(dW\right)^T\right)&=L\mathrm{E}\left(d\tilde{W}\left(d\tilde{W}\right)^T\right)L^T\\ &=L\mathrm{I}L^T\\ &=LL^T\\ &=Rdt \end{align}

where $\mathrm{I}$ is the identity matrix.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.