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Let the portfolio be given by: $$X=X_1+X_2$$ $(X_1,X_2)$ are dependent through a Copula function $C(u_1,u_2)$, such that the joint distribution is given by: $$F(x_1,x_2)=C(F(x_1),F(x_2))$$

What is the VaR of this portfolio?

Usually VaR is the inverse quantile: $VaR_\alpha=F^{-1}_X(\alpha)$.

I am not sure how to determine it in this multivariate case?

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You don't really have a multivariate case: we can only define VaR (in its usual sense) for a one-dimensional output. Recall that $$ \operatorname{VaR}_\alpha(X) = \inf\{v:F_X(v)\geq \alpha\} $$ and since in your case $X = X_1+X_2$ you just need to compute $F_X$ in terms of $X_1$ and $X_2$. For the notation of partial derivatives, I denote the generic variables of the copula function by $u_1$ and $u_2$. $$ F_X(v) = \mathbb P(X_1+X_2\leq v) = \int\limits_{-\infty}^\infty \frac{\partial C}{\partial u_1}\left(F_{X_1}(x_1),F_{X_2}(v-x_1)\right)\mathrm dF_{X_1}(x_1). \tag{1} $$ As long as you can compute/estimate this function, you can get a value/estimate for VaR. The formula $(1)$ can be obtained as follows: $$ \begin{align} F_X(v) &= \mathbb P(X_1+X_2\leq v) = \int\limits_{-\infty}^\infty \mathrm dx_1 \int\limits_{-\infty}^{v-x_1}\frac{\partial^2 C}{\partial u_1\partial u_2}\left(F_{X_1}(x_1),F_{X_2}(x_2)\right)F'_{X_1}(x_1)F'_{X_2}(x_2)\mathrm dx_2 \\ &= \int\limits_{-\infty}^\infty \mathrm dx_1 \int\limits_{-\infty}^{v-x_1}\frac{\partial}{\partial x_2}\left(\frac{\partial C}{\partial u_1}\left(F_{X_1}(x_1),F_{X_2}(x_2)\right)F'_{X_1}(x_1)\right)\mathrm dx_2 \\ &= \int\limits_{-\infty}^\infty \left.\frac{\partial C}{\partial u_1}\left(F_{X_1}(x_1),F_{X_2}(x_2)\right)F'_{X_1}(x_1)\right|_{x_2=-\infty}^{x_2=v-x_1}\mathrm dx_1 \\ &= \int\limits_{-\infty}^\infty \frac{\partial C}{\partial u_1}\left(F_{X_1}(x_1),F_{X_2}(v-x_1)\right)F'_{X_1}(x_1)\mathrm dx_1 \\ &= \int\limits_{-\infty}^\infty \frac{\partial C}{\partial u_1}\left(F_{X_1}(x_1),F_{X_2}(v-x_1)\right)\mathrm dF_{X_1}(x_1). \end{align} $$ For the partial derivatives notation, consider the following example. If $g(u_1,u_2) = u_1 + u_2$ then $$ \frac{\partial }{\partial u_1}g(x_1^2,x_2^2) = 1,$$ $$ \frac{\partial}{\partial x_1}g(x_1^2,x_2^2) = 2x_1. $$

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  • $\begingroup$ Can you explain how you get to this formula for $F_X(v)$? E.g. why is it $dF(x_1)$? I found this other link for the convolution pdf: stats.stackexchange.com/questions/21549/… $\endgroup$ – emcor Nov 24 '14 at 9:03
  • $\begingroup$ @emcor: added the derivation of the formula. You can of course obtain a similar one in terms of densities, however I was not sure whether your CDFs do have densities, so decided to go for a bit more general case. $\endgroup$ – Ulysses Nov 24 '14 at 9:17
  • $\begingroup$ Thank you, what is the upper limit for the first and second integrals? Can you please express $f(x)$ in pdf's, I cannot follow what is the difference between $F'(x_1)$ and $dF(x_1)$, and why in the second line $F'(x_2)$ suddenly vanishes? $\endgroup$ – emcor Nov 24 '14 at 9:22
  • $\begingroup$ @emcor: I've put the limits there, for some reason they have not appeared in the first version. Now, $\mathrm dg(x) = g'(x)\mathrm dx$ which is kinda fundamental theorem of calculus. Finally, $F'_{X_2}(x_2)$ has not vanished: it's just a chain rule for partial derivative. If $g(u,v)$ is some function, then $$ \frac{\partial}{\partial y}g(a(x),b(y)) =\frac{\partial g}{\partial v}(a(x),b(y))\cdot b'(y). $$ As often with partial derivatives, notation may be confusing, so let me make it more formal in my answer if that is confusing you. $\endgroup$ – Ulysses Nov 24 '14 at 9:38
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    $\begingroup$ @emcor: indeed, since $F_{X_2}(-\infty) = 0$ and $C(u_1,0) =0$ regardless of $u_1$ $\endgroup$ – Ulysses Nov 24 '14 at 13:16

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