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Hy I posted this question first at mathflow.net they suggested me this page, which I was not aware of.

Question

Let $(X_1,X_2)$ be a multivariate normal random vector ($X_1$ and $X_2$ need not be independent). Is it possible to calculate $$VaR_{\alpha}(e^{X_1}+e^{X_2})$$

analyticaly?

Or is it even possible to calulate it in terms of $VaR_{\alpha}(e^{X_1})$ and $VaR_{\alpha}(e^{X_2})$ i.e. is there a representation (a function $g(\cdot,\cdot)$ ) of the form $$VaR_{\alpha}(e^{X_1}+e^{X_2})=g(VaR_{\alpha}(e^{X_1}),VaR_{\alpha}(e^{X_2})).$$

The case where $X_1$ and $X_2$ are independent could be aproached in terms of convolution which dont give in my eyes any impression if it is analyticaly tractable.

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The answer is no, to my knowledge.

I suggest you take look at this thread on MO, about the sum of log normal random variables. A few of the articles mentioned there might help you.

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For independent random variables the variance of a sum is the sum of the variances. If the random variables are not independent, then there's a covariance term $$Var(X_1 + X_2) = VarX_1 + VarX_2 + 2*Cov(X_1, X_2)$$

Exponentiating doesn't change this relation; it just makes your random variable log-normal.

Maybe you're looking for Steins's lemma? Is there a further downstream question?

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    $\begingroup$ Hi i don t think that it is a question about Variance but about Value at Risk $\endgroup$ – TheBridge Mar 17 '11 at 20:25
  • $\begingroup$ hi richardh, what you're writing is true, but my operator $VaR_{\alpha}$ stands for the Value-at-Risk (or simply the $\alpha$-Quantile) and not for the Variance (Var). $\endgroup$ – bobbey Mar 17 '11 at 20:28
  • $\begingroup$ Ah! I knew it seemed to easy! :) But isn't VaR just a distribution of returns? You should still be able to add the variances of the returns, find the new variance and then use whatever portion of the tail, right? If these VaR are normal (or some other two parameter distribution), then it should work. $\endgroup$ – Richard Herron Mar 17 '11 at 20:36
  • $\begingroup$ Its ok ;), $VaR_{\alpha}$ is just the $\alpha$-quantile of a random variable. is the problem here seems to be that we dont know very much about the distribution of $\exp(X_1)+\exp(X_2)$ and especially about its quantiles (or for example of how the quaniles depend on the variance). Of course in Gaussian framework the problem reduces in calculating the mean and variance (and scaleing it) of the sum of random variables. $\endgroup$ – bobbey Mar 17 '11 at 21:37
  • $\begingroup$ The way how this question arised was indeed in connection with returns. If the logreturn is normally distributed then the simple-return is lognormal with shift -1 (i.e. simple return=lognormal random variable-1). Now if you have a portfolio of two risks then what is the VaR of the portfolios simple return if we model the logreturns to be normal randomvariables (which with fin market data is ok for longer time spans) $\endgroup$ – bobbey Mar 17 '11 at 22:07
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I think the closest to an "analytical" answer (and I do not mean in terms of the accuracy, but in the sense that you can solve it by using pen and paper instead of a computer) would be to use linearization. Consider $g(X_1,X_2)=e^{X_1}+e^{X_2}$, we'd now like to compute $VaR_\alpha(g(X_1,X_2))$.

Linearization of $g$ gives:$$g(X_1,X_2)\approx g(\mathbf{\mu}) + \nabla g^T(\mu)(\mathbf{X}-\mathbf{\mu})$$

Thus we have that $$VaR_\alpha(g(X_1,X_2)) \approx VaR_\alpha(g(\mathbf{\mu}) + \nabla g^T(\mu)(\mathbf{X}-\mathbf{\mu})) = VaR_\alpha(\nabla g^T(\mu)(\mathbf{X}-\mathbf{\mu})) - g(\mathbf{\mu})$$

Where the last equality holds due to the translation invariance of value-at-risk. Now we have that $$\nabla g^T(\mu)(\mathbf{X}-\mathbf{\mu})= e^{\mu_1}(X_1-\mu_1) + e^{\mu_2}(X_2-\mu_2) = e^{\mu_1}X_1+e^{\mu_2}X_2 -\mu_1e^{\mu_1} - \mu_2e^{\mu_2}$$

Once again we can move out the constants due to the translation invariance of value-at-risk. (Note that I skip the risk-free rate).

Now we have $$VaR_\alpha(g(X_1,X_2)) \approx VaR_\alpha(e^{\mu_1}X_1+e^{\mu_2}X_2)+e^{\mu_1} + e^{\mu_2} -\mu_1e^{\mu_1} - \mu_2e^{\mu_2}$$

Since $X_1,X_2$ are just normally distributed, so will their sum be. Thereafter its just straightforward computations.

Worth noting is that linear approximation only works well when the probability mass in centered closely around the mean, i.e. if tails are light and variance is small. It becomes particularly worrisome since in the value at risk we're measuring probabilities far out in the tail.

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