Approximating VaR for a Sum of Dependent Lognormal Variables

# How to compute the Value-at-Risk of the sum of two dependent lognormal random variables?


# How to compute the Value-at-Risk of the sum of two dependent lognormal random variables?












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.

## Answer by TheBridge (score 6, accepted)

https://quant.stackexchange.com/a/750

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.

## Answer by Richard Herron (score 1)

https://quant.stackexchange.com/a/749

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?

## Answer by Good Guy Mike (score 1)

https://quant.stackexchange.com/a/9881

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.