How to minimize the difference between a parametric VaR and a MC-VaR with lognormal assumption?

Question

Given that we want to find the Value at Risk for a portfolio of stocks only, there are two main methods to proceed. In the problem, we also assume that stocks follow a geometric Brownian motion.

A full scale simulation:

1. simulate normal deviates B;
2. plug into $S_t = S_0 e^{(\mu-0.5\sigma^2)t + \sigma t B_t}$;
3. pick the 1% highest portfolio value. Call this $V_p$ ;
4. VaR is then $V_t - V_p$.

where $V_t$ is the current value of portfolio.

This methodology is of course good, but can fail to give an accurate solution quickly. Therefore, one can use the quick fix of risk metrics and assume that aritmethic returns equals logreturns ( $\log(1+x) =x$ ) and one get their method:

Parametric version:

1. Return of portfolio is a linear combination of gaussians, and therefore univariat gaussian itself (Y).
2. VaR is then $\Phi(0.01)\sigma_Y V_t$

Both methods can be justified seperately, but my problem is that I want to alternate to use the two methods (due to time constraints in some cases). A bright observer of this VaR-measure will then see that the amount jumps up and down depending on the method used (especially for high-volatile stocks). Is it therefore possible in some way to modify the parametric VaR-measure to give better results?

Note:

• Assumed Gaussian for simplicity.
• The multivariate case is of course the important question, but you can do the generalization yourself).
• Of course, better simulation techniques can be used (this is not the question adressed here).
• Also, simulating aritmetic returns will neither address the problem.

Answer 1

A probabilistic view on your full scale simulation. In the steps 1-3 you calculate the 0.99 quantile of the lognormal distribution with parameters $\ln N(\ln S_0 +(\mu - \frac{\sigma^2}{2})t,\sigma^2 t^2)$.

The cdf of lognormal distribution is $\Phi(\frac{\ln x-\mu}{\sigma})$ Thus, you can calculate $V_p$ through $V_p=e^{\ln S_0 +(\mu - \frac{\sigma^2}{2})t + q_{0.99} \sigma t}$

where $q_{0.99}$ defined through $\Phi(q_{0.99})=0.99$

Replacing Monte-Carlo in steps 1-3 with this formula, you'll calculate VaR quick and accurate.

Answer 2

If you are in a non-gaussian situation, do not explicitely know the distribution of $S_t$, and have to resort to either approximations (gaussian, mixture of gaussians, etc.) or Monte Carlo simulations, you can remove the noise of the Monte Carlo simulations by reusing the same random numbers to generate the data. This can usually be achieved by explicitely setting the random seed before the computations.

But this is not a "better" result: the variation in the data has just been hidden, and a constant, unknown bias has been introduced. Ideally, you should provide some form of confidence interval on the VaR estimator: that will explain the variations. To have a "better" result, you should reduce that interval.

If you also want to hide the jumps resulting from the change in the method used to compute the VaR, you can estimate the difference between the two (by computing the VaR using both methods, when time is available), assume it is roughly constant, and add it to the estimator you want to "correct". (I would refrain from doing that if the bias is too large, when compared to the confidence interval.)