# Monte Carlo simulation

Discussion in 'P1.T2. Quantitative Methods (20%)' started by ldcn8023@yahoo.com, Nov 1, 2011.

1. ### ldcn8023@yahoo.comMember

Hi David,

Could you answer my following question? This is the handbook question 4.9.

A risk manager has been requested to provide some indication of the accuracy of a Monte Carlo simulation. Using 1000 replications of a normally distributed variable S, the relative error in the one-day 99% VAR is 5%. Under these conditions,

a. Using 1000 replications of a long option position on S should create a larger relative error.
b. Using 10000 replications should create a larger relative error.
c. Using another set of 1000 replications will create an exact measure of 5.0% for relative error.
d. Using 1000 replications of a short option position on S should create a larger relative error.

2. ### David Harper CFA FRM CIPMDavid Harper CFA FRM (test)

Hi ldcn,
Yep, hard question (at first glance, i thought it was flawed), quintessential FRM (and reflects why i'm, on balance, comfortable writing hard questions).

The relative error refers to the size of the standard error of the quantile (i.e., the VaR). Or, similarly the standard deviation of the VaR estimate. So, i would remind first what we are talking about: the MCS simulation in this case generates 1,000 trials and the 99% VaR is ~the 10th worst. We run another 1K trials and we can expect a different 99% VaR. Each 99% VaR is itself an estimate based on a sample. In short, this is a quantile estimate, where VaR is just a quantile and estimate refers to the fact that our VaR is a random variable itself (in theory it varies around some unknowable population VaR). "Greater accuracy" refers to a VaR estimate that has narrower sampling variation.

Here the variance/standard error of these estimate decreases (i.e., gets more accurate) as either/both (i) the number of trials increases (ii) as the distribution itself is tighter; on the latter point, we might be reminded of CLT: CLT tells us that the sample mean estimate becomes "more accurate" as the sample size increases; this is reflected in a tighter (less variance) distribution.

Given that, I'd summarize the answer as:

1a. I got stuck myself right at the start, figuring out what the question asking: it's asking about the accuracy (relative error) of the sample quantile (i.e., the 1% tail) COMPARED TO the accuracy of the 1% tail estimate of the underlying normal variable. The underlying (S) is normal, but the option on (S) is skewed.
1b. It's easy (IMO) to miss that because it's VaR we only care about the left/loss tail! Otherwise, (a) would be correct!
2. Then we just need to realize that the short option position has negative (left) skew, which creates more variance in the 1% tail (for any given trial count) ... think of this as a sort of variance in the long left tail: any sample drawn will vary widely with the long tail.
3. We can rule out (b) b/c more trials creates more accuracy; It matter to understand why we can rule out (c): it's because these estimates are samples, go back to get another one, and the estimate will be different. This is "sampling variation." And (a) is tricky b/c the long option has skew, too! But it isn't skew that impacts VaR.

I hope that helps, I think this is a CLASSIC FRM question, instructive and difficult, David

3. ### ldcn8023@yahoo.comMember

I see why we rule out b and c. I don't quite understand a and c. The graph of short option payoff is two straight lines. (X-S when S<X and 0 when X<S) Where is the left tail? The call option graph is also two straight lines. Why do we rule out a?

4. ### David Harper CFA FRM CIPMDavid Harper CFA FRM (test)

Hi ldcn
• You are correct that the option payoff lines are straight but please note:
1. While payoffs are a function only of future intrinsic value, returns are a function of the interim MTM value changes (option delta concerns a "mark-to-market" change in value without necessarily exercising the option, but
2. okay that isn't the key to your point. Your point can still be addressed w.r.t. the payout lines: although they are straight, the short option payoff is still skewed as the upside is capped at the preimum. A naked stock has a totally straight line, but even the short payoff is skewed per the truncated upside
• In regard to (a), that's IMO the "meanest" part of the question! You are correct, the quantile is similarly less accurate for the gain on the long option; quantile accuracy is basically weakened by skew on the long option like on the short option. However, the questions asks about value at risk (!), so it's asking only about quantile accuracy under the left-tail LOSS not a gain. (put another way, A would be correct too, if the question was along the lines of "the 5% or 95% confident quantile")
thanks, David

5. ### ldcn8023@yahoo.comMember

David, thanks a lot!
I see your point. I don't quite understand "the short option payoff is still skewed as the upside is capped at the preimum". I think the payoff of selling short option has a large downside and capped upside at the premium while long short option position has a large upside and its downside is capped at premium? In d, it says "a short option position" which I think means holding/long short position?

6. ### David Harper CFA FRM CIPMDavid Harper CFA FRM (test)

ldcn,

well, careful on over-analyzing the words. Short option = sell an option = write an option. There isn't here a long/short idea. If you write me an OTC option, you are short and i am long and, until maturity, you hold a short position (i.e., you sold it, and you are exposed to my gains) and i hold a long. So, i would just treat sell = short = write.

the skew of the short option is, like, say you write me an option collecting $5 premium we have many outcomes on one side of the distribution which are: S = x1, gain =$5
S = x2, gain = $5 S = x3, gain =$5
S = x4, gain = $5 ... then on the other side of the distribution S = x5, loss = -1$
S = x5, loss = -2$S = x5, loss = -3$
S = x5, loss = -4\$

that is very stubby on the right/gain and long/skewy on the left, thanks, David

7. ### ldcn8023@yahoo.comMember

I see! Thank you!