Hi guys,I wonder if anyone can enlighten me on simulating a local volatility process without resorting to small time step sizes.In a flat surface, we can simulate from time t1 to t2 directly by sqrt(int(sigma^2, t=t1... t2)) * Z as the random term because it has the same distribution as int(sigma * dW, t=t1... t2). Therefore we don't need to use small time step sizes to simulate a flat vol process. However, there doesn't seem to be a straightforward generalization when sigma depends on stock price.My input is a implied vol surface, which can be converted to local vol surface using some interpolation scheme. Thanks guys.

aren't Local Vols a kind of average of stoch vols ? you want to simulate average ?

QuoteOriginally posted by: BonHi guys,I wonder if anyone can enlighten me on simulating a local volatility process without resorting to small time step sizes.In a flat surface, we can simulate from time t1 to t2 directly by sqrt(int(sigma^2, t=t1... t2)) * Z as the random term because it has the same distribution as int(sigma * dW, t=t1... t2). Therefore we don't need to use small time step sizes to simulate a flat vol process. However, there doesn't seem to be a straightforward generalization when sigma depends on stock price.My input is a implied vol surface, which can be converted to local vol surface using some interpolation scheme. Thanks guys.Say you want to simulate stock prices S1 at T1 = first expiration. Since your input is an implied vol. surface, you havea complete set of option prices for T1, call them C(K) where K is the strike. So C''(K) at K=S1 gives you the probabilitydensity p(S1). Integrating it yields the cumulative distribution of S1 and you can make draws from it by the standard "inverse cumulative" method. Same thing works for any other fixed time T; just convert the implied vol. surface at that time slice to prices C(K),and proceed as before. The resulting stock prices are, of course, distributed randomly under "risk-adjusted" probabilities, not the real-worldprobabilities. So, you have to be careful about what you are doing with them. For example, if you used the drawsto estimate Prob(S1 > K), it would only be the risk-adjusted probability of being in the money at T1. Also, I wouldquestion the need for such a simulation, since any questions like this can be answered directly from p(S1) withoutthe need for random draws.

Would the knowledge of p(S1) and p(S2) be enough to price a barrier option with monitoring at t=1 and t=2 and t=1.5. Either of the A and B is required to price path dependentsA: data generating processes for stock(and vol)B: p(St) for all t, 0<=t<=T p(s1) and p(s2) give a sparse information set of B(between t=1 and t=2), is there a way to interpolate to get complete B(between t=1 and t=2) ?

My first thought was no, but I am still thinking about this good question -- see my answer here for what's generally wrong about local vol.I will rephrase your question to an even simpler one. Suppose there is a "complete set" of vanillas in the sense of Dupire/DermanThis means we know the vanilla prices for all strikes and for all expirations 0 < T <= T2.We have a barrier option which pays off on T2 and is monitored only at T1 (and, of course T2), where 0 < T1 < T2.The option knocks out if S1 > B, where B is the barrier, or if S2 > B.From the vanillas, we can get p(S1 | F0) and p(S2 | F0), each conditional on today's 'state of the world', call it F0.In fact, we can get p(St | F0) and the local vol sigma(t,S | F0), for 0 < t <= T2. Is this enough to simply price (but not necessarily hedge/replicate), this simple barrier option *today*?------------------------------------------------------------------------------------------------------------------Update: I think the answer is still "no". Here is my argument.To price this option today we need p(S2 | S1 < B, F0), where F0 is the state of the world at T0.But, p(S2 | S1 < B, F0) = p(S2, S1 < B | F0)/p(S1 < B | F0)We have available the denominator of the right hand side, but no way to get the numerator even witha complete set of vanillas and the deduced local vol sigma(t,S | F0). That's because F0 generally contains morethan just the stock price S0, and if any of these additional factors are stochastic, we can't get the numerator.Another way to say it: you need the full dynamics to get p(S2, S1 | F0).Who has thoughts?

I think the answer must be no for a general process: you need to know about the correlations of the marginal P(S_i) distributions to price a barrier option. Think about this example: suppose I have 1 observation time and a up-and-in binary call option with a barrier at S_0 at t_1=t_0+t. The final option is struck at S_0 and pays out at t_2=t_0+2t.Suppose S is regular GBM with zero drift until t_1, and then from t_1 to t_2 just repeats exactly the same incremental path so that:ln(S_1) = ln(S_0) -s^2t/2 + Xln(S_2) = ln(S_0) -2s^2t/2+ 2XX ~ N(s^2 t, 0)Both t_1 and t_2 have flat skew with volatility s. But, here, the price of the barrier binary option is the same as the price of the regular binary option expiring at t_2 because the option is in-the-money iff the barrier is hit.

This is an excellent example -- is it yours or, if not, whose?Also, didn't you mean to write, "the price of the barrier binary option is the same as the price of the regular binary option expiring at t_1"?

Thank you. It's mine, as far as I'm aware.I think I meant t_2. There is a problem with this, of course: it's not a martingale other than at t_0. In particular, at t_1, you know S_2 for sure. To get a martingale, you'll need stochastic volatility, I'm thinking. Then you're not going to set such a simple example (your skew won't be BS), but things should still work out. How about this:If the path is going to hit the barrier at t_1, have high vol first followed by low vol afterwards. If not; have it the other way around. In either case the total quadratic variation is going to be the same, so still get flat skew at t_1 and t_2. This is going to make your barrier option more valuable, though, the greater the difference between the vols since, after hitting, you know that you are going to have low vol so you will stay in the money.Of course, this one is a-causal and hence non-Markov unless you introduce a hidden vol state, which you might think is cheating. This is the cleanest s-vol example I can think of, though.

Help me understand what I am missing. Your up and in binary call, as I understand it,pays 1 if both S2 > S0 and S1 > S0, where S0 is both the "in" barrier and the final strike.But if S1 > S0, then S2 > S0 is guaranteed by your postulated process.So, the probability of receiving a payoff is the same as the probability of S1 > S0.But the probability of S1 > S0 is the value of a regular binary call expiring at t1.

Yes, that is true. But this price is also the same as the regular binary call paying at t_2 since t_1 -> t_2 adds no new information: S_1>=S_0 <=> S_2>=S_0.

I see, thanks.Your example raises the following questions, for you or anybody:Suppose we can observe, at t=0, a complete set of option prices for 0 < T < Tmax, in the Dupire/Derman sense.Assume an environment with zero interest rates, zero cost-of-carry parameters.Is it possible to create a continuous-time (local) martingale S(t) such that:(i) every option skew observed at t=0 and expiring at T in 0 < T < Tmax is flat with the same implied vol., yet(ii) the process, for 0 < t < Tmax, is *not* dS(t) = c S(t) dB(t), where c is a constant and B(t) is a Brownian motion Or, is it possible to create a continuous-time (local) martingale S(t) such that:(i) every option skew observed at t=0 and expiring at T in 0 < T < Tmax is flat with implied vols depending on T, yet(ii) the process, for 0 < t < Tmax, is *not* dS(t) = c(t) S(t) dB(t), where c(t) is a deterministic (non-random) function known at t=0. If "yes", what is an example, and if "no", what is the proof?

I was wondering that. I think the answer is still "yes", unless you require something conceptually close to iii) and S_T | F_t only depends on t and S_tOtherwise, I think you can use my forward-looking stochastic vol example, or probably all sorts of other cases which are non-Markov in the stock price.If we make condition iii) strong enough, I think we end up with the classic construction of a Brownian motion, so the answer definitely becomes "no" at some point. I wonder how weak one can make the condition? I may have to use this as an interview question this afternoon...

No, I definitely didn't want to have your (iii) as I expect any 'yes' answer to have other random factors like stochastic vol. or be non-markov in some similar way.I need to think about your particular stoch. vol. example.I was worried somebody might come up with a 'yes' example with a risk-adjusted process that was a local martingale, but perhaps the real world process offered an arbitrage opp. If so, I would want to exclude that with some appropriate condition (iv).

If the real-world prices are arbitrageable, then you don't get a risk-neutral measure at all. This assumes that everyone has the same information, though.My example is arbitrageable only if you have secret information about whether the price will hit the barrier or not: just go short or long vanilla options and delta-hedge.It's also only really a "yes" at t_0. As soon as you have any finite time-interval of high-enough frequency observations, you know the vol for sure and, so, know whether you are going to hit the barrier or not.

Does all these discussion imply that although we can find the law of S from T_0 to T_1, we still can't find the law of S from T_1 to T_2, given S_1? My information, of course, is all vanilla prices over all strikes and maturities.