Hi,This is a repost, as I haven't received any answers in the general forum. I would guess that the readers of this forum have probably less problems with the answer.Assumption:Imagine the following process: A standard geometric Brownian motion with constant volatility sigma and a superimposed discrete risk process that multiplies the share price from one (known) point in time T onwards with x>1 or y<1 (there is a probability p_x that it will be S*x and a probability p_y=1-p_x that it will be S*y. x=(1-p_y*y)/p_x should hold, but this is not the point here).Example:Take a R&D company, e.g. a biotech company that will release the success or failure of a trial. Usually the share price behaves "normally" with a geometric Brownian motion, but at the release of new information the share price takes a jump.Question:When we measure the historical volatility, will it be impacted by that jump?Reason for question:When we measure the historical volatility we typically choose some intervals (months, weeks, days) and calculate the standard error of the logreturns and adjust them for the time period, as the volatility should be annualised. The time period i with the jump in it, i.e. t_i<T<t_{i+1} naturally biases the volatility calculation, i.e. this time period increases the volatility. But the volatility is actually a measure of instantaneous change, so we should really choose very small time increments (like minutes or even less) to calculate the volatility. Then the time interval with the jump has a smaller weight. But on the same time the absolute size of the log returns gets smaller except for that time interval. So, after annualisation, will the volatility depend on the chosen time interval? Is it even possible that the volatility becomes sigma of the geometric Brownian motion by taking infinitesimal small time intervals for the measurement?Remark:The implied volatility of options on such stocks is little helpful, as it has to account for all risk processes, but the Black-Scholes model only accounts for the volatility of the geometric Brownian motion.A brief technical sketch of the answer would be most welcome (besides the actual answer, of course).Thanks.

You have postulated a process that behaves like GBM plus it has a single jump.Let's write for the log-return process(*) dX = a dt + sigma dB(t) + dJ(t).Here dJ(t) represents the jump (in log terms) and is zero everywhere, except at the single time t=TJ, the time of the jump.Your question was: will the historical volatility be impacted by the jump? Let's define the historical volatility H to be the annualized realized std. deviationover a period [0,T], based upon measurements of X(t), recorded at intervals DeltaT. So ... H depends on [0,T] and DeltaT, right?Then, 1. Take the case where sigma = 20%. If you take a period [0,T] that does not contain TJ, then H will converge to 20% as DeltaT->0. In the alternative, if you take a period [0,T] that does contain TJ, then H will converge to sqrt(sigma^2 + f J^2) as DeltaT->0,where J is the logarithmic jump size and f is the annualization factor. [This last formula follows pretty directly from the definition of H^2, breaking the sum of squares into the single jump term plus the rest.] For example, if [0,T] = 1 year and the (log-) jump was 40%, then f=1 and J = 0.40. So, youget H -> 0.20 in the first case and H -> sqrt(0.20^2 + 0.40^2) ~ 0.45 in the second. A different way to measure. Instead of fixing [0,T], you could take a smaller and smaller measurementperiod [0,T] surrounding each time t. In this case, H(t) will converge to sigma in the first instance and +infinity in the second.Here H(t) represents the realized historical volatility in the immediate neighbrhood of time t. Either way, this shows that the answer to your question is: yes, H will be strongly affected by whether or not TJ is in [0,T] or not.2. When I say "take a period that does not include TJ", this can also mean simply dropping the measurements of the log-returns immediately surrounding the jump event.3. If the purpose of measuring H is to estimate sigma, then the cure to this potential estimation problem isobvious: just exclude the jump event from the calculation of H.

Alan,That's extremely helpful, thank you. Putting it in math terms helps a lot.But actually this means that a GBM with a jump process has infinite volatility? Or depends on the measurement intervals and can be anything? This is quite disturbing and leads to another question:In CAPM beta is defined as sigma_S/sigma_M*corr(S,M). Since sigma_S-->infty and sigma_M and corr are limited this means that beta -->infty. On the other hand the jump process is independent of teh market M and should not have an impact in CAPM. There is a contradiction, but probably just because I overlook something. Can you help?

If there's just one jump, the measured historical vol is not going to be infinite, but it will be higher than the diffusive vol, sigma.If there are future jumps, your expected realized volatility may be infinite, depending on the structure of the jumps.There's a significant school of people (after Mandelbrot in the 1960s) who would argue that financial time-series do not have finite second moments at all, by the way, which is one of the many reasons not to like CAPM! Or, indeed, most of the Ito-calculus modeling we tend to use all the time.There is quite a large literature on identifying jumps and measuring diffusive vs jump volatility: try various papers by Ait-Sahalia, Jacod, Andersen, Bollerslev, Diebold and their friends. Generally (as Alan says), you need arbitrarily high-frequency to distinguish jumps perfectly. But you can also try and estimate the likelihood that a particular return is a jump using Markov-chain Monte-Carlo methods like in e.g. Eraker, Johannes & Polson 2003 "Impact of Jumps in Volatility and Returns". These methods are very model-dependent, though, and you need a pretty long time-series to have any confidence, though.

ok, I see.Just to finish this issue with the CAPM: Even though the volatility is finite, the independent jump process seems to impact beta (as it impacts sigma_S and beta=sigma_S/sigma_M*corr(S,M)). So my last question:Is the statement that beta only considers the amount of market risk of an asset just not exactly correct? I assume it is correct in absence of the jump process. But since the jump process has precisely nothing to do with the market (that's the assumption) the statement seems not quite right. Any thoughts?

You are correct to infer that if measuring volatility can be problematic with jumps, thenmeasuring cross-volatilities (beta, correlations, etc.) will also be problematic.The basic model for log-returns now becomes dX = a1 dt + sigma dB(t) + dJ1(t).dM = a2 dt + sigma_M dW(t) + dJ2(t)where dB(t) dW(t) = rho dt and the second equation is, say, for some broad-market index return. If there weren't any jumps, then the realized historical covariance would be a good estimatorfor the actual diffusion covariance (rate) as the measurement period DeltaT -> 0. Since the variance ratesare also efficiently estimated, this would lead to good estimates of rho, beta and so on. With the jumps, there are problems. Also, you generally have to worry about simultaneous/correlated jumps. With a biotech company jumping because of a clinical trial release, there isn't going to be much of a correlation. If INTC jumps because of improving semi-conductor sales, there will be a spill-over jump in QQQQ and probably SPX as well (actually, it would have to be the SP futures, since the regular session will be closed.) In general, the measured covariances will combine the diffusion and jump effects, just like in the single stock example.When the jumps are obvious and scheduled, like in the case of earnings or clinical trials, they can more or less be separated out.When the jumps occur at unpredictable times, it is much more difficult.Anyway, as crmorcom pointed out, there is a huge literature out there on questions of financial theory and econometrics with jumps.