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.