If returns are normal, then % changes are not.
Posted: February 27th, 2003, 3:46 am
1. If we assume that returns(defined as the log of price relatives) are normally distributed (with zero mean), we are required to accept that the probability of a return of .0953 is the same as the probability of a return of - .0953. We are also required to accept that the probability of a .0953 return does not equal the probability of a return of - .1053.2. The Geometric Brownian Motion requires that small percentage changes of the stock are normally distributed (at least that is what is claimed by many). This means that the probability of a .1% increase in the stock is equal to the probability of a .1% decrease.3. 1. above contradicts 2. above4. If the probability of percentage changes are equally (with zero mean) , then there will be a tendacy for the stock to go zero. With the higher volatile stocks going to zero faster. The stock will proceed to zero at a pace of vol^2/2 each year on average.5. Is there someone who will comment on how these options models incorporate two contradictory assumptions.Peace :Worldoptions