The delta will do the job of predicting how much the option price will change when the stock moves - BUT NEVER in an environment when implied volatility changes more than compensate for the effect of delta. Many times, in sharply falling markets, calls do not decline in price - due to the effect of IV. Many times, in rising markets, calls fail to move higher in price, due to IV getting crushed. Mark
Try to understand this because it will help you really understand the magniture of what a worse case can be under normal volty conditions. If realized vol will be the same as currently implied (42.??), and if the current price of SPY ($82.76) is the top price during the next 38 days, we can say this with ABSOLUTE certainty: SPY will reach a price equal to or LESS than $58.75. That is $24 down (with certainty under the above two conditions which is not unrealistic). So what people see as dramatic moves are normal moves. There is chance that what I wrote would be a shock to people. You need to understand why it is fundamental truth of how markets work. If you are shocked or your mind cannot accept it, all it says that they you have not fully put your mind around the market.
Oh no!!!! You are back! And just as ignorant as ever. I hope no one takes what you write seriously. One thing I can say with absolute certainty is that you are as dumb as a mud stump.
No, almost certainly not. Beau W. graduated from a community college in Tennessee. Unfortunately there was no room for courses in options theory in the ag dominated curriculum. All he knows about options he learned from RiskFreeTrading. There is a slight chance that he means that the relationship among relative deltas of options with the same K/S is a crude measure of relative volatility, but I doubt it.
If it is realized all in one side, could you give the numbers? If it is realized to the down side what are the numbers? PS: next time re-read what I write at least twice, because I wrote the conditions under which the result is valid.
If Pythagoras were to say " If we have a right angle triangle , then we have A^2+B^2 = C^2)." Some people would respond that it is not true and give as their reason that a triangle with no right angle does not possess the above property. I hope the triangle example will shed some light in the mind of people who makes similar mistakes. I would like somone to challenge the validity of the result I stated in the post I am quoting in this post.