On Mon, 30 Nov 2009 17:47:37 -0600, "mcatolico"
<mcatolico@mindsprin
>This captures things (I think). But it's really much more basic.
>
>At any point in the life of an option, that option has some equivalent delta
>value. As you approach expiration the delta value approaches either zero
>(otm) or 100 (itm). Traders choose to hedge or not hedge to some degree at
>any point but at expiration the impetuous is to hedge completely as time
>goes to zero. If the position has 100 long deltas, you sell 100 delta units.
>if it is short 100 deltas, you buy 100 delta units. If the position has no
>deltas, then any new delta exposure is pure risk absorption or speculation.
>
Since I have never understood the general connection between being
delta neutral and making money, your explanation doesn't help me at
all. In the exposition I gave of the scalping algorithm, it is clear
what the money making idea is. The owner of a slightly in the money
option, close to expiration, sells (for calls) or buys (for puts) the
amount of the underlying needed to guarantee that the current
intrinsic value will be locked in if the option expires ITM. If
enough traders do this, the underlying will go to the other side of
the strike where the traders, will lock in more cash by making the
reverse trade in their underlying positions so that the underlying
will again pass through the strike and the process will be repeated
and the trader(s) who are participating in the pendulum bobbing back
and forth will make dough that expiration day.
Now contrast this situation will the rationale for making a position
delta neutral at other times. I see, in this and other forums,
diagonal traders explain how they sell just enough ATM calls against
their LEAPS to make the total position DN. Of course the next day the
market moves and they are no longer DN. Why did they think that was
the best move in their quest to make money? So I don't think invoking
the magic mantra of "delta neutraility" to explain a strategy is "much
more basic" that relating it to a plausible scenario of market
behavior.
>-----Original Message-----
>From: OptionClub@yahoogro
>Behalf Of Ricky Jimenez
>Sent: Monday, November 30, 2009 12:05 PM
>To: OptionClub@yahoogro
>Subject: Re: [TheOptionClub.
>
>On Mon, 30 Nov 2009 11:00:26 -0500, I wrote:
>
>>On Mon, 30 Nov 2009 06:01:05 -0600, "Jack" <jack@jackcpa.
>>
>>>I understand it the underlying is up a quarter and you short it, you do
>not have any risk. If it goes up, you exercise, if it drops, you buy and
>close.
>>>
>>>
>>>
>>>
>>>
>>>Now, if it is down a quarter (you still own 100 calls but have no position
>in the stock), if you buy the stock ($775,000) you have no protection. If
>it continues to drop, you are long stock and long calls. All this risk to
>make $2,500?
>>
>>I hope both you and Michael will forgive me if I butt in here and try
>>to explicitly specify the algorithm in question.
>>
>>1. If you have N long option contracts (underlying multiple = 100)
>>ITM by M, sell 100*N shares of the underlying if they are calls, buy
>>100*N shares if they are puts. You forego further gains if the
>>underlying gets further in the money but lock in a payoff of M per
>>share. If you own N straddles, you can do this if either side is ITM
>>by M. Goto 2.
>>
>>2.a. If the options now are OTM and there is very little time before
>>expiration (probably broker and hardware dependent), remove the
>>position in the underlying. If you own N straddles, this step can be
>>ignored, assuming one side will be exercised.
>>
>>2.b.. If there is sufficient time before expiration and the options
>>are OTM by M, remove the position in the underlying. You lock in a
>>payoff of 2*M. Then Goto 1. Of course you can choose different Ms
>>for steps 1 and 2.
>>
>>I hope this doesn't lead to more confusion.
>
>The word "payoff" used above is misleading. What I meant was that by
>using the scalping algorithm for one iteration, you will have a M or
>2*M per share cash credit in your account at expiration. (Note than
>when you have a straddle, you start a new iteration when you are in
>step 2b of the prior iteration.) The payoff from the position will be
>0. The overall profit depends on the prior history before you entered
>the algorithm as well as transaction costs.
>
>
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