Pamela, here's a discussion of the push point. If
you've already figured it all out, ignore it.
<Warning, long boring post.> BTW, how do you make
bold and italics?
Let's look at what's being analyzed here. For reference, I've
downloaded the OI for AMD puts and calls as of last Friday:
AMD 9/5/97
Close 35 5/16
Closing Option Statistics
Strike Call Put
OI OI
------ ---- ----
27.5 0 0 (not a listed strike)
30 320 1122
32.5 364 1535
35 1143 2551
37.5 8913 3935
40 6635 2727
42.5 2319 2199
45 3489 421
47.5 0 0 (not a listed strike)
Most of these options were issued (i.e., written) by the market
maker(s) for the option, that is, they are short most of these.
(If there were no steves and Herms in the world to write options,
the MMs would be short all of them.) I don't have statistics
on the actual percent of options normally written by MMs, but I've
read often that it's by far the majority.
For discussion, assume the MM for AMD options has written all
of these, i.e., he is short all of the above OI. Let's further
assume that he owns all these naked (i.e., he hasn't hedged
anything). This isn't going to be true, but by and large what
the MM doesn't own or hasn't hedged is probably spread out
among the positions and won't change the conclusions.
Anyhow, this is all the info we have, so we'll use what we've got.
On witching day (9/19), the MM has to close his short positions,
which, under our unhedged assumption, means he has to buy all
these options back (here's the rule: "He who sells what isn't
his'n must buy it back or go to prison").
The price to buy all this back depends on the OI and the price
of the stock at expiration, because at that time all the options
will be at parity. (This statement isn't exactly true
because there are conditions that can keep an option from
parity at expiration, but they're rare and usually small. Also,
there's an assumption that the OI balance doesn't change before
expiration. Not true, but we can't predict that.
I'm ignoring both of these for discussion.)
Note that the price the MM pays to close isn't his PROFIT, because
he got that premie when he sold these options, which premie has all
eroded to parity (or nearly) at expiration. Nevertheless, the
less he has to pay to buy his position back, the more he will
profit (or the less he will lose if he's negative).
Here's a table of the OI value versus stock price, computed from
the above data (i.e., what the MM would have to pay to close
his position if the stock closes at the price in the left column
on expiration day):
Price Value of OI
----- -----------
27.5 14,340,250
30 10,717,750
32.5 7,455,750
35 4,668,500
37.5 2,804,750
40 4,153,000
42.5 7,841,750
45 12,660,000
47.5 18,455,750
You compute this by summing the value of each ITM option
at each strike.
I've included mythical downstrike and upstrike options of 27.5
and 47.5 which didn't actually exist at the time of the data to
illustrate what happens outside the options, though if the stock
actually traded there we'd see more options offered.
In between these strike prices, the OI value follows a straight
line segment drawn between the above points. The slope of the line
is just the difference between the total number of calls below
the price and the total number of puts above the price, that is,
the sum of the in-the-money options at that price.
Here's a rough and ready printer graph of the above OI value table:
Value = MM's Cost to
close all positions
(in millions)
| *
|
|
15 +
*
| *
|
| *
10 +
|
| *
| *
|
5 + *
| *
| *
|
|
0 +---+---+---+---+---+---+---+---+- Strike
27.5 30 32.5 35 37.5 40 42.5 45 47.5
If you print this out and draw straight lines between the * points
you will have the complete value graph. Extend the highest and
lowest lines as far as you want. (It may look like the low end
kinks backward, but that's an illusion due to rounding error
in the rough plot; it's not possible.)
Now if you're the MM, and you can influence the stock price, you
will try to drive it toward the minimum cost point on this graph,
because you can then close out your position for maximum profit
(minimum cost to close).
There's always one, or at most, two points at which the minimum
in the OI value graph occurs. If there are two points, they are
next to each other. If you figure out the dynam
points, the MM doesn't care where it lands as long as it's between
them, but forget this case for practical purposes.
Does this mean that the stock will end up at 37.5 on expiration day?
No, for a number of reasons, but it's where the MM's INTEREST lies,
which can definitely influence the stock in the absence of news or
other forces.
As option writers, we're in tune with the MM for our own distribution
of OI (our short option positions). This is why you want to have
your min-cost-to-close at the OI's min-cost strike (in the absence
of other information).
* * * * * * *
The OI Value table takes a lot of computation and is uneccessary
(though instructive) if all you want is the minimum cost point.
Since there's only one minimum (or at most two adjacent), you
can find the minimum by finding the strike which has a negative
(down) slope line segment to its left and a positive (up) slope
line to it's right.
The slope of a line segment is equal to the dollars the OI value
moves up or down for a one-dollar increase in the stock price.
This is the sum of the ITM calls (because they go up as the
stock goes up) minus the sum of the ITM puts (because they go
down as the stock goes up. The ITM calls at any point is the
sum of all lower calls, and the ITM puts is the sum of all
higher puts.
Exactly AT a strike, there are two slopes involved, the line
to the left and the line to the right. If you think about
moving up in price (left to right on the graph), as you move
past a strike the puts at the strike go OTM and the calls
at the strike go ITM. So if, at each strike, you sum up
the puts above the strike and the calls at OR below the strike,
you will have the slope of the line segment to the strike's
right. The strike with positive slope to its right and negative
slope to its left is the minimum OI value point on the graph.
Here is this computation from the above data:
Strike Call Put Calls at Puts Calls-Puts
----- OI OI and below above
30 320 1122 320 13368 -13048
32.5 364 1535 684 11833 -11149
35 1143 2551 1827 9282 -7455
37.5 8913 3935 10740 5347 5393 **
40 6635 2727 17375 2620 14755
42.5 2319 2199 19694 421 19273
45 3489 421 23183 0 23183
The number in the Calls-Puts column is the slope of the line segment
to the right of the corresponding strike in the Strike column.
The Calls-Puts column of the line above each strike is the slope
of the line segment to the immediate left. That is, for instance,
the slope to the right of 35 is -7455, and the slope to the left of
35 is -11149. (Actually the real slope is 100 times Calls-Puts,
since each contract represents a 100-dollar move for a one-dollar
stock move).
The line marked ** is the minimum value strike because it has -
(downward) slope to the left (-7455) and + (upward) slope to the
right (+5393). Of course we knew that from the value graph,
but this table is easier to compute.
Incidentally, if Calls-Puts turns out zero, then BOTH the
strike on the zero line and the following strike (line below in
the table) have the same minimum OI value, as do all points in
between. The MM is neutral in that range. Any price between gives
the same cost to close. |
