EA, I'm TA-challanged, but I got A's in math<ggg>. Maybe
you're not interested any more (and anyhow S Tops explained
it pretty well), but here's a brief but slightly more
general description of derivatives and Why They Are Interesting.
WARNING: Long Boring Post follows. Best skipped at a first
reading <G>.
If you take any curve, such as price versus time (just to pull one
out of the air<g>), its first derivative at any point is the slope
of a straight line drawn tangent to the curve at that point.
The value of the first derivative is the rate of change of the
curve at that point.
If P(t) is price at time t, then the first derivative is the
slope (in, say $$$/day) at time t, or, how many dollars per day
on this particular day t is P changing by. Often this
is written as P'(t) or dP/dt.
The second derivative of P is the first derivative of the
first derivative, that is, the rate of change of the first
derivative. In the case of P(t), the second derivative
would be in dollars per day per day, often written as P''(t)
or d2P/dt2 (where the 2's here are supposed to be super
scripts, as in squared, but I've forgotten the
superscript HTML tag).
What's interesting about P' and P'' are that together they
pretty well describe the shape of the P curve. When P'(t)
is zero, the P curve has zero slope which means it's
going through a (local) MAXIMUM or a MINIMUM, i.e., a
top or a bottom. When P' is positive, P is increasing
(stock moving up); when P' is negative, P is decreasing
(stock moving down).
Now if P'' is positive, that means the P curve is
CONCAVE UPWARD, meaning price P is accelerating upward.
Note that can happen, even if P' is negative meaning
price P is falling. In that case accelerating upward means
falling slower and slower each day. Similarly, if P'' is
negative, P is CONCAVE DOWN meaning prices are accelerating
downward (if P' is positive, P'' negative means the rate
of price increase every day is slowing).
Where P'' is zero, it means the P curve has an INFLECTION
POINT, that is, P is changing it's curvature from upward
(price accelerating upward) to downward (price accelerating
downward) or vice-versa. If P'' changes from + to -
(going through zero), it means the P curve's upward trend
is shifting to a downward trend.
By "trend" here I mean the direction price
is accelerating here - not the usual definition, sorry,
I don't know a better term. A stock accelerating down could
be SLOWING it's descent (P''>0) when suddenly it starts
falling by less each day. That would be a case where
P' < 0 (falling price) but P'' turns from <0 to >0
(was falling more each day but is now falling less each
day). That's what I mean. Maybe "tendency" would be a better
word.
Naturally with prices and days there's a lot of jumping around,
so even though derivatives are pointwise functions (you can
compute a first derivative with two days' data, a second
derivative with three days'), they
can be smoothed like any other function. Linear regression
(with moving endpoints) is one way of smoothing, as is
a moving average, ema, etc.
So, if tomorrow's price P depends on today's P' (>0 tomorrow's
is up) and today's P'' (>0 tomorrow's is up more than
today's was up), then you don't need to know the price at all<GG>.
I must try that <GGG>.
Sorry, a little wishful diversion there. The point is that
that the derivatives give you a quantitative measure of how
the trends are changing rather than depending on subjective
judgements.
This has dragged on too long and is no doubt boring, but
I just HAVE to make an analogy between physical motion and
price/time curves (I was educated as a physicist when I
was too young to know better, you see).
If you take a mass M at a position x(t) at time t,
(analogous to price at time T), then the first derivative
x'(t) is the mass's VELOCITY (different from speed in that
it includes the direction of motion, say + for up and
- for down), and the second derivative x''(t) is the
mass's ACCELERATION (again + for accelerating up and
- for accelerating down).
The mass times the acceleration
is the FORCE on the mass. The larger the mass the greater
the force for the same acceleration. I find that strongly
analogous to daily volume -- if there's a large volume
moving up, that's a large market force acting upward (and
conversely of course). If this analogy actually holds, then price
acceleration (p'') times volume is a good measure of the
market forces at work. It even holds (if the analogy is
good, of course) if the stock is moving DOWN on high volume,
but is decelerating its descent (accelerating upward).
Mass times velocity is MOMENTUM. If a mass has a large
momentum, it takes a large force to accelerate it OR
to decelerate it quickly. By what seems to me to be a strong
analogy, price change rate (P') times volume is a measure
of how hard a price trend is to accelerated or to reverse.
If a stock moves on high volume (big mass) OR moves very
rapidly (big P', or velocity), it requires larger forces
to accelerate or decelerate it.
Finally, I note that (in physics) FORCE is the first derivative
of MOMENTUM [Newton's second law: F = Ma, where a is the
acceleration, the first derivative of velocity, so
Force = Mass * dv/dt = d(Mv)/dt for constant mass =
d(Momentum)/dt]. That is, force is the time rate of change
of momentum.
Now if these analogies hold (or we can identify how they
don't and modify them accordingly), and learn how to measure
them, and analyze them quantitatively, then we would gain insights
not only into the descriptive behavior of markets (P, P'')
but also the DYNAMICS of the market (force, momentum, causes,
effects). If so, why, er, that would be good.
I must go and work on Spots's
theory of market physics ... <GGG> |
