Black-Scholes Option Pricing
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Here, we want to describe the logic behind the Black-Scholes Option Pricing Formula
which looks like this:
Figure 1
>What do all those symbols ...?
I think you may want to go back and peek at these introductory remarks concerning Probability
Distributions ...
here.
We're not going to derive the formula ... just give some indication of how it arises.
We'll be talking about a cumulative distribution function, F(x), and its derivative,
f(x), the density distribution.
Typical charts might look like this:
where the slope of the F-curve, namely dF/dx, is the f-curve and the area beneath the f-curve
is the F value.
If x is a random variable (distributed according to some distribution, as shown),
then the Mean or Expected value of x is E[x] =
Further, if we want to know the Mean or Expected value of some Quantity that depends upon x,
say Q(x), then it's E[Q(x)] =
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In particular, the squared deviation of the x's from their mean, namely (x-m)2,
has an Expected Value of
which we recognize as the square of the
Standard Deviation.
>We do?
Yes. In fact ...
>Just what's the purpose of all this?
Okay, let's just plunge right in:
We know that, at expiry of a Call Option, if S is the stock price and K is the Strike Price
then the Option is worth S - K provided S is greater than K ... or it's worth nothing
(if S is less than K). In other words, the Option Price (at expiry) is the maximum of
S - K or 0, namely:
(1) C = Max(S - K, 0)
Ah, but that's at expiry of the Option.
If the Option expires t years (or months or weeks) in the future, what's the Option
worth now?
Suppose that, t years in the future, the stock price is x.
Then the Option, at expiry, is worth Ct = Max(x - K, 0),
according to Equation (1).
The Expected Value of this Quantity, assuming some distribution of yearly (or monthly or weekly)
stock returns, is then
(2) E[Ct] = E[Max(x - K, 0)]
which we recognize as
=
where we integrate from K since Max(x - K,0) is zero when x < K.
>If that's the Expected value of the Option at expiry, then what's the expected value now, today, this very minute?
It's the Present Value of E[Ct], assuming a risk-free rate of, say, r.
That means it's worth
(3) C = e-rt E[Ct]
or simply e-rt
or
(4) C =
e-rt - K e-rt
= e-rt - K e-rt {1 - F(K)}
where F is the cumulative distribution.
>Where did that e-rt come from?
Well, if we were to put $A in the bank at an interest rate of r, then, after t years,
it'd be worth B = A{1+r}t so
A = B {1+r}-t gives the Present Value of $B.
>And where did e-rt come from?
We assume continuous compounding, so if we subdivide a year into N periods,
the interest rate for each period is r/N and the number of periods is N so we have (1+r/N)N
as the yearly growth, at the annual rate r ... if the compounding is continuous.
>And where did e-rt come from?
(1+r/N)N approaches the value er, as N approaches infinity - that's a
well known theorem. Anyway, that's the meaning of continuous compounding.
Here's an example with r = 0.05 (meaning 5%):
>A well known theorem?
Yes.
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Now, instead of {1+r} as the annual growth factor, we
use {er} and so {1+r}-t becomes
{er} -t which is e-rt, so the present Value
is written as Ct e-rt ... instead of Ct
= C{1+r}-t.
>Sounds like mumbo-jumbo to me.
The math is much, much nicer. Besides, this calculation of present value is what one
means by "risk-neutral": the value of an asset at time t discounted to its present value
using the risk-free rate.
The two pieces in Equation (4) will give rise to the two pieces of the Black-Scholes formula
in Figure 1.
Now we stare at the stock price at time t, namely St (which, as a random variable,
we're calling x). If the returns over each time period (a year, a month, a week) are
r1, r2, r3, etc. then we write
1 + rk = exp(gk), where exp(x) means = ex.
The cumulative gain over t time periods, namely
(1+r1)(1+r2)...(1+rt)
can now be written more simply as:
exp(g1)exp(g2)exp(g3)...exp(gt) =
exp(g1+g2+...+gt) =
exp(Σgk) =
exp({(1/t)Σgk } t) =
exp(Mt)
where M =
{(1/n)Σgk }
is the Mean value of the g's (over t time periods).
Of course, the set of g's (namely g1, g2, ... gt),
are randomly distributed, so M is a random variable
... but what's the distribution function?
Here's where we make a simplifying assumption: we assume that these
g's are Normally distributed and, since 1+r = eg, this means that we're assuming
that the returns r1, r2, etc. are Log-normally distributed. This
identifies the functions f(x) and F(x) and makes possible the evaluation of the integrals
in Equation (4).
>zzzZZZ
Don't worry, I don't intend to evaluate any integrals. It's much too scary for me.
Instead, we'll take another, more elegant tack:
for Part II
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