Okay, and now I think we have sort of the
the, tools we needed, to get to the the
important point.
Which was differentiating improper
integrals.
So we're going to start out with a
function F.
That's continuous, and we're going to
assume that the integral of f over
the entire real number line, so integral
from minus infinity to infinity, exists.
So, in particular, that means that it's
finite.
And now I want to consider a function b of
t.
Its a differentiable function and I am
going to
define a new function g of t which is
going to be the integral of my function f
from minus infinity up to b of t,
and I want to be able to compute g prime
of t.
And so so, sort of like taking the
derivative of a definite integral.
It's actually, doesn't end up being too
difficult.
So all we're going to do, because we
assumed that this integral
exists, that means that I can pick a point
somewhere in between.
And I can write g
of t as an integral from minus infinity up
to that
point plus the integral from that point up
to b of t.
And then because I'm free to choose that
point I might as
well choose something that's easy to work
with so I chose 0.
And so now I just want to take the
derivative of G of T, but
since the first term,
so if I look at this closely now, this is
a, it's an indefinite integral.
So the integral from minus infinity up to
zero.
But there's no t's in this first term.
So as far as I'm concerned,
when I'm taking a derivative with respect
to the variable t, this is a constant.
And so, when I take the derivative of that
with respect to t, when I take its
derivative.
I'm just going to get zero.
So, to compute the derivative with respect
to t of my function
g of t, I can just ignore this first term
and all I
have to do is compute the derivative with
respect to t of the
integral from zero to b of t of my
function f of x.
But notice what happened now.
So the, the problem that we were trying to
deal
with is that we had an, we had an improper
integral.
But because of this the limits here are
both finite values.
So, this is integral that we already know
how to take the derivative of.
This is just a, take the, we use exactly
the same rule
I talked about in I guess it would be
number seven today.
How to take the derivative of a definite
integral.
And, on top of that, we're lucky that the
bottom is
zero so we only have to use half of the
formula.
So, g prime of t is just going to be the
derivative with respect to t of the
anti derivative of f evaluated at b of t
minus the anti derivative of f evaluated
at zero.
And this is just the result from
differentiating.
a definite integral.
So I know that that's f of b of t times
the derivative of b of t.
So f of b of t times b prime of t.
And then to do the same thing for the
lower limit.
Or we can do the same thing for the lower
limit.
So I'm just going to instead of g of t,
I'm going to have h of t, just exactly the
same thing, except that now it's my lower
limits
that's the function of t rather than the
upper limit,
[NOISE].
And so I'm still going to consider
integrating this function from a of t to
zero, and then adding on the integral from
zero to infinity.
And then when I take the derivative of
that, this
second term now is going to be the
constant, so that's
just going to go away so the, the answer
here
is just going to be the derivative of this
first term.
And so the only difference we're going to
see, is just because
when I do this part of the fundamental
theorem of Calculus.
Now my function a of t is on the negative
side, so I get
the same answer as I got for the, for the
upper, limit being the function.
Except now it's just going to be minus.
F of a of t times a prime of t.
So now we have a, a very simple
rule for taking the derivative of an
indefinite integral.
And let's see if we can put that to work.
So,
[NOISE]
I'm going to look at something called the
delta of a European Call Option.
That I'm going to price using the
Black-Scholes formula.
And so the Black-Scholes price, for a
European call option,
so this is the simplest possible one to
look at.
On a non-dividend paying asset, so this
could just be
a stock for instance, with a strike price
of 1.
Has a, a volatility of 20%, so this means
the volatility of
the returns on my asset, and I'm looking
at three months to maturity.
And so again, the reason I'm, I'm putting
all of these real numbers in
here is just because the Black-Scholes
pricing
formula, it's a function of several
variables.
I'm going to make it into a function of a
single
variable s which is the, the current price
of the asset,
just by plugging in actual numeric values
to all of the rest of the inputs.
And so this is the, the pricing formula.
So I have s.
Times this function phi, evaluated at d
1, minus e to the minus 0.01,
times this function phi again, evaluated
at d 2.
Where d 1 of S is just going to equal 10
times the log of the, the current stock
price, so 10 times the log of S.
Plus 0.015.
D2 of s is equal to d1 of s minus 0.1.
And this function phi.
So this is where the, all of the math
we've been doing
for the, for the last two lectures is
going to come into play.
This function phi.
Is defined as the integral.
So this is a, a capital phi, wherein the
font
I'm using, it's drawn just with a vertical
middle line.
The lowercase phi is its derivative, which
has a, a slightly slanted line.
So I'm going to integrate lowercase phi
from minus infinity up to z.
And lower case v this is probability
density function for a
standard normal distribution and this is
the mathematical formula it has.
So in order to calculate the price of my
European call
option, I am going to need to be able to
do mathematics.
With this function capital phi.
But this is a function that's sort of
famous
for not having an antiderivative.
So you can't actually write down a formula
for the value of this integral.
This is the simplest that I can write it
down.
So in order to actually price these
things, you have
to use a computer to approximate the value
of this integral.
But when we do mathematics with the
Black-Scholes formula,
we still want to be able to get analytical
results.
You know?
If, if I try to do everything numerically.
Then you run into problems of, you have to
be very careful of, you know?
As you continue to do calculations, are
you
accumulating any round off error in your
compu
[INAUDIBLE].
In your sums things like that, so what we
would like
to be able to do is actually compute
derivatives of the
Black-Scholes pricing formula and get the
correct answer theoretically before we
actually have to start plugging in numbers
to the to the formula.
And so, the delta of a European call
option is just the derivative of the
price, so the derivative of the pricing
formula, so in this case c of s with
respect to the asset price s.
And it's denoted by
this Equilateral triangle is capital Greek
letter delta.
And so this is the derivative that I'm
going to try and compute.
So, it's already kind of complicated and
it's much, much
more complicated if I don't plug numbers
into it, because
as soon as I plug the numbers into it, I
can do some math and a lot of them combine
together.
but it's still complicated enough that I'm
going to try
and compute the derivative of this term by
term.
So first I'll look at this term and then
we'll
look at this term and we'll try and
simplify our answers
as much as we can.
So for the derivative of the first term I
have S times a function of S.
So I'm going to need to use the product
rule.
So that's just going to tell me that the
derivative of this
function here is going to be the
derivative of s times this.
Plus the derivative of this times s.
So here, I have the derivative of s is
just, oops.
Derivative of s is just 1.
So I just have the second piece here.
Then plus s times the derivative of, this
guy.
And I'm just going to replace that by its
definition.
So that's the, the definition of capital
phi of d one of
s is the integral from minus infinity up
to d one of s.
Of little phi of x.
And so we have a formula now for
evaluating this derivative.
So, if I'm taking the derivative of an
indefinite
integral, I know that it's just going to
be.
The integrand evaluated at the upper limit
times the derivative of
the upper limit, with respect to the
variable that I'm, I'm computing.
So I need to use the chain rule here.
And so I'm going to end up with s, that's
this
s here, times lower case phi of d 1 of s.
So that's this phi evaluated at the upper
end point of the integral
and then by the chain rule this is phi of
d of s.
So I need to also multiply that by the
derivative of d1 of s.
So now remember that d1 of s is equal to
10 times the log of s.
Plus 0.015, so the derivative with respect
to s of d1 of s
is just going to be 10 divided by s, so 10
times this here that's just a
constant so the derivative of the constant
is zero, the derivative of the log of
s with respect to s is just 1 over s so I
end up with
10 divided by s.
So that tells me that the derivative of
the first term now, this capital phi
of d1 of s plus s times lower-case phi of
d1 of s,
times 10 over s, and luckily one of the
nice things about Black-Scholes is
usually when you're doing it right stuff
kind of beneficially cancels out for you.
So here, the s and 1 over s are going to
cancel each other out.
So for the first term, the derivative is
capital phi of d
1 of S, plus 10 times little phi of d 1 of
S.
Okay.
So now let's look at the derivative of the
second term.
So here it's a little bit easier, so e to
the minus 0.01.
Looks complicated but that's just a
constant.
[COUGH]
So I'm going to take that outside of
my derivative, so the derivative was
linear remember.
And I'm left with the derivative of
capital phi of d2 of s.
And I'm just going to, so in the square
brackets is just the definition of that.
So that's the integral from minus infinity
to d2 of s, of little phi of x dx.
And so I'm going to use my pricing rule
sorry my im.
Improper integral rule
again.
So that's just going to be phi evaluated
at
the upper endpoint, and then by the chain
rule,
I need to divide that, or, sorry, I need
to multiply that by the derivative of d2
of S.
And so remember, d2 of s was just equal to
d1 of s minus 0.1.
So
this is just a constant, so the derivative
of d2 with respect
to s has to be the equal to the derivative
of d1
with respect to s, and we already computed
that, and that was
10 over s, so we can recycle that result a
little bit.
So that's this parked over here I can
replace with 10 over S and
that gives me e to the minus 0.01 times
phi of
d 2 of S times 10 over S.
And
so if I put all of that together.
The derivative of the first term
was capital phi of d 1 plus
10 times lower case phi of d 1 of s.
And the derivative of the second term so I
was
subtracting the second term so it's
going to be subtracted here.
I have10 over s.
E to the minus 0.01 times phi of d 2 of S.
So sorry, I should, this d 1 is a function
of
S, so it should have an S in the
parentheses, as well.
I forgot to put that there.
[COUGH]
But unfortunately, we're not finished yet.
So remember that little phi of x is equal
to 1 over the square root of 2 pi times
e to the minus x squared over 2.
So
what I'd like to do to simplify this.
Instead of having a d1 and a d2 in here.
I want to see if I can rewrite, just
remember d,
d1 and d2 weren't that different, d2 was
just d1 minus 0.1.
So maybe
I can make this simpler by writing writing
this term
in terms of d1 of S rather than d2 of S.
So I'm just going to start plugging things
into this definition of lower case phi.
And so first of all, I'm just going to
evaluate phi
at d 2 so I have d 2 of s squared.
Where they used to just be the argument
here.
But I want to have a d 1 there instead.
So I'm going to replace my d 2 with d 1 of
S, minus 0.1.
So I have e to the minus phi 1 of S minus
0.1, that quantity squared.
And now I'm going to try and to, to make a
little bit of sense of this.
It would be nice if I had this just in
terms of e to the
d1 of S squared because that, that would
allow me to write it in terms
of phi of d1 of S which would make it
match what's on in this second term here.
And so now what I want to do is sort of my
completing the squares trick backwards,
so let's just start by doing the math.
So the first thing I'm going to do is
expand the, the squared term here.
So I had.
D1 of S minus 0.1 squared.
So that ends up being d1 squared of S
minus 2 times 0.1 times d1 of S, so I have
0.2 d1 of S.
Then,
plus negative 0.1 squared, so that becomes
positive 0.01.
Let's see.
Did anything change?
No.
I think I made a cutting and pasting error
in this one, so it's hope, hopefully, that
doesn't last too long.
Okay, so essentially what was
supposed to happen in this line,
and I guess I somehow missed it.
Here I have 0.01 divided by 2, so that's
going to be 0.005 and remember when
I'm adding things together in an exponent,
I can also think of that as a product
of e to each one of those terms.
So what I meant to
do in this line here, I wanted to have e
to
the minus d1 squared over 2.
That's what's going to end up here.
I'm going to have minus times minus, so.
E to the minus, or e to the positive 0.2 d
one but
I'm dividing that by two so I get e to the
0.1 d one.
And
then, 0.01 divided by two was 0.05.
And that was just going to end up also
over here and remember there's the minus
sign multiplying it here so I have minus.
0.01 minus 0.005.
And that's what gives me this term here.
So hopefully, that would have been a
little bit
more clear if I had, not messed up this
line.
But,
okay.
And now I've, I've written this bit.
The one over 2 pi here.
E to the minus d 1 of s squared over 2,
well that's exactly what phi of d 1 of s
would be.
So hopefully that's starting to look a
little bit simpler now.
And now remember I had d 1 was just 10
times.
The log of, 10 times the log of S,
hopefully that'll show
up here.
10 times the log of S plus
0.15.
And so
here I have e to the log of S.
That's just going to be S.
Because this is a sum I could also think
of this as e to the positive 0.015
so that's going to cancel out my e to the
negative 0.015 or you could also think
of it as just I'm adding those two numbers
together getting e to the 0 and that's 1.
So this bit here is going to cancel out
this bit here.
This e to the log of s, that's going to be
just s, so that's going to cancel out my s
down here,
so what I am left with is this second term
the, the
magnitude of it being equal to 10 times
phi of d1 of s.
[INAUDIBLE]
.
So now, let's go back and look at my, my
formula for the delta of c.
And so the delta of c was originally phi
of d1,
plus 10 times lower-case phi of d1.
And then minus.
And we've just showed that this thing, 10
over s.
Times e to the minus 0.01 times phi of d2
of s.
That was the same thing as
10 times phi of d1 of s.
So what
I end up with is capital phi of d1 plus 10
lower-case phi minus 10 lower-case phi.
And in the end we find that the delta for
this call option, is actually just capital
phi of D 1.