Now we get to the, the fun part which is
actually taking derivitaives.
So we'll start off by calculating delta.
So the delta of a European call option is
the rate of change, oops.
Of c of s t, so again remember this is the
actual Black-Scholes pricing formula.
So, even though I've written it here as c
of s
comma t it's really a function of all of
the seven variables
that I had on the, the first set of
slides.
And I'm going to take that, it says the
rate
of change with respect to the asset price
s.
So, how how much does the value of the
call option change?
As the price of the underlying asset
changes.
And so mathematically I can write that
just as, delta is equal to the partial
derivative with respect to s of the
Black-Scholes pricing formula for a
European call option.
And so this is the pricing formula for a
European
call option that I had in the first set of
slides.
And, I'm just going to go ahead and start
using the rules we've
been developing er, earlier in this course
to start taking the dcerivative here.
So the first thing I'll, I'll notice.
Is that the pricing formula has two terms.
So I can use the linearity property of the
derivative to
write this as the, so this is a, a sum
essentially.
And so the derivative of a sum is the sum
of the derivatives.
And then I'm also in, in this one step
going to combine the.
Scaling properties, so when I'm taking the
derivative with respect to s.
So, I am taking a partial derivative with
respect to s.
Anything that isn't a function of s, I can
treat as
a constant, and any constant I can move
across the derivative operator.
So, in the first case here.
This part the e to the minus qt minus t.
That doesn't depend on s so I'm going to
take
that out and I end up with e to the minus
qt minus t times the partial derivative
with respect to
s of the product s times capital phi of d
plus.
And on, on the right hand side I'm
going to do the same thing, except here
I have two parts that don't depend on s,
so I can take those across the derivative
operator.
If I have minus ke to the r.
Sorry.
Minus ke to the minus rt minus t, times
the
partial derivative with respect to s of
phi of d minus.
And so, the next step here is both phi of
d plus.
D plus depends on all of the arguments
of the Black Scholes pricing formula
including s.
And so does d minus.
So, this is, I'm going to look at it
and say, that's the product of two
functions of s.
That's s, which is a very simple function
of s.
And phi of capitol d plus.
So I'll have to use the product rule on
the first term.
And so the product rule says if I have the
product of two
functions, I take the derivative of the
first function times the second function.
So that gives me just a capital phi of d
plus Plus the first function, s.
So this s comes from right here.
Times the derivative of the second
function.
So times the derivative with respect to s
of capital phi of d plus.
So now I have a pretty complicated formula
so what
I'm going to do is kind of attack pieces
of it
and then we'll be able to put all of those
together to get an expression for delta in
the end.
So the first thing I'm going to look at.
Are we have these partial derivatives with
respect to s of phi of, and
so I've used this plus, minus sign
here just because the functions are very
similar.
So quite often when
I take derivatives of that I'm going to
get the same answer.
if not, I'll put a plus, minus.
And where it's important, and then if I'm
just talking
about d plus, you'd think of this as a
plus.
If I'm talking about d minus, you think
about this as a minus.
So the chain rule says that if I want to
take the derivative of capital phi of d
plus with respect to s, that's going to be
equal to the derivative with respect to
its argument.
So with respect to this d plus or minus,
of fee of d plus or minus,
times the partial derivative of d with
respect to s.
And
now remember the definition of capital
fee.
As just we were integrating from minus
infinity up to
d of this function, lower case phi of x,
dx,
and so, I, I like to use the shorthand
notation little phi of x for this function
here.
Just it saves me writing some stuff.
Also, it's less likely that I'm going to
make a
mistake if I just have to write that down
rather than.
This whole expression.
And we saw in the, in the second week how
I could take the derivative of, of
something like that.
So it's just the derivative of an
indefinite integral.
And so, again, I'm just, I'm doing the
same calculation two times.
So it's either, I'm treating it as phi of
x.
Or I'm treating it as.
The actual expression for that function,
and
then when I evaluate these it turns out
that the derivative is either, is just
lower
case phi of either d plus or d minus.
And then it has this functional form on
the right hand side here.
So that gets me this part the derivative.
The partial derivative of capital phi with
respect to s.
So, what I have so far, I have this
expression for delta,
and then we just evaluated this thing here
and this thing here.
So I have this expression for the
derivative of capital fee,
and I still need this part here so the
derivative of,
the partial derivative of d plus or minus
with respect to s.
So
I have the first term here, is this guy.
And now to get the full derivative I still
need to evaluate the
derivative of phi plus, that kind of came
out by the chain rule.
So, I'll move one step ahead with our
formula.
So I'm going to substitute this result
here.
Oops.
This result in the second line, into the
first line, to update my expression for
delta.
But I still have this partial of d plus
with respect
to s, and partial of d minus with respect
to s.
So that's what I'm going to do in the next
slide.
So I need to compute the partial
derivatives
of d plus and d minus.
So remember from the, from the first set
of slides
I said, I gave definitions for what d plus
and
minus look like and so now I can write
them
as, just one, formula, using this plus
minus notation, so.
The only difference between d plus and d
minus was this plus
or minus sign here, so I've just replaced
it with this little placeholder.
And now I want to take the derivative of
this with respect to s.
And so I'm going to make a little
simplification here, so If I have a sum in
the numerator, I can write this as log of
s over k divided by
sigma times the square root of t minus t,
so that'll be one fraction, and then plus.
R minus q plus or minus sigma squared over
2 times
t minus t divided by sigma, square root of
t minus t.
So that's a second fraction, I'm just
adding those two fractions together.
And so the reason I'm going to want to do
that is because after, this
s here only appears in the, in the first
one of those fractions.
So what I'm going to end up with is.
I'll want to take the partial derivative
of the leftover part with respect to s but
there is no s's in this part so I can
treat that as a constant.
So this term in the right hand side that's
just
going to go away that will be equal to
zero.
And then I'm left with 1 over sigma times
the square root of t minus
t, times the partial derivative with
respect to s of the log of s over k.
So I'll do a little substitution here.
So I'll let u equal s over k.
Then I can rewrite this Here, as the
partial
derivative of, with respect to s, of log
of u.
And so by the chain rule, that's going to
be the derivative
of the log times the derivative of u with
respect to s.
So, the derivative of log of u is just 1
over u.
And then I have this partial u, partial s.
And I'm going to calculate that just by
taking the
partial derivative with respect to s of u
here.
So, that's going to be 1 over k.
So what I'm left with is 1 over sigma,
square root of t minus t.
K over s.
So, u was s over k.
If I take one over that, I get k over s.
And then times 1 over k, which is just
this partial derivative, partial
u, partial s, which I, which popped out
because of the chain rule.
So, we'll see that the k's are going to
cancel here.
And I end up with partial derivative with
respect to s of either d plus or d
minus is just 1 over s times sigma times
the square root of t minus t.
So in this case I was able to get both of
those calculations.
Because the, the plus or minus bit didn't
depend on
s, so that ended up being zero, so it
wouldn't
matter if I was taking the derivative of d
plus
or d minus, I'm going to get the same
expression here.
So, putting everything together, all I'm ,
all I need to do now is, I have this.
This is my most up to date expression for
delta, but it still
has this partial derivative of d plus and
d minus with respect to s.
But that's exactly what we calculated on
the previous slide here, so I just
have to put a 1 over S sigma squared to t
minus t into here.
Oops, I guess I had it on the next line.
And so, that's going to give the following
expression for delta.
So,
I'm going to get e to the minus q, t minus
t, times capital phi of d plus.
And I have the rest of this in light gray,
because
if you look in the text book, this is the
entire.
Oops, maybe I got it backwards.
Nope.
just the part in the dark black is the
entire expression for delta.
But we still have these other terms left
over.
And so in the next set of slides
I'm going to go through an argument about
why that's going to be equal to zero.