So moving on for the next lesson.
We'll talk about gamma, which is the
derivative of delta with respect
to s, or the second derivative of c with
respect to s.
So gamma is the rate of change of delta.
So I can write that as the partial
derivative with respect to s of delta.
And delta is the partial derivative of c
with respect to s.
So this is just the second derivative of c
with respect to s.
But again, the way I'm going to compute
it, since I've already computed this
derivative for delta.
I'm going to use that as a starting point
rather than starting from scratch.
So all we have to do, since we already
have an expression for
delta, is just take the derivative with
respect to s of that again.
But that turns out to be pretty difficult,
because in the last in
the last set of slides, we, we had this
three term expression for s.
But luckily, there's a shortcut.
So like I mentioned in the previous set,
if
we look at the second two terms in the
square
brackets here, I said that those were
going to be equal to zero.
But that's something that we have to show
mathematically before we can do it.
We did this.
I had one example of this before where I
had actually plugged in numbers for the
Black-Scholes pricing
formula, and we went through the, the work
of
showing that this part ends up being equal
to zero.
And now we'll do it in general so we never
have to
do it again.
So the textbook says that delta is just
equal to
this much simpler expression than what
we've come up with.
And finding an expression for gamma is
certainly
going to be a lot easier if I only
have to take the derivative of this,
rather
than taking the derivative of all three
terms.
So what I'd like to do is figure
out a way to show that this expression
here is equal to zero.
And so when I, when I write something like
this, I'm not sure
that it's equal to zero, I want to show
that it's equal to zero.
So you don't want to just write down equal
to zero
and then start plugging in things and
showing that it's true.
What I want to do is start with this
without saying it's equal to zero.
And just do algebraic
manipulations to it, until it finally is
equal to zero.
And then what we'll have is, a line of
work that starts
out from what I'm given, and ends with
what I want, basically zero.
And that'll be an argument that anytime I
see this.
Expression in a calculation, I can then
replace that with zero.
And that's going to save us a lot of work
in,
the next several lessons.
And so the strategy I'm going to use is I
have a, one term that has
a d plus in it and one term that has a d
minus in it.
But remember d minus could be written in
terms of
d plus just by adding a little bit to it.
And so what I'm going to do is try and
manipulate the
d minus into a d plus, and then there'll
be some stuff
left over.
And hopefully when I factor out the stuff
that's left over, it's going to.
Turn this second term into the first term,
and if both of those terms are
the same, then when I subtract one from
the other, I'm going to get zero.
So phi of d minus is equal to 1 over the
square root of 2 pi e.
So I'm using some different notation here
just because it,
it gets really small if I have everything
in the exponent.
So, exp as a function, so exp and then
something in parentheses that means
exactly the same thing as e to the power
of what's ever in the parentheses here.
So this is 1 over the square root of 2 pi
Square root of 2 pi times
e to the minus d minus squared over 2.
And that's just the,
the exact definition of phi of d minus.
And
now I'm going to replace d minus in that
expression,
with d plus minus.
Sigma times the square root of t minus t.
And that was just my expression for d
minus.
And now I have a, this quadratic, form
here.
So I'm going to expand that.
The, the first term, I'm going to end up
with a d plus squared over 2.
So it's pretty clear that that's going to,
that's going to get me a phi of d plus.
And then hopefully the stuff that's left
over is going to be useful to get this
result that I'm after, showing that the
second
term here is actually equal to the first.
So this is just what I had on the last bit
of the, of the previous slide.
So now what I'm going to do is just go
ahead and expand this quadratic
form here, and so I'll end up with d plus
squared minus 2d plus sigma
square root of t minus t plus sigma
squared times t minus t all over 2.
And now I'm going to regroup
the terms like this.
So basically, I'm just taking, ooh, got
the whole thing that time.
I'm just taking this d plus squared over 2
and putting that by itself.
Because this is now an expression, the 1
over the square root
of 2 pi E to the minus d plus squared over
two.
That's just and expression from phi of d
plus.
And then these two terms
are what's left over.
So the middle term that has
a two in the numerator and in the
denominator,
so those are going to cancel each other
out.
So I've, I've got from my, oh I can't
highlight anything.
I have my phi of d plus from this portion
of the previous line.
I have e to the d plus times sigma square
root of t minus t from this portion.
And then I have e, this hasn't changed
I've just written it a little bit
differently.
e to the minus sigma squared t minus t
divided by 2.
And now, just a quick reminder, of what d
plus is equal to so what I'm going to.
What
I have here is d plus times sigma square
root of
t minus t and so all these sigma square
root t
minus t is going to do is cancel out the
sigma
square root of t minus t in the
denominator of d plus.
So I, what I have is E to the numberator
of the definition of d plus here.
So, I'll do that in the next line.
So, all that's
happened is this d plus sigma square root
of t minus t has
just become the numerator of the
definition of d plus.
And then the bit on the right hand term is
still stayed
the same, just sigma squared, t minus t
over 2 and it's negative.
So, what's going to happen, I have e to
the log of s over k.
So the exponential and the logarithm, they
are.
Inverse functions of one another it's the,
the exponential undoes what the logarithm
did,
so this is just going to allow me to take
an s over k out.
And then when I have an exponential of a
sum, that's
equal to the, product, so this is the same
thing as e
to the log.
S over k times e to the r minus q plus
sigma squared over 2, t minus t.
So this bit here is going to stay in my
exponential function
but I'm going to factor out an s over k
and then also using
the same property So here I had a product
of exponentials.
So this is just going to be the sum of
this bit that I've
got highlighted and what's inside the
square brackets in the lower right here.
So that gives this s over k out of my
exponential
bit, and then leaves me with what's inside
the exponential bit here.
But now look what's happened, so this,
sigma squared t minus t
over two that I've been carrying down the
right side of my calculation.
I have a plus sigma squared over 2 times t
minus t minus sigma squared over 2 times t
minus t.
So all
of that's going to cancel out, and I'm
just going to have r minus q.
That quantity times t minus t left.
And so now I'm going to go back to
using the notation of writing this as
exponential.
So, this is e to the r times t minus t.
So that's where this bit came from.
And then this is e to the minus q.
Times t minus t, so that's
where this bit came from and now I have a,
an expression for phi of d minus
in terms of phi of d plus and this is the
expression that I am trying to show is
equal to zero again.
So I am going to substitute.
My expression here for phi of d minus into
the second term
here and I get this expression here.
And so the
first term has stayed the same but now I
have k times e to the minus r t minus t.
So that carried over from before, and now
I have what's
coming from my new expression, fee of d
plus, s divided
by k, e to the rt minus t, and e to the
minus qt minus t.
But here I have a k And
s over k, so that's going to be just s,
and I have
an s in the numerator, and an s in the
denominator here.
So all of the strike prices, and the, the
asset
prices are going to cancel each other out
in this term.
Then I have e to the minus rt minus t, and
e to the rt minus t.
So when I add those together I have
minus rt minus t plus rt minus t.
That's the same things as e to the zero
which is just one.
So, those two parts are going to cancel
each other out.
So, all I'm left with Is phi of d plus.
And e to the minus q t minus t.
Which is exactly what I had in the
numerator over here.
And now in the denominator, remember I've
canceled out this s here with this s here.
So I have the same denominator as well.
So when I do all of the canceling here,
I've made the second term into the first
term.
And I'm subtracting them from one another,
so I end up having zero.
So that means when it's time to actually
go ahead and calculate gamma.
Instead of having to take the derivative
with respect to s of this
great big formula, I can rewrite that as
just the derivative
with respect to s of e to the minus q t
minus t times capital phi of d plus.
And so when I'm taking the partial
derivative with respect
to S, e to the minus q t minus t.
There's no S in that, so I can treat that
as a constant.
So I can move that across the derivative
operator.
And so all I need to do is take the
derivative with respect to s of capital
phi of d plus.
And I'll go through this again, but we
actually did this already when we were
computing delta.
So, I'm taking the derivative of a
function of d plus
with respect to s so I have to use the
chain rule.
And that tells me that I can take the
derivative of capital phi of d plus.
And then I have to multiply that by the
derivative of d plus with respect to s.
And we've all ready calculated that as
well so.
partial derivative of either d plus or d
minus with respect to s.
It's just one over s times sigma times the
square root of t minus t
And then this we've already seen before
let's say this is the derivative of our
indefinite integral so this is just
going to to
be equal to lowercase phi of d plus.
So what I'm left with is that e to the
minus qt minus t that I had out here.
The 1 over s times sigma times the square
root of t minus t that came
from the chain rule so that's in the
denominator in the front here.
And, then, the lower case fee of d plus,
which was just a derivative that I got
here.
And so, we now have an expression for
gamma, which is the partial
derivative with respect to S of, sorry, of
delta with respect to S.
And this is exactly the same as the line
above it.
I've just replaced phi of d plus with the
actual definition of phi of d plus.
So, it's 1 over the square root of 2 pi.
E to the minus d plus squared over 2, so.
Looks like I missed a little bit because I
forgot the square
there, but this should be d plus squared
over 2.