So the last Greek that I want to go
through the calculation through is theta,
and so theta is the rate of change of
the value of the portfolio with respect to
time t.
And again, our portfolio that we're
considering
just contains a single call option, so the
price of our portfolio is going to be
the Black-Scholes price for a European
call option.
And we just start by taking he partial
derivative of the Black Scholes
price formula with respect to t and that's
what we're going to call theta.
And so here its going to be a little bit
more complicated.
Because now have this e to the minus qt
minus t.
That terms has a t in it.
E to the minus or rt minus t, that has
term in it, sorry has a t in it.
Phi of d plus is a function of d plus and
d plus depends on t.
And so does d minus.
So I have to use the product rule on each
term here.
So what I'll, I'll get is the derivative
of the first term.
So here is where the minus signs cancel
out, that
I confused myself with in the last set of
slides.
So here, it's minus.
I really should have done this with a
substitution.
I'm even confusing myself now.
So, what's going to happen, I have a minus
sign here, that's going to pop
out in front, and a minus sign here,
that's going to also pop out in front.
And I think maybe the easiest way to look
at this
is I have e to the minus q times capital
t.
That doesn't depend on little t.
And then, I have e to the minus q times
minus t.
So, when I take the derivative, it's e to
the qt that's kind of.
Playing the important part of that.
So I just get a q popping out in front.
So it's q times t.
And there really is no minus sign here.
It's just because of the actual way the
formula's
written down that it looks like there's
two minus signs.
But with respect to the variable I'm
taking
the derivative of, there's, it's e to the
qt.
So what I'm going to end up with is
derivative of the first function times the
second function.
Plus derivative of the second function.
So that's going to be the partial
derivative with
respect to t of phi of d plus.
Times the first function, that's just the
product rule.
And then this subtraction is coming from
here just because it's
the, the first term minus the second term
in the pricing formula.
And
then the same thing is going to happen.
This really is e to the rt times e to the
minus r capitol t.
So when I take the derivative, it's just
going to be one little r that pops out.
So I have derivative of the first term,
times the second term,
plus derivative of the second term, sorry,
second function, times the first function.
So it's just the product rule again, on
the second term.
Of the pricing formula.
And so now I'm going to, look at this and,
maybe we can squeeze
just a little bit more mileage out of our,
our, kind of intermediate result.
but what I want to do is, right now I
have d minuses in the square brackets, and
d pluses.
In, in the first line.
And that result ended up working out well,
when I had
an s in one term, and a k in the second
term.
So, what I'm going to do is just rewrite
these so that I have the, the two q's.
Now which ones are together, plus, plus,
minus, minus.
Phi to d t.
So when I have the two
functions that have the partial derivative
with respect to
t of the phi, those will go in one group.
And then the other two will go in the
second group.
So I end up with this expression here.
And
now I have this.
Oops.
So, now, it's this one that I can get rid
of.
So, I have a way to replace this with
something that depends on s.
So, actually something that's going to be
equal to this again.
So, let's see.
So, here I'm just working on the, the part
in the square brackets.
So, here I've used the chain rule so the
derivative, with respect to t of capital
phi of
d plus, that's become lower case phi of d
plus times partial d plus with respect to
t.
And, then, the same thing with the second
term the.
Derivative with respect to t of capital
phi has become
lowercase phi of d minus times the partial
of d
minus with respect to t.
And
now I can do the same trick that I had
done before.
I have now partial derivative.
So I can replace this guy here
with this.
That was that intermediate result that I
had before.
And so I can rewrite this bit in the
square brackets like this.
And now last time I had.
The derivative with respect to sigma, this
time it's
the derivative with respect to t, but I'm
still going
to have this d, d plus minus d minus,
which is going to simplify that expression
quite a bit.
So I'm going to look at this quantity in a
little bit closer detail.
So I can rewrite that again as the partial
derivative
with respect to t just a d plus minus d
minus.
And so in, in light gray here just, I tend
to confuse
myself if I try to take a derivative of
something with a square
root just Like when I have a, a rational
function, if I have
1 over sigma, I'd rather write that as
Sigma to the minus one.
Here I have square root of t minus t, I'd
rather
write that as the quantity t minus t to
the one half.
And so now I'm set up to use the power
rule to take the derivative here.
So what I'm going to end up with, I have
one half.
So that's where this two comes from.
Then I'll have the quantity, t minus t to
the negative one half.
That's going to be where the 1 over square
root of t minus t comes from.
And then I have this sigma left over in
front.
And that's where the sigma in the top
comes from.
And then by the chain rule, so the
derivative
that I'm taking here is with respect to t,
and
in the inside function, t is negative, so
I have
to multiply this by negative one by the
chain rule.
So if I'd made a u substitution, and said
u is equal to capital t
minus little t, then when I took the
derivative of u, I would get minus one.
And so this is the expression for theta
that I had before.
So, it's the second term that I'm going to
be able to simplify using my, my trick.
And I now have an expression for this
here, which is just
minus sigma divided by 2 times the square
root of t minus t.
And so that's going to finally give me my
expression for theta.
So this one, unfortunately, doesn't
simplify much further than this.
And so all I've done is just substitute
the
expression we had last time into the first
term here.
And then left the other two terms alone.
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