We should all be familiar with the Taylor Series (about $a$) of a function of a single variable $f(x)$: $\displaystyle f(x) = f(a) + \frac{(x-a)}{1!}f^{\prime}(a) + \frac{(x-a)^2}{2!}f^{\prime\prime}(0) + \frac{(x-a)^3}{3!}f^{\prime\prime\prime}(0) + \dots$

But did you know that this is also just Integration by Parts?

Recall that integration by parts (or IBP) says that \displaystyle \begin{aligned}\int_a^b u(y)v'(y)\textup{d}y &= \left[u(y)v(y)\right]_a^b - \int_a^b u'(y)v(y)\textup{d}y \\ &= u(b)v(b) - u(a)v(a) - \int_a^b u'(y)v(y)\textup{d}y\end{aligned}

Right, so with that in mind, let us generate the left hand side of the above Taylor series. We do this by a trivial application of the main corollary to the Fundamental Theorem of Calculus which states: \displaystyle \begin{aligned}\int_a^x f(y)\textup{d}y &= f(x) - f(a) \\ \Rightarrow f(x) &= f(a) + \int_a^x f'(y)\textup{d}y \end{aligned}

So we have our $f(x)$ given by the sum of $f(a)$ and this integral.

The integral can be evaluated by using IBP. Here’s how:

Let $u$ and $\textup{d}v$ be defined like so: \displaystyle\begin{aligned} u = f'(y) &\Rightarrow \textup{d}u = f''(y) \\ \textup{d}v = 1 &\Rightarrow v = t + c \end{aligned}

Now the trick is to set $c = -y$ so that $v = -(x - y)$.

Then, by IBPs we have \displaystyle\begin{aligned}\int_a^x f'(y)\textup{d}y &= f(a) + \left[-(x-y)f'(y)\right]_a^x - \int_a^x -(x - y)f''(y)\textup{d}y \\ &= f(a) -(x-x)f'(x) + (x-a)f'(a) + \int_a^x (x-y)f''(y)\textup{d}y \\ &= f(a) + \frac{(x-a)}{1!}f'(a) + \int_a^x (x-y)f''(y)\textup{d}y\end{aligned}

and we now have the first two terms of the Taylor series!

Continuing in this manner, except now we let $u$ and $\textup{d}v$ be defined like so: \displaystyle\begin{aligned} u = f''(y) &\Rightarrow \textup{d}u = f'''(y) \\ \textup{d}v = (x-y) &\Rightarrow v = \frac{-(x-y)^2}{2!} \end{aligned}

we get: $\displaystyle f(x) = f(a) + \frac{(x-a)}{1!}f'(a) + \frac{(x-a)^2}{2!}f''(a) + \int_a^x \frac{(x-y)^2}{2!}f'''(y)\textup{d}y$

and it’s hopefully now pretty clear that we could keep going to an accuracy of $n$ iterations of this procedure to obtain the general formula: $\displaystyle f(x) = \sum_{k=0}^n \frac{(x-a)^k}{k!}f^{(k)}(a) + \int_a^x \frac{(x-y)^n}{n!}f^{(n+1)}(y)\textup{d}y$

### Option Prices

So where do option prices come in to this?

Well, let us first go back to the Taylor series expansion of the function $f(x)$ to 2nd order accuracy, and center the expansion about the real number $a = 0$: $\displaystyle f(x) = f(0) + (x)f'(0) + \int_0^x (x-y)f''(y)\textup{d}y$

Now we need to do a little bit of geometry to motivate the next trick.

Observe precisely what this 2nd order integral is saying: Fig 1 – The shaded area is the integral in the 2nd order Taylor series expansion.

Now let us see what the same integral is over a slightly different function $\max(x-y)$: Fig 2 – The maximum function goes up to a point and then jumps down to zero for values of y beyond x. So actually, the value of this integral is exactly the same!

But the trick goes further!

Since we can keep adding $0$ to this integral value we do not change it. So we can actually extend the bounds of integration all the way up to infinity and still have the same integral value. Fig 3 – The upper bound of integral can easily be set to infinity because the maximum function (beyond x) contributes precisely zero to the integral value. Hence the area under this curve (between 0 and infinity) is exactly the same as the one from performing Integration by Parts.

This means we can write our Taylor series expansion as $\displaystyle f(x) = f(0) + (x)f'(0) + \int_0^{\infty} (x-y)^+f''(y)\textup{d}y$

This is called writing the series in plus notation.

Let us change notation on the dummy variable $y \rightarrow K$, and change the state variable $x \rightarrow S_T$, and now an option payoff becomes clearly visible! $\displaystyle f(S_T) = f(0) + S_T f'(0) + \int_0^{\infty} \underbrace{(S_T-K)^+}_{\textup{Call Payoff}}f''(K)\textup{d}K$

Note that we cannot jump straight away and say that this integral is the call option price because that expectation (and integral) must be computed with respect to $x$; clearly, this integral is being done with respect to strike $K$.

But look at what happens when we take the discounted expectation (under the risk-neutral measure) of both sides! In what follows, all expectations are conditional, and taken under the risk-neutral probability measure, i.e. where I will write $\mathbb{E}[\cdot]$ I actually mean: $\mathbb{E}_t^{\mathbb{Q}}[\cdot | S_t]$.

First, the left-hand side: $\displaystyle \textsf{DF}(t,T)\mathbb{E}\left[f(S_T)|S_t\right] =: V(t,T)$

This is just the net present value of an option $V(t,T)$ expiring at some future time $T$, with discount factor $\textsf{DF}(t,T)$.

Second, the right-hand side: $\displaystyle \textsf{DF}(t,T)\mathbb{E}[f(0)] + \textsf{DF}(t,T)\mathbb{E}[\underbrace{S_T}_{\textup{Stock}} f'(0)] + \textsf{DF}(t,T)\mathbb{E}\left[\int_0^{\infty} \underbrace{(S_T-K)^+}_{\textup{Call Payoff}} f''(K)\textup{d}K\right]$

where we have used the linearity of the expectation operator to split up the operations in to separate additions.

The expectation of the first term (which is constant) is just the constant $f(0)$.

For the second term, the discounted expected stock price under the risk-neutral expectation is just the current stock price, hence: $\textsf{DF}(t,T)\mathbb{E}[S_T f'(0)] = S_t f'(0)$, this also follows from the fundamental theorem of asset pricing, with the replicating portfolio of the stock being just long the stock.

For the final term we actually take the discount factor and the expectation inside the integral: \displaystyle\begin{aligned}\textsf{DF}(t,T)\mathbb{E}\left[\int_0^{\infty} (S_T-K)^+ f''(K)\textup{d}K\right] &= \int_0^{\infty}\left[\underbrace{\textsf{DF}(t,T)\mathbb{E}\left[ (S_T-K)^+\right]}_{\textup{Call Price}}\right]f''(K)\textup{d}K \\ &= \int_0^{\infty}C_K(t,T) f''(K)\textup{d}K \end{aligned}

where $C_K(t,T)$ is the future value at time $T$ of a call option struck at $K$.

Putting this all together we have that the value $V(t,T)$ at pricing time $t$ of an arbitrary derivative, maturing at $T$, on a stock $S_t$, with payoff functional $f(S_T)$, is given by: $\displaystyle V(t,T) = \textsf{DF}(t,T) f(0) + S_t f'(0) + \int_0^{\infty} C_K(t,T)f''(K)\textup{d}K$

What does this mean?

Well, it means that a replicating portfolio (attempting to replicate the value of this arbitrary derivative) will be made up of $f(0)$ amount of zero-coupon bonds, $f'(0)$ amount of the stock, and an (infinite) linear combination of European call options: $\displaystyle V(t,T) =\underbrace{f(0)\cdot\textsf{DF}(t,T)}_{\textup{Zero-Coupon Bonds}} + \underbrace{f'(0)\cdot S_t}_{\textup{Stock}} + \underbrace{\int_0^{\infty} C_K(t,T)f''(K)\textup{d}K}_{\textup{Linear Combination of Calls}}$

## Conclusion

We saw in a previous blog that the set of call prices $\{C_K(t, T),\,\forall K\}$ spans the space of all derivative prices. We saw that this is true because derivative prices are nothing but percentiles from probability distributions at terminal time $T$; and all probability distributions can be approximated from (suitably scaled by $\lambda$ spreads) market traded butterfy prices $\textup{BF}_{K,\lambda}(t,T)$; which is just the difference in price between two call spreads (which is a position consisting of two call options, so a total of 4 call options), giving: $\displaystyle \lambda \textup{BF}_{K,\lambda}(t,T) \approx \frac{\partial^2 C_K(t,T)}{\partial K^2}$

with equality as $\lambda\rightarrow\infty$ (which implies a thinner and thinner butterfly centered at $K$).

And we say that by the Breeden-Litzenberger result, the value of such a Butterfly is approximately equal to (a suitable scaled) terminal probability density function $f(x)$.

Then the Fundamental Theorem of Asset Pricing lets you go from a probability density function to a price of a derivative: $\displaystyle \textup{FTAP}\,:\, f(x) \longrightarrow V(t,T)$

In particular, with knowledge of all call prices $\{C_K(t,T),\,\forall K\}$ we can determine the value any derivative payoff.

Call option prices span the space of all possible derivative prices (with payoffs at maturity $T$).

So really, a replicating portfolio of an arbitrary derivative such as this, will always be some amount of zero-coupon bonds (where we extract the information required for discounting), some amount of the underlying asset (where we ignore its drift, but the difference between its naturally drifting process and the one which doesn’t dertemines the information required for the change of measure), and finally now, some amount of some (infinite) linear combination of elements (call prices) from the spanning set of all possible derivative prices.

## References

1. Prof. Stephen Blyth, MIT Open Course Ware, Lecture 18.S096 (2013) – https://www.youtube.com/watch?v=eG_aRPy1KVE&t=1935s&ab_channel=MITOpenCourseWare.