5 Fundamental Theorem of Asset pricing

Change of measure, Radon-Nikodym Derivative. Martingale Representation Theorem. Fundamental Theorem of Asset Pricing. Option pricing using the FTAP.

The method described above allows us to price derivatives in a multi-period Binomial model, but it is a bit time-consuming, and if we let \(N\) get big, it can be lengthy to compute prices. Can we find a quicker way?

The answer lies in the expression: \[\frac{1}{1+r}\left[ c_T \frac{u-(1+r)}{u-d} + c_H \frac{(1+r)-d}{u-d} \right].\] which we can rewrite as: \[\frac{1}{1+r}\left[ c_T \left(1-\frac{(1+r)-d}{u-d}\right) + c_H \frac{(1+r)-d}{u-d} \right].\] or: \[\frac{1}{1+r}\left[ c_T \left(1-q\right) + c_H q \right].\] where \(q := \frac{(1+r)-d}{u-d} \in (0,1)\) (by Theorem29). This looks like the average payoff, where we do not use the real probability \(p\) of a move up, but a new probability \(q\). Observe that the actual value of \(p\) plays no role in computing the price of the derivative. We will call \(q\) the risk-neutral probability.

So (as suggested in the first lecture), it may make sense to think about probabilistic models for the prices which are determined not by the real world, but by the model for the underlying asset.

Example 35. Recall Example27, we had \(u=\frac{3}{2}, d = \frac{4}{5}, r = \frac{1}{5}\), so \[q = \frac{(1+r)-d}{u-d} = \frac{\frac{6}{5}-\frac{4}{5}}{\frac{3}{2} - \frac{4}{5}} = \frac{4}{7},\] and we can write \[\frac{1}{1+r}\widetilde{\mathbb{E}} \left[ C_1 \right] = \frac{1}{1+r}\left[ \left(1-q\right) C_1(T) + q C_1(H) \right] = \frac{1}{\frac{6}{5}} \left[\frac{3}{7} \times 0 + \frac{4}{7} \times 4\right] = \frac{5}{6} \times \frac{16}{7} = \frac{40}{21}\]

as expected.

From now on, we will use the notation \(\widetilde{\mathbb{E}} \left[ X \right]\) and \(\widetilde{\mathbb{P}}(A)\) to denote expectation and probability taken using the risk-neutral probabilities.

We can in fact make this a bit more rigorous on the coin-flipping probability space.

Definition 36. Let \(H_n(\omega)\) be the number of heads we flip on the first \(n\) coin flips on the outcome \(\omega \in \Omega\). Then we can define \[Z_n(\omega) = \left( \frac{q}{p}\right)^{H_n(\omega)} \left(\frac{1-q}{1-p}\right)^{n-H_n(\omega)}.\] We say that \(\widetilde{\mathbb{E}} \left[ \cdot \right]\) is the change of measure of \(\mathbb{E}\left[ \cdot \right]\) induced by the Radon-Nikodym derivative⁷ \(Z_N\) by: \[\widetilde{\mathbb{E}} \left[ Y \right] = \mathbb{E}\left[ Z_NY \right]\] for any random variable.

Observe that we can express the probability of a given sequence of heads and tails in terms of the number of heads, so if \(\overline{\omega} \in \Omega\) is a sequence of \(N\) heads and tails, with \(k = H_N(\overline{\omega})\) heads, then \[\mathbb{P}(\{\overline{\omega}\}) = p^k (1-p)^{N-k} = p^{H_N(\overline{\omega})} (1-p)^{N-H_N(\overline{\omega})}.\]

Observe also that we can deduce statements about probability from statements about expectation using: \[\widetilde{\mathbb{P}}(A) = \widetilde{\mathbb{E}} \left[ \boldsymbol{1}_A \right] = \mathbb{E}\left[ Z_N \boldsymbol{1}_A \right].\] In particular, the probability of the path \(\overline{\omega} = \omega_1 \omega_2 \omega_3 \cdots \omega_N\), where \(\omega_i \in \{H,T\}\) is given by: \[\begin{align} \widetilde{\mathbb{P}}(\{\overline{\omega}\}) & = & \widetilde{\mathbb{E}} \left[ \boldsymbol{1}_{\{\overline{\omega}\}} \right] = \mathbb{E}\left[ Z_N \boldsymbol{1}_{\{\overline{\omega}\}} \right] \\ & = & \mathbb{E}\left[ \left( \frac{q}{p}\right)^{H_N(\omega)} \left(\frac{1-q}{1-p}\right)^{N-H_N(\omega)} \boldsymbol{1}_{\overline{\omega}}(\omega) \right] \\ & = & \left( \frac{q}{p}\right)^{H_N(\overline{\omega})} \left(\frac{1-q}{1-p}\right)^{N-H_N(\overline{\omega})} \mathbb{P}(\{\overline{\omega}\})\\ & = & \left( \frac{q}{p}\right)^{H_N(\overline{\omega})} \left(\frac{1-q}{1-p}\right)^{N-H_N(\overline{\omega})} p^{H_N(\overline{\omega})} (1-p)^{N-H_N(\overline{\omega})} \\ & =& q^{H_N(\overline{\omega})} (1-q)^{N-H_N(\overline{\omega})}\end{align}\]

Since the probability of any event on a coin-flipping probability space can be built up from the probabilities of individual outcomes, it follows from this computation that \(\widetilde{\mathbb{P}}(\cdot)\) really does correspond to a Binomial model with up-probability \(q\) instead of \(p\).

Observe also that for all \(\omega \in \Omega\), \(0 < Z_n(\omega) < \infty\), so for \(A \subseteq \Omega\): \[\begin{align} \mathbb{P}(A) >0 & \iff \mathbb{E}\left[ \boldsymbol{1}_A \right] >0 \iff \mathbb{E}\left[ Z_N \cdot Z_N^{-1} \boldsymbol{1}_A \right] >0 \iff \widetilde{\mathbb{E}} \left[ Z_N^{-1} \boldsymbol{1}_A \right] >0 \\ & \iff \widetilde{\mathbb{P}}(A) >0. \tag{5.1}\end{align}\]

Why is this probability space important? Let’s consider the discounted asset price — that is, the price of the asset discounted into today’s money by dividing by the value of 1 invested in the bank: \[S_n^* = \frac{S_n}{(1+r)^n}.\] This has some nice consequences for the value of a self-financing portfolio.

Lemma 37. Let \(V_n\) be the value of a self-financing portfolio in the assets \(S_n^1, S_n^2, \dots, S_n^K\) and the bank account. If we write \[V_n^* = \frac{V_n}{(1+r)^n},\] for the discounted portfolio value, and \[S_n^{i,*} = \frac{S_n^i}{(1+r)^{n}}\] then \[\begin{align} \tag{5.2} V_{n+1}^* = V_n^* + \sum_{i=1}^K \phi_n^i(S_{n+1}^{i,*}- S_{n}^{i,*}).\end{align}\]

In particular, we only need to specify the portfolio of risky assets to compute \(V_{n+1}^*\) given \(V_n^*\) — the bank account ‘takes care of itself’.

Proof. We have \[\begin{align} V_{n+1} & = & \phi_n^0(1+r) + \sum_{i=1}^K \phi_n^i S_{n+1}^i \\ & = & (1+r) \left[ \phi_n^0 + \sum_{i=1}^K \phi_n^i S_{n}^i\right] + \sum_{i=1}^K \phi_n^i (S_{n+1}^i - (1+r) S_{n}^i) \\ & = & (1+r) V_n + \sum_{i=1}^K \phi_n^i ((1+r)^{n+1}S_{n+1}^{i,*} - (1+r)\cdot (1+r)^n S_{n}^{i,*}),\end{align}\] and so dividing through by \((1+r)^{n+1}\) we get (5.2) .

The discounted asset prices also have another nice property:

Theorem 38. Under the probability measure \(\widetilde{\mathbb{P}}\), the discounted asset price \(S_n^*\) is a martingale.

Because of this property, we will often call \(\widetilde{\mathbb{P}}\) the risk-neutral probability measure.

Solution: Need to show:

1. \(\tilde\mathbb{E}|S_n^*| < \infty\) for all \(n\): Note \(0\le S_n^*\le S_0u^n\), so \(\tilde\mathbb{E}|S_n^*| \le S_0u^n< \infty\)

2. The process \(S_n^*\) is adapted: \(S_n^*\) depends only on the first \(n\) price moves, so it is adapted.

3. \(\tilde\mathbb{E}_n[S_{n+1}^*] = S_n^*\)

   Write \(S_{n+1}^*=\frac{(1+r)^n}{(1+r)^{n+1}}S_n^*Y_{n+1}\) where \[Y_{n+1}= \begin{cases} u&,\ S_{n+1}=uS_n\\ d&,\ S_{n+1}=dS_n \end{cases}\] Then \[\begin{align} \tilde\mathbb{E}_n[S_{n+1}^*]= &\tilde\mathbb{E}_n[(1+r)^{-1}S_n^*Y_{n+1}]=(1+r)^{-1}S_n^*\tilde\mathbb{E}_n[Y_{n+1}]=(1+r)^{-1}S_n^*\tilde\mathbb{E}[Y_{n+1}]\end{align}\] Now \[\begin{align} \tilde\mathbb{E}[Y_{n+1}]=&qu+(1-q)d=u\frac{(1+r)-d}{u-d}+d\frac{u-(1+r)}{u-d}=1+r\end{align}\] Finally \[\begin{align} \tilde\mathbb{E}_n[S_{n+1}^*]=&(1+r)^{-1}S_n^*\tilde\mathbb{E}[Y_{n+1}]=S_n^*\end{align}\]

\(\square\)

We can use both these results to see the following:

Lemma 39. Suppose that the discounted asset prices are martingales. Then the discounted value of any bounded self-financing portfolio \(\boldsymbol{\phi}\) is a martingale.

Proof. If \(S_n^{i,*}\) is a martingale for each \(i\), then so too is \[M_n = \sum_{k=1}^n \phi_{k-1}^i (S_{k}^{i,*} - S_{k-1}^{i,*})\] by Theorem26. The sum of martingales is a martingale (check the definition!), so we also see that \[V_n^* = V_0^* + \sum_{k=1}^n \sum_{i=1}^K \phi_{k-1}^i (S_{k}^{i,*} - S_{k-1}^{i,*})\] is a martingale.

These results say that self-financing portfolios (discounted) are martingales — the next result is a sort of converse. As we shall see, it says essentially that martingales (under \(\widetilde{\mathbb{P}}\)) are self-financing portfolios!

Theorem 40 (Martingale Representation Theorem). Let \(M_n\) be a martingale under \(\widetilde{\mathbb{P}}\). Then there exists a process \(\phi_n\) which depends only on \(S_1, \dots, S_n\) such that \[M_n = \widetilde{\mathbb{E}} \left[ M_N \right] + \sum_{k=1}^{n} \phi_{k-1} (S_k^*-S_{k-1}^*).\]

Solution: Notation: \(\omega=\omega_1\dots\omega_{n-1}\) (history), at time \(n\): \(\omega H\) and \(\omega T\) possible given \(\omega\).

Martingale: \(M_{n-1}(\omega)=qM_n(\omega H)+(1-q)M_n(\omega T)\) and \(\widetilde{\mathbb{E}} \left[ M_N \right]=M_0\)

Now suppose \(M_{n-1} = \widetilde{\mathbb{E}} \left[ M_N \right] + \sum_{k=1}^{n-1} \phi_{k-1} (S_k^*-S_{k-1}^*)\) holds, need to show it is also true for \(M_n\), i.e., find \(\phi_{n-1}\) such that \(M_n=M_{n-1}+\phi_{n-1}(S_n^*-S_{n-1}^*)\).

Note: \(S_n^*(\omega H)-S_{n-1}^*(\omega)=S_{n-1}^*(\omega)\left(\frac{u}{1+r}-1\right)\) and \(S_n^*(\omega T)-S_{n-1}^*(\omega)=S_{n-1}^*(\omega)\left(\frac{d}{1+r}-1\right)\).

Solve \(M_n=M_{n-1}+\phi_{n-1}(S_n^*-S_{n-1}^*)\) for \(H\) case: \(\phi_{n-1}(\omega)=\frac{M_n(\omega H)-M_{n-1}(\omega)}{S_{n-1}^*(\omega)\left(\frac{u}{1+r}-1\right)}\).

Need to show: \(T\) case holds also. We have: \[\begin{align} M_n(\omega T) =&M_{n-1}(\omega)+\phi_{n-1}(\omega)(S_n^*(\omega T)-S_{n-1}^*(\omega))\\ =&M_{n-1}(\omega)+\frac{M_n(\omega H)-M_{n-1}(\omega)}{S_{n-1}^*(\omega)\frac{u-(1+r)}{1+r}}S_{n-1}^*(\omega)\frac{d-(1+r)}{1+r}\\ =&M_{n-1}(\omega)+\frac{M_n(\omega H)-M_{n-1}(\omega)}{u-(1+r)}(d-(1+r))\\ \iff M_{n-1}(\omega) =&M_n(\omega T)\frac{u-(1+r)}{u-d}+M_n(\omega H)\frac{d-(1+r)}{u-d}\\ =&qM_n(\omega H)+(1-q)M_n(\omega T)\end{align}\] which is true since \(M_n\) is a martingale. \(\square\)

Theorem 41 (Fundamental Theorem of Asset Pricing). Consider a derivative contract on the \(N\)-period Binomial model with payoff \(C_N\) at time \(N\). Then there is no arbitrage if and only if the price of the contract at time \(n\) is: \[\begin{align} \tag{5.3} C_n = \frac{1}{(1+r)^{N-n}}\widetilde{\mathbb{E}}_{n} \left[ C_N \right], \quad \text{ for } n = 0, 1, \dots, N.\end{align}\] In particular, \(C_0 = \frac{1}{(1+r)^N} \widetilde{\mathbb{E}} \left[ C_N \right]\).

Solution: Step 1: show no arbitrage if \(C_n = \frac{1}{(1+r)^{N-n}}\widetilde{\mathbb{E}}_n[C_N]\).
Step 2: show arbitrage exists if \(C_n \ne \frac{1}{(1+r)^{N-n}}\widetilde{\mathbb{E}}_n[C_N]\).

Step 1.: no arbitrage if \(C_n = \frac{1}{(1+r)^{N-n}}\widetilde{\mathbb{E}}_n[C_N]\):

By Exercise 2.2(b): \(C_n^*:=\widetilde{\mathbb{E}}_n[(1+r)^{-N}C_N]\) is a martingale under \(\widetilde{\mathbb{P}}\). By Lemma 4.5: the discounted value of any self-financing portfolio trading in the asset and the derivative is also a martingale, so \(0=V_0^*=(1+r)^{-N}\widetilde{\mathbb{E}} \left[ V_N \right]\).

Now suppose an arbitrage exists, i.e., \(\mathbb{P}(V_N\ge 0)=1\) and \(\mathbb{P}(V_N>0)>0\), then \(\widetilde{\mathbb{P}}(V_N\ge 0)=1\) and \(\widetilde{\mathbb{P}}(V_N>0)>0\), i.e., \(\widetilde{\mathbb{E}} \left[ V_N \right]>0\), giving a contradiction.

Step 2.: arbitrage if \(C_n \ne \frac{1}{(1+r)^{N-n}}\widetilde{\mathbb{E}}_n[C_N]\).

• consider derivative as second asset \(S^2\) and assume there exists a time \(n\) with \(S_n^2\ne C_n\)

• consider case \(S_n^2>C_n\) only (\(S_n^2<C_n\) analogous)

• \(M_k:=\widetilde{\mathbb{E}}_k[(1+r)^{-N}C_N]\) is a \(\widetilde{\mathbb{P}}\)-martingale

• Theorem40: can find \(\phi_k^1\) such that \(M_N=M_n+\sum_{k=n+1}^N\phi_{k-1}^1(S_k^{1,*}-S_{k-1}^{1,*})\)

• Strategy to construct arbitrage:

  • do nothing before \(n\)

  • hold \(\phi_k^1\) units of \(S_k^1\) at times \(k=n,\dots, N\), sell derivative at time \(n\) for \(S_n^2\), and invest/borrow in bank account (\(V_n^*=0\))

  • at time \(N\): \(V_N=(1+r)^{N-n}(S_n^2-C_n)>0\ \implies\ \)arbitrage

Set \(V_k^*=V_k(1+r)^{-k}\), so \(V_{k+1}^*=V_k^*+\phi_k^1(S_{k+1}^{1,*}-S_{k}^{1,*})+(-1)(S_{k+1}^{2,*}-S_{k}^{2,*})\), \(V_n^*=0\) \[\begin{align} V_N^*=&V_n^*+(V_{n+1}^*-V_n^*)+\dots+(V_{N}^*-V_{N-1}^*)\\ =&\sum_{k=n+1}^N\phi_{k-1}^1(S_{k}^{1,*}-S_{k-1}^{1,*})+(-1)(S_{k}^{2,*}-S_{k-1}^{2,*})\\ =&M_N-M_n-(S_N^{2,*}-S_n^{2,*})\end{align}\] where \(S^2\) is derivative, i.e., \(S_N^{2,*}=(1+r)^{-N}C_N=M_N\) and \(S_n^{2,*}=(1+r)^{-n}S_n^2\), so \[\begin{align} V_N^*=&(1+r)^{-n}S_n^2-\widetilde{\mathbb{E}}_n[(1+r)^{-N}C_N]=(1+r)^{-n}(S_n^2-C_n)>0.\end{align}\] \(\square\)

This allows us to make some quick observations about how to price options. For example, suppose that we have a derivative whose payoff depends only on the value of the asset at time \(N\), say \(C_N(\omega) = f(S_N(\omega))\) for some function \(f\). We know (under \(\widetilde{\mathbb{P}}\)) that the probability of a single ‘up’ move is \(q\), and that each step is independent of the previous steps, so \[\begin{align} \tag{5.4} \widetilde{\mathbb{P}}(S_N = S_0 u^k d^{N-k}) = \binom{N}{k} q^k (1-q)^{N-k}, \quad \text{ for } k = 0, \dots, N.\end{align}\]

Therefore \[\begin{align} \widetilde{\mathbb{E}} \left[ f(S_N) \right] & = & \sum_{k=0}^N f(S_0 u^k d^{N-k}) \widetilde{\mathbb{P}}(S_N = S_0 u^k d^{N-k})\\ & = & \sum_{k=0}^N f(S_0 u^k d^{N-k}) \binom{N}{k} q^k (1-q)^{N-k}.\end{align}\] If follows from Theorem41 that the arbitrage-free price of a derivative with path-independent payoff \(C_N = f(S_N)\) is: \[\begin{align} \tag{5.5} \frac{\widetilde{\mathbb{E}} \left[ C_N \right]}{(1+r)^N} = \frac{1}{(1+r)^N}\sum_{k=0}^N f(S_0 u^k d^{N-k}) \binom{N}{k} q^k (1-q)^{N-k}\end{align}\]

More generally, for path-dependent options, we can quickly compute the expected payoff of an option:

Example 42. An Asian option is a derivative which pays the holder \[(A_N-K)_+\] where \(A_N\) is the average of the asset price between time \(0\) and time \(N\): \[A_N = \frac{1}{N+1} \sum_{n=0}^{N} S_n.\] Suppose \(N=3, u = 2, d = 1/2, r = 1/4, S_0 = 1, K = 1\). What is the price of the Asian option?

Solution:

Path	\(A_N\)	\((A_N-K)_+\)	\(\widetilde{\mathbb{P}}(\)Path\()\)
\(HHH\)	\(15/4\)	\(11/4\)	\(1/8\)
\(HHT\)	\(9/4\)	\(5/4\)	\(1/8\)
\(HTH\)	\(6/4\)	\(1/2\)	\(1/8\)
\(HTT\)	\(9/8\)	\(1/8\)	\(1/8\)
\(THH\)	\(9/8\)	\(1/8\)	\(1/8\)
\(THT\)	\(<1\)	\(0\)	\(1/8\)
\(TTH\)	\(<1\)	\(0\)	\(1/8\)
\(TTT\)	\(<1\)	\(0\)	\(1/8\)

• \(\widetilde{\mathbb{E}} \left[ [(A_N-K)_+] \right]=\frac 18\left(\frac{11}{4}+\frac{5}{4}+\frac{1}{2}+\frac{1}{8}+\frac{1}{8}\right)=\frac{19}{32}\)

• FTAP: Price = \((1+r)^{-3}\widetilde{\mathbb{E}} \left[ (A_N-K)_+ \right]=\left(\frac 54\right)^{-3}\frac{19}{32}=\frac{38}{125}\)

\(\square\)

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7. This is not a financial derivative, nor really a normal ‘mathematical derivative’, but really a quotient…↩︎