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Multiperiod Market

Generic

Exercise

We consider a multi-period financial market with a bank account \(B\) and one stock \(S^1 = S\).
Which of the following processes \(\eta = (\eta_t)_{1 \leq t \leq T}\), where \(\eta_t : \Omega \to \mathbb{R}\), are predictable?

  1. \(\eta_t = 1_{\{S_t > S_{t-1}\}}\) for \(t = 1, \ldots, T\);
  2. \(\eta_1 = 1\) and \(\eta_t = 1_{\{S_{t-1} > S_{t-2}\}}\) for \(t = 2, \ldots, T\);
  3. \(\eta_t = 1_A 1_{\{t > t_0\}}\) for some \(A \in \mathcal{F}_{t_0}\) and some \(t_0 \in \{0, \ldots, T-1\}\);
  4. \(\eta_t = 1_{\{S_t > S_{S_0}\}}\);
  5. \(\eta_1 = 1\) and \(\eta_t = 2 \eta_{t-1} 1_{\{S_t < 1\}}\) for \(t \geq 2\).

Insider Problem

Let \(Y_1, \ldots, Y_T\) be independent identically distributed random variables such that \(E[Y_t] = 0\) for every \(t\) and not identically constant on some probability space \((\Omega, \mathcal{F}, P)\). We consider the filtration \(\mathcal{F}_0 = \{\emptyset, \Omega\}\) and \(\mathcal{F}_t = \sigma(Y_1, \ldots, Y_t)\) and the discounted price process:

\[ X_0 := 0, \quad X_t := X_0 + \sum_{s=1}^t Y_s. \]

We extend the filtration with the information provided by \(X_T\), that is:

\[ \tilde{\mathcal{F}}_t = \sigma(\mathcal{F}_t, X_T), \quad t = 0, \ldots, T \]

This is the information that an insider knows, for whatever reason, about the terminal value of the discounted price at time \(T\). We denote the non-insider filtration \(\mathbb{F}\) and the insider filtration \(\tilde{\mathbb{F}}\).

Show that:

  1. If \(Z_1\) and \(Z_2\) are identically distributed integrable random variables, then it holds

    \[ E[Z_1 |\sigma(Z_1+Z_2)] = \frac{Z_1 + Z_2}{2} \]

  2. \(X\) is a martingale under the filtration \(\mathbb{F}\). Show that \(X\) cannot be a martingale under the insider filtration \(\tilde{\mathbb{F}}\). However, the process:

    \[ \tilde{X}_0 = X_0, \quad \text{and} \quad \tilde{X}_t = X_t - \sum_{s=0}^{t-1} \frac{X_T - X_s}{T - s}, \quad t = 1, \ldots, T \]

    is a martingale under \(\tilde{\mathbb{F}}\).

  3. With the information about the terminal discounted value \(X_T\) of the stock, it is possible to realize arbitrage gains. Find a strategy \(\eta\), predictable with respect to \(\tilde{\mathbb{F}}\), such that among all other strategies \(\mu\), predictable with respect to \(\tilde{\mathbb{F}}\) with \(|\mu_t| \leq 1\), the expected gain \(E[G_T]\) is maximal, where:

    \[ G_T = \sum_{s=1}^T \eta_s \Delta X_s. \]

Exercise

In an arbitrage-free market with a bank account:

\[ B_t = (1 + r)^t, \quad r > -1, \, t = 0, \ldots, T \]

Let \(C_t^{call}\) and \(C_t^{put}\) be the price processes of a call and put on a self-financed portfolio \(V\) with strike \(K\), that is:

\[ C^{call} = (V_T - K)^+ \quad \text{and} \quad C^{put} = (K - V_T)^+. \]

Show the put-call parity:

\[ C_t^{call} - C_t^{put} = V_t - K(1 + r)^{t-T}, \quad t = 0, \ldots, T. \]

American Option

Exercise

Let \(H\) be the discounted price of an American option in a complete financial market, and denote by \(U\) the Snell envelope. Denote by \(\tilde{H} := H_T\) the corresponding European option whose price process is given by:

\[ V_t = E_{P^\ast}\left[ H_T \mid \mathcal{F}_t \right], \quad t = 0, \ldots, T \]

Show that:

  1. \(U_t \geq V_t\) for every \(t\).
  2. If \(V_t \geq H_t\) for all \(t\), then \(U_t = V_t\) for every \(t\).
  3. If \(H\) is a \(P^\ast\)-sub-martingale, then \(U = V\).
  4. Show that if \(H_t = f(X_t)\) for some convex function \(f\) for every \(t\), for instance, an American call option, then \(H\) is a \(P^\ast\)-sub-martingale.

Exercise (Difficult)

In a multi-period arbitrage-free and complete financial market with one stock \(S\), consider the following American put option:

\[ H_t^K = \frac{(K - S_t)^+}{(1 + r)^t}, \quad t = 0, 1, \ldots, T \]

where \(r > 0\) and \(K\) is a positive strike. We denote by \(\tau^K_{\min}\) the smallest optimal stopping time for the buyer, that is:

\[ \tau^K_{\min}(\omega) = \inf\{t : U_t^K(\omega) = H_t^K(\omega)\}, \]

where \(U^K\) is the corresponding Snell envelope of \(H^K\).

Show that:

  1. \(\tau^K_{\min} \geq \tau^{K^\prime}_{\min}\) whenever \(K \leq K^\prime\).
  2. For \(K\) large enough, show that ultimately \(\tau^K_{\min} = 0\).

Default Probability

Exercise: First part is easy, second part less so

Let \((\Omega, \mathcal{F}, P)\) be a probability space, \(Y_1, Y_2, \ldots\) a sequence of independent identically distributed random variables taking values \(\pm 1\) and such that:

\[ P[Y_t = 1] = p, \quad \text{and} \quad P[Y_t = -1] = 1 - p = q \]

for \(1 > p \geq 1/2\). Consider the filtration \(\mathcal{F}_0 = \{\emptyset, \Omega\}\) and \(\mathcal{F}_t = \sigma(Y_1, \ldots, Y_t)\), \(t \geq 1\). The random walk \(Z\) is given by:

\[ Z_0 = 0 \quad \text{and} \quad Z_t = \sum_{s=1}^t Y_s, \quad t \geq 1. \]

Finally, for two integers \(a\) and \(b\), we define the stopping time:

\[ \tau_a = \inf\{ t : Z_t = a \}, \quad \tau_b = \inf\{ t : Z_t = -b \}, \quad \text{and} \quad \tau = \tau_a \wedge \tau_b. \]

You can assume that for any value of \(a\) and \(b\), it holds \(\tau(\omega) < \infty\) for almost all \(\omega\).

  1. Since \(1 > p \geq 1/2\), show that \(Z\) is a sub-martingale and a martingale if and only if \(p = 1/2\).

  2. Suppose that \(p = 1/2\), and show that for every time horizon \(T\), it holds:

    \[ E\left[ Z_{T \wedge \tau} \right] = 0 \]

    Deduce that:

    \[ P[Z_\tau = a] = \frac{b}{a + b}. \]

  3. Always for \(p = 1/2\), show that the process \(Z^2_t - t\) is a martingale. Show that \(E[Z_\tau^2 - \tau] = 0\), and using the previous point, show that the expected time to reach \(a\) or \(-b\) for the first time is given by:

    \[ E\left[ \tau \right] = ab. \]

  4. For \(1 > p > 1/2\), show that the process:

    \[ M_0 = 1, \quad M_t = \left( \frac{q}{p} \right)^{Z_t}, \quad t = 1, 2, \ldots \]

    is a martingale.

  5. For \(1 > p > 1/2\), show that:

    \[ P\left[ Z_\tau = a \right] = \frac{(q/p)^b - 1}{(q/p)^{a + b} - 1}. \]

From now on, suppose that \(p > 1/2\) and for \(\lambda \in \mathbb{R}\), define:

\[ \phi(\lambda) = \ln(E[\exp(\lambda Y_t)]) = \ln(E[\exp(\lambda Y_1)]) = \ln(pe^{\lambda} + qe^{-\lambda}) \]

as well as the process \(L(\lambda)\) by:

\[ L_t(\lambda) = \exp\left( \lambda Z_t - t \phi(\lambda) \right), \quad t = 0, 1, 2, \ldots \]

  1. Show that:

    \[ \phi^\prime(0) = E\left[ Y_1 \right] \]

    and (with differentiation done with respect to \(\lambda\)!)

    \[ L^\prime_t := \frac{d L_t}{d\lambda}(0) = Z_t - t \phi^\prime(0). \]

  2. Show that \(L(\lambda)\) is a martingale. Sketch the reason why the process \(L^\prime\) is then also a martingale.

  3. From the two previous points, show that:

    \[ E\left[ Z_{t \wedge \tau} - t \wedge \tau \phi^\prime(0) \right] = 0, \quad \text{for every time } t. \]

  4. From equation (4), deduce that the average time before reaching either \(a\) or \(-b\) is given by:

    \[ E[\tau] = \frac{a + b}{p - q} \frac{(q/p)^b - 1}{(q/p)^{a + b} - 1} - \frac{b}{p - q}. \]