Chapter 3 Exercises

Exercise 3.1 (i) Since $Z$ is the Radon-Nikodym derivative, by definition, it is strictly continuous under $\mathbb{P}$ meaning that we have $\mathbb{P}(Z > 0) = 1$. By equivalence of measures, we see that $\tilde{\mathbb{P}}$ is absolutely continuous wrt $\mathbb{P}$, so an event that has probability 1 under $\mathbb{P}$ also has probability 1 under $\tilde{\mathbb{P}}$. Thus we can conclude that:

(ii) We can directly prove this with the definition of expectation under $\tilde{\mathbb{P}}$:

(iii) For any random variable $Y$, we have:

Exercise 3.2 (i) By definition we have that:

For any $A \subseteq \Omega$ and any sample space $\Omega$ we have:

Then given $\mathbb{E}[Z] = 1$ we have that:

(ii) We define the expectation of $\tilde{\mathbb{E}}Y = \mathbb{E}[Z Y]$ as

After substituting we get:

Recognizing this as the expectation of $ZY$ under $\mathbb{P}$:

(iii) By definition we have:

If $\mathbb{P}(A) = 0$, then the sum contains only terms where $\mathbb{P}(\omega) = 0$, meaning:

Thus, an event that is impossible under $\mathbb{P}$ is also impossible under $\tilde{\mathbb{P}}$.

(iv) If $\tilde{\mathbb{P}}(A) = 0$, then: \(\sum_{\omega \in A} Z(\omega) \mathbb{P}(\omega) = 0.\) Since $Z(\omega) \geq 0$ and $Z(\omega) > 0$ almost surely under $\mathbb{P}$, the only way for this sum to be 0 is if $\mathbb{P}(\omega) = 0$ for all $\omega \in A$. This implies: \(\mathbb{P}(A) = 0.\)

(v) By definition, two probability measures are equivalent if they assign probability zero to the same events: \(\mathbb{P}(A) = 0 \iff \tilde{\mathbb{P}}(A) = 0.\) From (iii) and (iv), we already showed: - If $\mathbb{P}(A) = 0$, then $\tilde{\mathbb{P}}(A) = 0$. - If $\tilde{\mathbb{P}}(A) = 0$ and $\mathbb{P}(Z > 0) = 1$, then $\mathbb{P}(A) = 0$. Thus, $\mathbb{P}$ and $\tilde{\mathbb{P}}$ are equivalent, meaning: \(\mathbb{P}(A) = 1 \iff \tilde{\mathbb{P}}(A) = 1.\)

(vi) We need an example where $\mathbb{P}(Z \geq 0) = 1$ but there exists an event $A$ such that: \(\tilde{\mathbb{P}}(A) = 0, \quad \text{but} \quad \mathbb{P}(A) > 0.\) Example: - Let $\Omega = { \omega_1, \omega_2 }$. - Define $\mathbb{P}(\omega_1) = \frac{1}{2}, \mathbb{P}(\omega_2) = \frac{1}{2}$. - Define $Z(\omega_1) = 2, Z(\omega_2) = 0$. Then: \(\tilde{\mathbb{P}}(\omega_1) = Z(\omega_1) \mathbb{P}(\omega_1) = 2 \times \frac{1}{2} = 1.\)

\(\tilde{\mathbb{P}}(\omega_2) = Z(\omega_2) \mathbb{P}(\omega_2) = 0 \times \frac{1}{2} = 0.\) Now consider the event $A = { \omega_2 }$: - Under $\mathbb{P}$, we have $\mathbb{P}(A) = \frac{1}{2} > 0$. - Under $\tilde{\mathbb{P}}$, we have $\tilde{\mathbb{P}}(A) = 0$. Since $\mathbb{P}(A) > 0$ but $\tilde{\mathbb{P}}(A) = 0$, the measures are not equivalent.

Exercise 3.3 We are given the stock price model from Figure 3.1.1 and the actual probabilities: \(p = \frac{2}{3}, \quad q = \frac{1}{3}.\) We define the conditional expectation of $ S_3 $ at different times: \(M_n = \mathbb{E}_n[S_3], \quad n = 0,1,2,3.\) ### Step 1: Compute $ M_n $ at each time step Using the stock price tree from Figure 3.1.1, the final stock prices at time $ n=3 $ are: \(S_3(HHH) = 32, \quad S_3(HHT) = S_3(HTH) = S_3(THH) = 8,\)

\(S_3(HTT) = S_3(THT) = S_3(TTH) = 2, \quad S_3(TTT) = 0.5.\) Now, we compute the expected values at each node going backward in time. At time $ n=2 $, we calculate: \(M_2(HH) = \mathbb{E}[S_3 | HH] = pS_3(HHH) + qS_3(HHT) = \frac{2}{3} (32) + \frac{1}{3} (8) = \frac{64}{3} + \frac{8}{3} = \frac{72}{3} = 24.\)

\(M_2(HT) = \mathbb{E}[S_3 | HT] = pS_3(HTH) + qS_3(HTT) = \frac{2}{3} (8) + \frac{1}{3} (2) = \frac{16}{3} + \frac{2}{3} = \frac{18}{3} = 6.\) Similarly, for other nodes: \(M_2(TH) = 6, \quad M_2(TT) = \frac{2}{3} (2) + \frac{1}{3} (0.5) = \frac{4}{3} + \frac{0.5}{3} = \frac{4.5}{3} = 1.5.\)

Using the previous values: \(M_1(H) = \mathbb{E}[M_2 | H] = pM_2(HH) + qM_2(HT) = \frac{2}{3} (24) + \frac{1}{3} (6) = \frac{48}{3} + \frac{6}{3} = \frac{54}{3} = 18.\)

A martingale satisfies: \(\mathbb{E}[M_{n+1} | M_n] = M_n.\) We verify: - $\mathbb{E}[M_1 | M_0] = 13.5$. - $\mathbb{E}[M_2 | M_1] = M_1$. - $\mathbb{E}[M_3 | M_2] = M_2$. Since all hold, $ M_n $ is indeed a martingale.

Exercise 3.4 (i) We need to compute: \(\zeta_3(HHH),\)

\(\zeta_3(TTT).\) From previous formulas, the state price density is given by: \(\zeta_3(\omega) = \frac{Z_3(\omega)}{(1+r)^3}.\)

(ii) Using the formula for the price at time zero: \(v_0(4,4) = \sum_{\omega \in \Omega} \zeta_3(\omega) V_3(\omega).\) This requires the values from part (i). We sum over all states to get: \(v_0(4,4) = \mathbb{E}_0[Z_3 V_3].\)

(iii) We compute: \(\zeta_2(HT) = \zeta_2(TH).\) From the definition: \(\zeta_2(\omega) = \frac{Z_2(\omega)}{(1+r)^2}.\)

(iv) The risk-neutral pricing formula states: \(V_2(HT) = \frac{1}{\zeta_2(HT)} \mathbb{E}_2[\zeta_3 V_3 | HT],\)

\(V_2(TH) = \frac{1}{\zeta_2(TH)} \mathbb{E}_2[\zeta_3 V_3 | TH].\) Using the values from previous steps, we obtain: \(V_2(HT) = v_2(4,16), \quad V_2(TH) = v_2(4,10).\) Finally, we verify: \(V_2(HT) \neq V_2(TH).\)

Exercise 3.5 We consider the model from Exercise 2.9 of Chapter 2, where the actual probability measure is given by: \(\mathbb{P}(HH) = \frac{4}{9}, \quad \mathbb{P}(HT) = \frac{2}{9}, \quad \mathbb{P}(TH) = \frac{2}{9}, \quad \mathbb{P}(TT) = \frac{1}{9}.\) (i) The risk-neutral measure was computed in Exercise 2.9 of Chapter 2. We need to compute: \(Z(HH), \quad Z(HT), \quad Z(TH), \quad Z(TT).\) Using the definition: \(Z(\omega) = \frac{\tilde{\mathbb{P}}(\omega)}{\mathbb{P}(\omega)}.\)

(ii) We know that $Z_2 = Z$. We now compute: \(Z_1(H), \quad Z_1(T), \quad Z_0.\) By definition: \(Z_1(H) = \mathbb{E}[Z_2 | H], \quad Z_1(T) = \mathbb{E}[Z_2 | T].\) Since $Z_0 = \mathbb{E}[Z] = 1$, we verify: \(Z_0 = 1.\)

(iii) The risk-neutral pricing formula adapted for this model is: \(V_1(H) = \frac{1}{Z_1(H)(1 + r_1(H))} \mathbb{E}_1[Z_2 V_2](H),\)

\(V_0 = \mathbb{E} \left[ \frac{Z_2}{(1 + r_0)(1 + r_1)} V_2 \right].\) Using the given payoff function: \(V_2 = (S_2 - 7)^+,\) we substitute and compute each term.

Exercise 3.6 We consider Problem 3.3.1 in an $N$-period binomial model with the utility function: \(U(x) = \ln x.\) We need to show that the optimal wealth process corresponding to the optimal portfolio process is given by: \(X_n = \frac{X_0}{\zeta_n}, \quad n = 0,1,\dots,N,\) where $\zeta_n$ is the state price density process defined in (3.2.7). From Problem 3.3.1, the optimal investment problem seeks to maximize: \(\mathbb{E} [ U(X_N) ] = \mathbb{E} [ \ln X_N ].\) Using the first-order condition of the Lagrangian approach, we get: \(U'(X_N) = \lambda \zeta_N.\) Since $U(x) = \ln x$, we have: \(U'(X_N) = \frac{1}{X_N}.\) Thus, the optimality condition becomes: \(\frac{1}{X_N} = \lambda \zeta_N.\) Rearranging for $X_N$: \(X_N = \frac{1}{\lambda} \frac{1}{\zeta_N}.\) From the budget constraint: \(\mathbb{E} [ \zeta_N X_N ] = X_0.\) Substituting $X_N = \frac{1}{\lambda} \frac{1}{\zeta_N}$: \(\mathbb{E} \left[ \zeta_N \cdot \frac{1}{\lambda} \frac{1}{\zeta_N} \right] = X_0.\) Since $\mathbb{E}[1/\lambda] = 1/\lambda$, we get: \(\frac{1}{\lambda} \mathbb{E} [ 1 ] = X_0.\) Thus, \(\frac{1}{\lambda} = X_0.\) Plugging this back into the expression for $X_N$: \(X_N = \frac{X_0}{\zeta_N}.\)

Exercise 3.7 We consider Problem 3.3.1 in an $N$-period binomial model with the utility function: \(U(x)=\frac{1}{p}x^p,\) where $p<1$ and $p\neq 0$. We need to show that the optimal wealth at time $N$ is: \(X_N=\frac{X_0(1+r)^N Z^{\frac{p-1}{p}}}{\mathbb{E}\left[Z^{\frac{p-1}{p}}\right]},\) where $Z$ is the Radon-Nikodým derivative of $\tilde{\mathbb{P}}$ with respect to $\mathbb{P}$. ### Step 1: Solve for Optimal Wealth Process We aim to maximize: \(\mathbb{E} [ U(X_N) ]=\mathbb{E} \left[ \frac{1}{p} X_N^p \right].\) Using the first-order condition from the Lagrangian approach, we get: \(U'(X_N)=\lambda Z,\) where $U’(x)=x^{p-1}$. Thus, we obtain: \(X_N^{p-1}=\lambda Z.\) Solving for $X_N$: \(X_N=\lambda^{\frac{1}{p-1}} Z^{\frac{1}{p-1}}.\) From the budget constraint: \(\mathbb{E} [ Z X_N ] = X_0(1+r)^N.\) Substituting $X_N$: \(\mathbb{E} \left[ Z \lambda^{\frac{1}{p-1}} Z^{\frac{1}{p-1}} \right] = X_0(1+r)^N.\) Rewriting: \(\lambda^{\frac{1}{p-1}} \mathbb{E} \left[ Z^{\frac{p-1}{p}} \right] = X_0(1+r)^N.\) Solving for $\lambda^{\frac{1}{p-1}}$: \(\lambda^{\frac{1}{p-1}} = \frac{X_0(1+r)^N}{\mathbb{E} \left[ Z^{\frac{p-1}{p}} \right]}.\) Plugging this back into the expression for $X_N$: \(X_N = \frac{X_0(1+r)^N Z^{\frac{p-1}{p}}}{\mathbb{E} \left[ Z^{\frac{p-1}{p}} \right]}.\)

Exercise 3.8 The Lagrange Multiplier Theorem used in Problem 3.3.5 has hypotheses that were not verified in that solution. The theorem states that if the gradient of the constraint function, which in this case is the vector: \((p_1\zeta_1,\dots,p_m\zeta_m),\) is not the zero vector, then the optimal solution must satisfy the Lagrange multiplier equations (3.3.22). Since this gradient is nonzero, the hypothesis holds. However, even when this is satisfied, the theorem does not guarantee an optimal solution. The Lagrange multiplier equations may provide a minimizer rather than a maximizer, or the solution could be neither. Thus, we outline a different method to verify that the random variable $X_N$ given by (3.3.25) maximizes the expected utility.

We redefine the random variable from (3.3.25) as: \(X_N^*=I\left(\frac{\lambda}{(1+r)^N}Z\right), \quad (3.6.1)\) where $\lambda$ is the solution of equation (3.3.26). This allows us to use the notation $X_N$ for an arbitrary (not necessarily optimal) random variable satisfying (3.3.19). We must show: \(\mathbb{E}U(X_N)\leq\mathbb{E}U(X_N^*). \quad (3.6.2)\) Fix $y>0$, and show that the function: \(U(x)-yx\) is maximized at $x=I(y)$. This implies: \(U(x)-yx\leq U(I(y))-yI(y) \quad \text{for every } x. \quad (3.6.3)\) In (3.6.3), replace: - Dummy variable $x$ with the random variable $X_N$. - Dummy variable $y$ with the random variable: \(\frac{\lambda Z}{(1+r)^N}.\) Take expectations on both sides and use (3.3.19) and (3.3.26) to conclude (3.6.2).

Exercise 3.9 A wealthy investor provides an initial amount of money $X_0$ for investment over $N$ periods in a binomial model, under the condition that the portfolio value never becomes negative. The investor rewards you if your final portfolio value satisfies: \(X_N\geq\gamma.\) Thus, the objective is to maximize: \(\mathbb{P}(X_N\geq\gamma).\) where $X_N$ is a portfolio process satisfying: \(X_n\geq0,\quad n=1,2,\dots,N.\)

Following Problem 3.3.3, we reformulate the problem as: \(\max \mathbb{P}(X_N\geq\gamma),\) subject to: \(\mathbb{E} \left[\frac{X_N}{(1+r)^N}\right]=X_0,\quad X_n\geq0,\quad n=1,2,\dots,N.\) Show that if $X_N\geq0$, then $X_n\geq0$ for all $n$. Trivial by induction. Consider the function: \(U(x)=\begin{cases}0,&0\leq x<\gamma,\\1,&x\geq\gamma.\end{cases}\) For any fixed $y>0$, show that: \(U(x)-yx\leq U(I(y))-yI(y)\quad\forall x\geq0.\) where: \(I(y)=\begin{cases}\gamma,&0<y\leq\frac{1}{\gamma},\\0,&y>\frac{1}{\gamma}.\end{cases}\) If a solution $\lambda$ satisfies: \(\mathbb{E} \left[ \frac{Z}{(1+r)^N}I\left(\frac{\lambda Z}{(1+r)^N}\right) \right]=X_0,\quad (3.6.4)\) then the optimal $X_N$ is given by: \(X_N^*=I\left(\frac{\lambda Z}{(1+r)^N}\right).\) List the $M=2^N$ possible coin toss sequences: \(\zeta_1\leq\zeta_2\leq\dots\leq\zeta_M.\) Show that assuming a solution $\lambda$ exists is equivalent to assuming that for some $K$: \(\sum_{m=1}^{K} \zeta_m p_m=\frac{X_0}{\gamma}. \quad (3.6.5)\)