Incomplete Markets (Utility over Terminal Wealth)

Abstract

This chapter studies the investor’s portfolio optimization problem in an incomplete market. The solution in this chapter parallels the solution for the complete market setting in the portfolio optimization Chap. 1.

Appendix

Lemma 11.3 (Existence of a Saddle Point)

Assume that

• (1) a solution $\hat {X}_{T}$ exists to $v(x)=\underset {X_{T}\in \mathcal {C}^{e}(x)}{\sup }E\left [U(X_{T})\right ]$ and

• (2) a solution $\hat {Y}_{T}$ exists to $\tilde {v}(y)=\underset {Y_{T}\in D_{s}}{\inf }E[\tilde {U}(yY_{T})]$.

Then, v and $\tilde {v}$ are in conjugate duality, i.e.

$$\displaystyle \begin{aligned} v(x)=\underset{y>0}{\inf}\left(\tilde{v}(y)+xy\right),\:\forall x>0{} \end{aligned} $$

(11.30)

$$\displaystyle \begin{aligned} \tilde{v}(y)=\underset{x>0}{\sup}\left(v(x)-xy\right),\:\forall y>0. \end{aligned} $$

(11.31)

In addition,

• (i) $\tilde {v}$ is strictly convex, decreasing, and differentiable on (0, ∞),

• (ii) v is strictly concave, increasing, and differentiable on (0, ∞),

• (iii) defining $\hat {y}$ to be where the infimum is attained in expression (11.30),

  $$\displaystyle \begin{aligned} v(x)=\tilde{v}(\hat{y})+x\hat{y}. \end{aligned}$$

  And, $(\hat {X}_{T},\hat {Y}_{T},\hat {y})$ is a saddle point of $\mathcal {L}(\hat {X}_{T},\hat {Y}_{T},\hat {y})$.

Proof (Part 1)

If a solution $\hat {X}_{T}$ exists to $v(x)=\underset {X_{T}\in \mathcal {C}^{e}(x)}{\sup }E\left [U(X_{T})\right ]$, then by the free disposal Lemma 11.1, it has the identical solution and value function as $v(x)=\underset {X_{T}\in \mathcal {C}(x)}{\sup }E\left [U(X_{T})\right ]$.

Given v(x) define $v^{*}(y)=\underset {x>0}{\sup }\left (v(x)-xy\right ),\:\forall y>0$.

First, because v(x) < ∞ for some x > 0, v is proper. v is increasing and strictly concave because $v(x)=\underset {X_{T}\in \mathcal {C}(x)}{\sup }E\left [U(X_{T})\right ]$ and U is increasing and strictly concave. This implies v(x) < ∞ for all x > 0 (see Pham [149, p. 181]). v(x) is strictly concave on (0, ∞), hence continuous on (0, ∞), and therefore upper semi-continuous.

By Pham [149, Theorem B.2.3, p. 219], v and v ^∗ are in conjugate duality, where $v(x)=\underset {y>0}{\inf }\left (v^{*}(y)+xy\right )$.

By Pham [149, Proposition B.2.4, p. 219], we get that v ^∗(y) is differentiable on $\text{int}(\text{dom}(\tilde {v}))$.

By Pham [149, Proposition B.3.5, p. 219], since v ^∗ is strictly convex, we get v is differential on (0, ∞).

To complete the proof, we need to show that $v^{*}(y)=\tilde {v}(y)$.

(Step 1) By the definition of $\tilde {U}$ we have $U(x)\leq \tilde {U}(y)+xy$ for all x > 0 and y > 0.

Thus, $U(X_{T})\leq \tilde {U}(yY_{T})+X_{T}yY_{T}$ for $X_{T}\in \mathcal {C}(x)$. Taking expectations yields

$$\displaystyle \begin{aligned}E\left[U(X_{T})\right]\leq E[\tilde{U}(yY_{T})]+yE\left[X_{T}Y_{T}\right]\leq E[\tilde{U}(yY_{T})]+yx.\end{aligned}$$

The last inequality uses $X_{T}\in \mathcal {C}(x)$.

Taking the supremum on the left side, the infimum on the right side, and using the definition of $\tilde {v}(y)=\underset {Y_{T}\in D_{s}}{\inf }E[\tilde {U}(yY_{T})]$, we get:

$$\displaystyle \begin{aligned}v(x)\leq\tilde{v}(y)+xy\; \text{for all}\; y>0,\; \text{or}\; v(x)\leq\underset{y>0}{\inf}\left(\tilde{v}(y)+xy\right).\end{aligned}$$

(Step 2) We have $\tilde {U}(y)=U(I(y))-yI(y)$ for all y > 0.

Hence, $\tilde {U}(y\hat {Y}_{T})=U(I(y\hat {Y}_{T}))-yY_{T}I(y\hat {Y}_{T})$ where $\hat {Y}_{T}$ attains $\tilde {v}(y)$.

Take expectations to get

$$\displaystyle \begin{aligned}\tilde{v}(y)=E\left[U(I(y\hat{Y}_{T}))\right]-yE\left[\hat{Y}_{T}I(y\hat{Y}_{T})\right].\end{aligned}$$

Choose y such that $X_{T}=I(y\hat {Y}_{T})\in \mathcal {C}(x)$. Then, this equals $\tilde {v}(y)+xy=E\left [U(X_{T})\right ]$.

Taking the supremum on the right side yields

$\tilde {v}(y)+xy\leq v(x)$. The infimum of the left side gives $\underset {y>0}{\inf }\left (\tilde {v}(y)+xy\right )\leq v(x)$.

Combined steps 1 and 2 show $\underset {y>0}{\inf }\left (\tilde {v}(y)+xy\right )=v(x)$.

Because $\tilde {v}(x)<\infty$ for some x > 0, $\tilde {v}$ is proper.

Because $\tilde {v}(y)=\underset {Y_{T}\in D_{s}}{\inf }E[\tilde {U}(yY_{T})]$, we see that $\tilde {v}$ is strictly convex, hence continuous on $\text{int}(\text{dom}(\tilde {v}))$.

By Pham [126, Theorem B.2.3, p. 219], $\tilde {v}(y)=\underset {x>0}{\sup }\left (v(x)-xy\right ),\:\forall y>0$.

Define $\hat {y}$ to be where the infimum is attained in $\underset {y>0}{\inf }\left (\tilde {v}(y)+xy\right )=v(x)$. It exists because v(x) < ∞ and $\tilde {v}$ is strictly convex. Then, $v(x)=\tilde {v}(\hat {y})+x\hat {y}.$

(Part 2) By Guler [66, p. 278], $\mathcal {L}(\hat {X}_{T},\hat {Y}_{T},\hat {y})$ is a saddle point if and only if the solution to the primal problem equals the solution to the dual problem, i.e.

$$\displaystyle \begin{aligned} v(x)=\underset{X_{T}\in L_{+}^{0}}{\sup}\left(\underset{Y_{T}\in D_{s},\,y>0}{\inf}\mathcal{L}(X_{T},Y_{T},y)\right) =\underset{Y_{T}\in D_{s},\,y>0}{\inf}\left(\underset{X_{T}\in L_{+}^{0}}{\sup}\,\mathcal{L}(X_{T},Y_{T},y)\right). \end{aligned}$$

We show this later condition.

First, $v(x)=\underset {y>0}{\inf }\left (\tilde {v}(y)+xy\right )$. Using the definitions, this is equivalent to

$$\displaystyle \begin{aligned} \begin{array}{rcl} v(x)&\displaystyle =&\displaystyle \underset{y>0}{\inf}\left(\underset{Y_{T}\in D_{s}}{\inf}E[\tilde{U}(yY_{T})]+xy\right)\\ &\displaystyle =&\displaystyle \underset{y>0}{\inf}\left(\underset{Y_{T}\in D_{s}}{\inf}E\left[\underset{X_{T}\in L_{+}^{0}}{\sup}U(X_{T})-yX_{T}Y_{T}\right]+xy\right). \end{array} \end{aligned} $$

Exchange the sup and E[⋅] operator, which is justified by the same proof as in the appendix to the portfolio optimization Chap. 10.

$$\displaystyle \begin{aligned} \begin{array}{rcl} v(x)&\displaystyle =&\displaystyle \underset{y>0}{\inf}\left(\underset{Y_{T}\in D_{s}}{\inf}\left(\underset{X_{T}\in L_{+}^{0}}{\sup}E\left[U(X_{T})-yX_{T}Y_{T}\right]\right)+xy\right)\\ &\displaystyle =&\displaystyle \underset{y>0}{\inf}\left(\underset{Y_{T}\in D_{s}}{\inf}\left(\underset{X_{T}\in L_{+}^{0}}{\sup}E\left[U(X_{T})\right]-y\left(E\left[X_{T}Y_{T}\right]-x\right)\right)\right)\\ &\displaystyle =&\displaystyle \underset{y>0}{\inf}\left(\underset{Y_{T}\in D_{s}}{\inf}\left(\underset{X_{T}\in L_{+}^{0}}{\sup}\,\mathcal{L}(X_{T},Y_{T},y)\right)\right).\end{array} \end{aligned} $$

This completes the proof.

Remark 11.7 (Extension to Chap.12)

This same proof works with $X_{T}\in \mathcal {C}(x)\subset L_{+}^{0}$ replaced by $(c,X_{T})\in \mathcal {C}(x)\subset \mathcal {L}_{+}^{0}\times L_{+}^{0}$ and $Y\in \mathcal {D}_{s}\subset \mathcal {L}_{+}^{0}$, where $v(x)=\underset {(c,X_{T})\in \mathcal {C}(x)}{\sup }E\left [\int _{0}^{T}U_{1}(c_{t})dt+U_{2}\left (X_{T}\right )\right ]$ and $\tilde {v}(y)=\underset {Y\in \mathcal {D}_{s}}{\inf }\:E\left [\int _{0}^{T}\tilde {U}_{1}(yY_{t})dt+\tilde {U}_{2}(yY_{T})\right ]$. This completes the remark.