Complete Markets (Utility over Terminal Wealth)

Abstract

This chapter studies an individual’s portfolio optimization problem. In this optimization, the solution differs depending on whether the market is complete or incomplete. This chapter investigates the optimization problem in a complete markets setting, and the next chapter analyzes incomplete markets.

Appendix

1.1 Portfolio Weights

Define the portfolio weights as

$$\displaystyle \begin{aligned} \pi_{0}(t)=\frac{\alpha_{0}(t)\mathbb{B}_{t}}{\mathbb{X}_{t}}=\frac{\alpha_{0}(t)}{X_{t}}\qquad \mathrm{and} \end{aligned}$$

$$\displaystyle \begin{aligned} \pi_{i}(t)=\frac{\alpha_{i}(t)\mathbb{S}_{i}(t)}{\mathbb{X}_{t}}=\frac{\alpha_{i}(t)\left(\frac{\mathbb{S}_{i}(t)}{\mathbb{B}_{t}}\right)}{\frac{\mathbb{X}_{t}}{\mathbb{B}_{t}}}=\frac{\alpha_{i}(t)S_{i}(t)}{X_{t}} \end{aligned}$$

for all i = 1, …, n.

Note that in percentage holdings, the normalization versus non-normalization definitions are identical. Of course, all quantities in the denominator as assumed to be nonzero.

Define $\pi _{t}=\pi (t)=(\pi _{1}(t),\ldots ,\pi _{n}(t))'\in \mathbb {R}^{n}$.

Note that π ₀(t) + π _t  ⋅1 = 1 where $\mathbf {1}=(1,\ldots ,1)'\in \mathbb {R}^{n}.$

The following gives the correspondences between the two formulations

$$\displaystyle \begin{aligned} \begin{array}{ccc} \pi_{i}(t)=\frac{\alpha_{i}(t)S_{i}(t)}{X_{t}} & \Leftrightarrow & \alpha_{i}(t)=\frac{\pi_{i}(t)X_{t}}{S_{i}(t)}\\ X_{t}=x+\int_{0}^{t}\frac{\pi_{u}}{S_{u}}X_{u}\cdot dS_{u} & \Leftrightarrow & X_{t}=x+\int_{0}^{t}\alpha_{u}\cdot dS_{u}\\ 1=\pi_{0}(t)+\pi_{t}\cdot\mathbf{1} & \Leftrightarrow & X_{t}=\alpha_{0}(t)+\alpha_{t}\cdot S_{t} \end{array} \end{aligned}$$

where $\frac {\pi _{t}}{S_{t}}=\left (\frac {\pi _{1}(t)}{S_{1}(t)},\ldots ,\frac {\pi _{n}(t)}{S_{n}(t)}\right )'\in \mathbb {R}^{n}$ and (for later use) $\frac {dS_{t}}{S_{t}}=\left (\frac {dS_{1}(t)}{S_{1}(t)},\ldots ,\frac {dS_{n}(t)}{S_{n}(t)}\right )' \in \mathbb {R}^{n}$.

Proof

The proof of the last identification is

$$\displaystyle \begin{aligned} \begin{array}{rcl}X_{t}&\displaystyle =&\displaystyle \alpha_{0}(t)+\alpha_{t}\cdot S(t)\quad \Leftrightarrow\\ \frac{X_{t}}{X_{t}}&\displaystyle =&\displaystyle \frac{\alpha_{0}(t)}{X_{t}}+\frac{\alpha_{t}\cdot S(t)}{X_{t}}\quad \Leftrightarrow\\ 1&\displaystyle =&\displaystyle \pi_{0}(t)+\pi_{t}\cdot1.\end{array} \end{aligned} $$

This completes the proof.

1.2 Proof of Expression (10.7)

For the use of this result in an incomplete market, the portfolio optimization Chap. 11, we note that the following proof holds for any Y _T  ∈ D _s as well.

Proof (Exchange of sup and E[ ⋅ ] Operator)

It is trivial that

$$\displaystyle \begin{aligned} \underset{X_{T}\in L_{+}^{0}}{\sup}\,E\left[U(X_{T})-yX_{T}Y_{T}\right]\leq E[\underset{X_{T}\in L_{+}^{0}}{\sup}\,(U(X_{T})-yX_{T}Y_{T})]. \end{aligned}$$

We want to prove the opposite inequality.

Since U is strictly concave, there exists a unique solution $X_{T}^{*}$ to

$$\displaystyle \begin{aligned} \underset{X_{T}\in L_{+}^{0}}{\sup}\left[U(X_{T})-yX_{T}Y_{T}\right]=\left[U(X_{T}^{*})-yX_{T}^{*}Y_{T}\right]. \end{aligned}$$

But,

$$\displaystyle \begin{aligned} \underset{X_{T}\in L_{+}^{0}}{\sup}\,E\left[U(X_{T})-yX_{T}Y_{T}\right]\geq E\left[U(X_{T}^{*})-yX_{T}^{*}Y_{T}\right] \end{aligned}$$

$$\displaystyle \begin{aligned} =E[\underset{X_{T}\in L_{+}^{0}}{\sup}\,(U(X_{T})-yX_{T}Y_{T})], \end{aligned}$$

which completes the proof.

1.3 Proof of Expression (10.8)

For the use of this result in an incomplete market, the portfolio optimization Chap. 11, we note that the following proof holds for any Y _T  ∈ D _s as well.

Proof (Exchange of E[ ⋅ ] and Derivative)

Since the derivative exists, we use the left derivative.

$$\displaystyle \begin{aligned} \begin{array}{rcl} \underset{\triangle\rightarrow0}{\lim}\frac{\tilde{U}((y{+}\triangle)Y_{T}){-}\tilde{U}(yY_{T})}{\triangle}&\displaystyle =&\displaystyle \underset{\triangle Y_{T}\rightarrow0}{\lim}\frac{\tilde{U}((y{+}\triangle)Y_{T}){-}\tilde{U}(yY_{T})}{(y{+}\triangle)Y_{T}{-}yY_{T}}\cdot\underset{\triangle\rightarrow0}{\lim}\frac{(y{+}\triangle)Y_{T}{-}yY_{T}}{\triangle}\\ &\displaystyle =&\displaystyle \underset{\triangle Y_{T}\rightarrow0}{\lim}\frac{\tilde{U}((y+\triangle)Y_{T})-\tilde{U}(yY_{T})}{(y+\triangle)Y_{T}-yY_{T}}\cdot Y_{T}\\ &\displaystyle =&\displaystyle \underset{\triangle\rightarrow0}{\lim}\tilde{U}'(\xi Y_{T})Y_{T}\ \text{a.s.}\ \mathbb{P},\ \text{where}\ \varDelta<0. \end{array} \end{aligned} $$

The last equality follows from the mean value theorem (Guler [66, p. 3]), i.e. there exists an ξ ∈ [y + Δ, y] such that

$$\displaystyle \begin{aligned}\tilde{U}((y+\triangle)Y_{T})-\tilde{U}(yY_{T})=\tilde{U}'(\xi Y_{T})\left[(y+\triangle)Y_{T}-yY_{T}\right].\end{aligned}$$

Thus, $\frac {\partial E[\tilde {U}(yY_{T})]}{\partial y}=\underset {\triangle \rightarrow 0}{\lim }\frac {E\left [\tilde {U}((y+\triangle )Y_{T})-\tilde {U}(yY_{T})\right ]}{\triangle }=\underset {\triangle \rightarrow 0}{\lim }E[\tilde {U}'(\xi Y_{T})Y_{T}]$.

Now $E[\tilde {U}(yY_{T})]<\infty$ because $\tilde {v}(y)<\infty$ and Y _T is the supermartingale deflator such that $\tilde {v}(y)=E[\tilde {U}(yY_{T})]$.

By Kramkov and Schachermayer [126, Lemma 6.3 (iv) and (iii), p. 944], $AE\left (U\right )<1$ implies there exists a constant C and z ₀ > 0 such that

$$\displaystyle \begin{aligned}-\tilde{U}'(z)z<C\tilde{U}(z)\ \text{for}\ 0<z\leq z_{0},\end{aligned}$$

and

$$\displaystyle \begin{aligned}\tilde{U}(\mu z)<K(\mu)\tilde{U}(z)\ \text{for}\ 0<\mu<1\ \text{and}\ 0<z\leq z_{0},\end{aligned}$$

where K(μ) is a constant depending upon μ.

Combined, $-\tilde {U}'(\mu z)\mu z<C\tilde {U}(\mu z)<CK(\mu )\tilde {U}(z)$ implies that there exists a z ₀ > 0 such that

$$\displaystyle \begin{aligned}-\tilde{U}'(\mu z)\mu z<\bar{K}(\mu)\tilde{U}(z)\ \text{for}\ 0<z\leq z_{0}\end{aligned}$$

where $\bar {K}(\mu )$ is a constant depending upon μ for 0 < μ < 1.

Letting z = yY _T and $\mu =\frac {\xi }{y}<1$, because Δ < 0 so that ξ < y. Then,

$$\displaystyle \begin{aligned} -\tilde{U}'(\xi Y_{T})Y_{T}<\frac{1}{\xi}\bar{K}(\frac{\xi}{y})\tilde{U}(yY_{T}). \end{aligned}$$

Since the right side is $\mathbb {P}$-integrable, using the dominated convergence theorem,

$$\displaystyle \begin{aligned} \underset{\triangle\rightarrow0}{\lim}E[\tilde{U}'(\xi Y_{T})Y_{T}]=E[\underset{\triangle\rightarrow0}{\lim}\tilde{U}'(\xi Y_{T})Y_{T}]=E[\tilde{U}'(yY_{T})Y_{T}]. \end{aligned}$$

The last equality follows from the continuity of $\tilde {U}'(\cdot )$. The continuity of $\tilde {U}'(\cdot )$ follows because $\tilde {U}(\cdot )$ is strictly convex, hence $\tilde {U}'(\cdot )$ is a strictly increasing function, which is therefore differentiable a.s. $\mathbb {P}$ (see Royden [160, Theorem 2, p. 96]), and hence continuous. This completes the proof.