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 Proof of Expression (10.7)

To use this result in an incomplete market, 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

We want to prove the opposite inequality.

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

But,

which completes the proof.

1.2 Proof of Expression (10.8)

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

Proof

(Exchange of E[⋅] and Derivative)

for δ > 0.

Since the derivative exists, we use the left derivative.

a.s. $\mathbb {P}$ where Δ < 0.

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

$\tilde {U}((y+\varDelta )Y_{T})-\tilde {U}(yY_{T})=\tilde {U}'(\xi Y_{T})\left [(y+\varDelta )Y_{T}-yY_{T}\right ]$.

Thus, .

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 [132, 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

$-\tilde {U}'(z)z<C\tilde {U}(z)$ for 0 < z ≤ z ₀, and

$\tilde {U}(\mu z)<K(\mu )\tilde {U}(z)$ for 0 < μ < 1 and 0 < z ≤ z ₀ 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

$-\tilde {U}'(\mu z)\mu z<\bar {K}(\mu )\tilde {U}(z)$ for 0 < z ≤ z ₀ 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,

$-\tilde {U}'(\xi Y_{T})Y_{T}<\frac {1}{\xi }\bar {K}(\frac {\xi }{y})\tilde {U}(yY_{T})$. Since the right side is $\mathbb {P}$ integrable, using the dominated convergence theorem,

.

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 [167, Theorem 2, p. 96]), and hence continuous. This completes the proof.