A General Maximum Principle for Anticipative Stochastic Control and Applications to Insider Trading

Abstract

In this paper we suggest a general stochastic maximum principle for optimal control of anticipating stochastic differential equations driven by a Lévy-type noise. We use techniques of Malliavin calculus and forward integration. We apply our results to study a general optimal portfolio problem of an insider. In particular, we find conditions on the insider information filtration which are sufficient to give the insider an infinite wealth. We also apply the results to find the optimal consumption rate for an insider.

The research leading to these results has received funding from the European Research Council under the European Community’s Seventh Framework Programme (FP7/2007-2013)/ERC grant agreement no. [228087].

Appendix: Proof of Theorem 7.13

Proof

1. Since $\widehat{u}\in\mathcal{A}_{\mathbb{G}}$ is a critical point for J(u), there exists a δ>0 as in (7.24) for all bounded $\beta\in\mathcal{A}_{\mathbb{G}}$. Thus,

(7.77)

where $\widehat{Y}=Y_{\beta}^{\widehat{u}}$ is as defined in (7.25).

We study the two summands separately. By Corollaries 7.5 and 7.12 and the product rule, we get

Similarly, we have using both Fubini and duality theorems,

Changing the notation s→t, this becomes

(7.78)

Recall that

and combining (7.33)–(7.78), it follows that

(7.79)

We observe that $\mathcal{A}_{\mathbb{G}}$ contains all β _α given as β _α (s):=αχ _[t,t+h](s) for some t,h∈(0,T),t+h≤T, where α=α(ω) is bounded and $\mathcal{G}_{t}$-measurable. Then $Y^{(\beta_{\alpha })}(s)=0$ for 0≤s≤t, and hence (7.79) becomes

(7.80)

where

Note that by the definition of Y with $Y(s)=Y^{(\beta_{\alpha })}(s)$ and s≥t+h, the process Y(s) follows the dynamics

(7.81)

for s≥t+h with initial condition Y(t+h) at time t+h. By Itô’s formula for forward integral, this equation can be solved explicitly, and we get

(7.82)

where, in general, for s≥t,

Note that G(t,s) does not depend on h, but Y(s) does. Defining H ₀ as in (7.27), it follows that

$$A_1=E\biggl[\int_t^T \frac{\partial H_0}{\partial x }(s)Y(s)\,ds\biggr].$$

Differentiating with respect to h at h=0, we get

$$\begin{aligned}\frac{ d}{ dh }A_1\bigg\vert_{h=0}&=\frac{ d}{ dh }E\biggl[\int_t^{t+h} \frac{\partial H_0}{ \partial x }(s)Y(s)\,ds\biggr]_{h=0}\\&\quad {}+ \frac{ d}{ dh }E\biggl[\int _{t+h}^T \frac{ \partial H_0}{ \partial x }(s)Y(s)\,ds\biggr]_{h=0}.\end{aligned}$$

Since Y(t)=0, we see that

$$\frac{ d}{ dh }E\biggl[\int_t^{t+h} \frac{\partial H_0}{\partial x }(s)Y(s)\,ds\biggr]_{h=0}=0 .$$

Therefore, by (7.82),

where Y(t+h) is given by

Therefore, by the two preceding equalities,

where

and

Applying again the duality formula, we have

where we have put

$$F(t,s)=\frac{\partial H_0}{\partial x }(s)G(t,s) .$$

Since Y(t)=0, we see that

$$A_{1,2}=0.$$

We conclude that

(7.83)

Moreover, we see that

(7.84)

(7.85)

(7.86)

On the other hand, by differentiating A ₃ with respect to h at h=0, we get

Since Y(t)=0, we see that

Using the definition of $\widehat{p}$ and $\widehat{H}$ given respectively by (7.36) and (7.35) in the theorem, it follows by (7.80) that

$$E\biggl[ \frac{\partial}{\partial u}\widehat{H}\bigl(t,\widehat{X }(t),\widehat {u}(t)\bigr)\bigg\vert \mathcal{G}_{t}\biggr] + E[A]=0\quad \text{a.e. in}\ (t,\omega),$$

(7.87)

where

(7.88)

2. Conversely, suppose that there exists $\widehat{u} \in\mathcal {A}_{\mathbb{G}}$ such that (7.34) holds. Then by reversing the previous arguments, we obtain that (7.80) holds for all $\beta_{\alpha}(s):=\alpha\chi_{[ t,t+h]}(s) \in\mathcal {A}_{\mathbb{G}}$, where

for some t,h∈(0,T),t+h≤T, where α=α(ω) is bounded and $\mathcal{G}_{t}$-measurable. Hence, these equalities hold for all linear combinations of β _α . Since all bounded $\beta\in\mathcal{A}_{\mathbb{G}}$ can be approximated pointwise boundedly in (t,ω) by such linear combinations, it follows that (7.80) holds for all bounded $\beta\in\mathcal{A}_{\mathbb{G}}$. Hence, by reversing the remaining part of the previous proof, we conclude that

$$\frac{d }{d y}J_1(\widehat{u}+y\beta) \bigg\vert_{y=0}=0\quad \text{for all }\beta,$$

and then $\widehat{u}$ satisfies (7.33). □