FTAP in finite discrete time with transaction costs by utility maximization

Abstract

The aim of this paper is to prove the fundamental theorem of asset pricing (FTAP) in finite discrete time with proportional transaction costs by utility maximization. The idea goes back to L.C.G. Rogers’ proof of the classical FTAP for a model without transaction costs. We consider one risky asset and show that under the robust no-arbitrage condition, the investor can maximize his expected utility. Using the optimal portfolio, a consistent price system is derived.

Appendix

We want to prepare the proof of Lemma 3.3. Given two sub-σ-fields $\mathcal{A}_{0} \subset\mathcal{A}_{1}$ and $0 < \underline{X}_{0} \le\overline{X}_{0}$ $\mathcal{A}_{0}$-measurable as well as $0 < \underline{X}_{1} \le\overline{X}_{1}$ $\mathcal{A}_{1}$-measurable, we assume that

$$\mathbf{L}^0\big({-}K(\underline{X}_0, \overline{X}_0), \mathcal{A}_0\big) \cap\mathbf{L}^0\big(K(\underline{X}_1, \overline{X}_1), \mathcal {A}_1\big) $$

is a vector space. It is straightforward to check that under this assumption

$$\big\{ \mathbf {E}\big[ |\underline{X}_1 - \overline{X}_0 | \,\big\vert \, \mathcal{A}_0 \big] = 0 \big\} = \big\{ \mathbf {E}\big[ |\underline{X}_0 - \overline{X}_1 | \,\big\vert \,\mathcal{A}_0 \big] = 0 \big\} , $$

and $A:=\{ \mathbf {E}[ |\underline{X}_{1} - \overline{X}_{0} | \,\vert\, \mathcal {A}_{0} ] = 0 \}$ is the biggest $\mathcal{A}_{0}$-measurable set such that

$$\underline{X}_0=\overline{X}_0=\underline{X}_1=\overline{X}_1 \quad\hbox{on}\ A. $$

For ω∈A, we define P ₀(ω) to be the orthogonal projection on the linear space generated by ; for ω∉A, we set P ₀(ω):=0. Then we have P ₀ v=v iff $v \in\mathbf{L}^{0}(-K(\underline{X}_{0}, \overline{X}_{0}), \mathcal{A}_{0}) \cap\mathbf{L}^{0}(K(\underline{X}_{1}, \overline{X}_{1}), \mathcal{A}_{1})$.

The following lemma is a version of Proposition 3.3 from [15]. We write $\ell^{X_{1}}(v)$ for the liquidation value of v with respect to the bid and ask prices $\underline{X}_{1}, \overline {X}_{1}$, i.e.

$$\ell^{X_1}(v) = \begin{cases} v^0 + v^1 \underline{X}_1, & v^1 \ge0, \\ v^0 + v^1 \overline{X}_1, & v^1 \le0. \end{cases} $$

The liquidation value is continuous in v and satisfies ℓ(αv)=αℓ(v) for α≥0. This is why the proof from [15] also works (with the obvious changes) in our model with transaction costs.

Lemma A.1

If $\mathbf{L}^{0}(-K(\underline{X}_{0}, \overline{X}_{0}), \mathcal{A}_{0}) \cap\mathbf{L}^{0}(K(\underline{X}_{1}, \overline{X}_{1}), \mathcal{A}_{1})$ is a vector space, then there exists a strictly positive $\mathcal {A}_{0}$-measurable γ ₀ such that

$$\mathbf {P}[ \ell^{X_1} (v) < -\gamma_0 \,\vert\, \mathcal{A}_0 ] > 0 $$

for every $v \in\mathbf{L}^{0}( -K (\underline{X}_{0}, \overline{X}_{0}), \mathcal{A}_{0})$ which satisfies P ₀ v=0 and |v|=1.

Proof of Lemma 3.3

For t=0,1, we pick an $\mathcal{A}_{t}$-measurable $S_{t} \in \operatorname {ri}[\underline{X}_{t}, \overline{X}_{t}]$ and set for α _t ∈(0,1)

$$\underline{V}_t := (1-\alpha_t) S_t + \alpha_t \underline{X}_t, \qquad \overline{V}_t := (1-\alpha_t) S_t + \alpha_t \overline{X}_t. $$

We fix α ₁∈(0,1) such that $2 (1-\alpha_{1}) (1+\overline {X}_{1}^{2}) < \gamma_{0}/2$. It follows from

and

for every |v|=1 that

$$\mathbf {P}[\ell^{V_1}(v) < -\gamma_0/2 \,\vert\, \mathcal{A}_0 ] \ge \mathbf {P}[\ell ^{X_1}(v) < - \gamma_0 \,\vert\, \mathcal{A}_0] > 0 $$

for every $v \in\mathbf{L}^{0} (-K(\underline{X}_{0}, \overline{X}_{0}), \mathcal{A}_{0})$ which satisfies P ₀ v=0 and |v|=1.

We have $v - P_{0}v \in\mathbf{L}^{0} (-K(\underline{V}_{0}, \overline {V}_{0}), \mathcal{A}_{0})$ and $\ell^{V_{1}}(v)=\ell^{V_{1}}(v-P_{0}v)$ for every $v \in\mathbf{L}^{0} (-K(\underline{V}_{0}, \overline{V}_{0}), \mathcal {A}_{0})$. Thus, it is enough to find an appropriate α ₀ such that

$$\mathbf {P}[\ell^{V_1}(v) < 0 \,\vert\, \mathcal{A}_0] > 0 \quad\hbox{on}\ \{v \neq0 \} $$

for every $v \in\mathbf{L}^{0} (-K(\underline{V}_{0}, \overline{V}_{0}), \mathcal{A}_{0})$ with P ₀ v=0. We can assume that |v|=1 on {v≠0}. It is straightforward to check that then

where $c(\alpha_{0}):=(1/\alpha_{0} - 1) (1 + S_{0}^{2})$. Note that (id−P ₀)(v−c(α ₀)e ⁰)≠0 a.s. and c(α ₀)↓0 as α ₀↑1. Now

$$\ell^{V_1}\bigg( \frac{v - c(\alpha_0)e^0}{|(\operatorname {id}-P_0) (v - c(\alpha_0)e^0)|} \bigg) = \ell^{V_1}\bigg( \frac{(\operatorname {id}-P_0)(v - c(\alpha_0)e^0)}{|(\operatorname {id}-P_0) (v - c(\alpha_0)e^0)|} \bigg) <-\gamma_0/2 $$

implies on {v≠0} that

$$\ell^{V_1} (v) \le-(\gamma_0/2) \big(1-|(\operatorname {id}- P_0)c(\alpha _0)e^0|\big) + c (\alpha_0). $$

We fix α ₀ near 1 such that the last expression is strictly negative. It follows that

$$\mathbf {P}[\ell^{V_1}(v) < 0 \,\vert\, \mathcal{A}_0] > 0 \quad\hbox{on}\ \{v \neq0 \} $$

for every $v \in\mathbf{L}^{0} (-K(\underline{V}_{0}, \overline{V}_{0}), \mathcal{A}_{0})$ with P ₀ v=0. □

Proof of Lemma 3.5

Claim (a). We show that on $A=\{ \operatorname {inf}_{\mathcal{H}} \underline{X} =\mathbf {E}_{\mathbf{Q}} [ Z \,\vert\, \mathcal{H} ] \}$, we have $\inf_{\mathcal{H}} \underline{X} = \sup_{\mathcal{H}} \overline{X}$. Indeed, from $\operatorname {inf}_{\mathcal{H}} \underline{X} \le\underline{X} \le Z$, we get that $\operatorname {inf}_{\mathcal{H}} \underline{X} = \underline{X} = Z$ and thus $\underline{X}=\overline{X}$ on A, since $Z \in \operatorname {ri}[ \underline{X}, \overline{X} ]$. Then $\operatorname {inf}_{\mathcal{H}} \underline{X}= \overline{X}$ and $\operatorname {inf}_{\mathcal{H}} \underline{X} = \operatorname {sup}_{\mathcal{H}} \overline{X}$ on A. Similarly, we get that $\inf_{\mathcal{H}} \underline{X}$ and $\sup_{\mathcal{H}} \overline{X}$ coincide on $\{ \mathbf {E}_{\mathbf{Q}}[ Z \,\vert\, \mathcal{H} ] = \operatorname {sup}_{\mathcal {H}} \overline{X} \}$.

Claim (b). For $n \in\mathbb{N}$, put

$$\begin{aligned} \tilde{a}_n &:= \frac{\mathbf{1}_{ \{ \underline{X} < \operatorname {inf}_\mathcal{H} \underline{X} + \frac{1}{n} \}}}{\mathbf {P}[ \underline{X} < \operatorname {inf}_\mathcal{H} \underline{X} + \frac{1}{n} \,\vert\, \mathcal{H}] } + \frac{1}{n} \frac {\mathbf{1}_{ \{ \underline{X} \ge \operatorname {inf}_\mathcal{H} \underline{X} + \frac{1}{n} \}}}{1+\underline{X}}, \\ \tilde{b}_n &:= \frac{\mathbf{1}_{ \{ \overline{X} > f_n \}}}{\mathbf {P}[ \overline{X} > f_n \,\vert\, \mathcal{H}] } + \frac{1}{n} \frac {\mathbf {1}_{ \{ \overline{X} \le f_n \}}}{1+\overline{X}}, \end{aligned}$$

where

$$f_n := \mathbf{1}_ {\{ \operatorname {sup}_{\mathcal{H}} \overline{X} < \infty\}} \bigg(\operatorname {sup}_{\mathcal{H}} \overline{X} - \frac{1}{n}\bigg) + 1_{ \{ \operatorname {sup}_{\mathcal{H} } (\overline{X} ) =\infty\} } n. $$

Then $\tilde{a}_{n}, \tilde{b}_{n} > 0$ and $\mathbf {E}[\tilde{a}_{n} \underline{X} \,\vert\, \mathcal{H}] \rightarrow \operatorname {inf}_{\mathcal{H}} \underline{X}$, $\mathbf {E}[\tilde{b}_{n} \overline{X} \,\vert\, \mathcal{H}] \rightarrow \operatorname {sup}_{\mathcal{H}} \overline{X} \$a.s. as n→∞. Since $1 \le \mathbf {E}[\tilde{a}_{n} \,\vert\, \mathcal{H}] \le1 + \frac{1}{n}$ and $1 \le \mathbf {E}[\tilde{b}_{n} \,\vert\, \mathcal{H}] \le1 + \frac{1}{n}$, we normalize $\tilde{a}_{n}$ and $\tilde{b}_{n}$, i.e. replace $\tilde{a}_{n}$ by $\frac{\tilde{a}_{n}}{\mathbf {E}[ \tilde{a}_{n} \,\vert\, \mathcal{H} ]}$ and $\tilde{b}_{n}$ by $\frac{\tilde{b}_{n}}{\mathbf {E}[ \tilde{b}_{n} \,\vert\, \mathcal {H} ]}$, and still have $\mathbf {E}[\tilde{a}_{n} \underline{X} \,\vert\, \mathcal {H}] \to \operatorname {inf}_{\mathcal{H}} \underline{X}$, $\mathbf {E}[\tilde{b}_{n} \overline {X} \,\vert\, \mathcal{H}] \rightarrow \operatorname {sup}_{\mathcal{H}} \overline{X} \$a.s. as n→∞.

Now we set $\varOmega_{1} := \{ \operatorname {inf}_{\mathcal{H}} \underline{X} < \operatorname {sup}_{\mathcal{H}} \overline{X} \}$ and $\varOmega_{2} := \{ \operatorname {inf}_{\mathcal{H}} \underline{X} = \operatorname {sup}_{\mathcal{H}} \overline{X} \}$; thus Ω=Ω ₁∪Ω ₂. For ω∈Ω ₁ (up to a null set), we choose $n=n( \omega) \in\mathbb{N}$ minimal such that $(\operatorname {inf}_{\mathcal{H}} \underline{X}) ( \omega) \le \mathbf {E}[\tilde{a}_{n} \underline{X} \,\vert\, \mathcal{H}](\omega) < c (\omega)$. Then ω↦n(ω) is $\mathcal{H}$-measurable on Ω ₁, and $\tilde{a}$ defined as $\tilde{a} ( \omega) := \tilde{a}_{n (\omega)} (\omega)$ on Ω ₁ and $\tilde{a} ( \omega) = 1$ on Ω ₂ is strictly positive and satisfies $\mathbf {E}[\tilde{a} \,\vert \, \mathcal{H}] = 1$, $\mathbf {E}[\tilde{a} \underline{X} \,\vert\, \mathcal {H}] < c \ \mbox{on}\ \varOmega_{1}$. Similarly, $\tilde{b}$ defined by $\tilde{b}(\omega) := \tilde{b}_{m ( \omega)} ( \omega)$ on Ω ₁, where $m( \omega) \in\mathbb{N}$ is minimal such that $\mathbf {E}[ \tilde{b}_{m} \overline{X} \,\vert\, \mathcal{H} ] ( \omega)> c( \omega)$, and $\tilde{b} ( \omega) = 1$ on Ω ₂ is strictly positive and satisfies $\mathbf {E}[\tilde{b} \,\vert\, \mathcal{H} ] = 1$, $\mathbf {E}[ \tilde{b} \overline{X} \,\vert\, \mathcal{H}] > c \ \mbox{on}\ \varOmega_{1}$. We put

$$\begin{aligned} a& := \frac{c-\mathbf {E}[ \tilde{a} \underline{X} \,\vert\, \mathcal{H} ]}{\mathbf {E}[\tilde{b} \overline{X} \,\vert\, \mathcal{H} ] - \mathbf {E}[ \tilde{a} \underline{X} \,\vert\, \mathcal{H} ]} \tilde{b} \mathbf {1}_{\varOmega_1} + \frac{1}{2} \mathbf{1}_{\varOmega_2}, \\ b&:= \frac{\mathbf {E}[\tilde{b} \overline{X} \,\vert\, \mathcal{H}]-c}{\mathbf {E}[\tilde{b} \overline{X} \,\vert\, \mathcal{H}] - \mathbf {E}[ \tilde{a} \underline {X} \,\vert\, \mathcal{H}]} \tilde{a} \mathbf{1}_{\varOmega_1}+\frac{1}{2} \mathbf {1}_{\varOmega_2}. \end{aligned}$$

Then we have $\mathbf {E}_{\mathbf{Q}} [Z \,\vert\, \mathcal{H} ] = c$, where $Z:=\frac{a\underline{X}+b\overline{X}}{a+b} \in \operatorname {ri}[\underline{X},\overline{X}]$ and Q is defined by $\frac {\mathrm{d} \mathbf{Q}}{\mathrm{d}\mathbf {P}} := a+b$. □

Proof of Lemma 3.6

Closedness of F follows from the fact that φ(0)≤φ(ta) for a∈F, t>0, due to concavity. Fix a closed $K \subset \{x \in\mathbb{R}^{d} \, : \, |x|=1\}$ and choose a sequence $(x_{j})_{j=1}^{\infty}$ of $\mathcal{H}$-measurable random variables with values in K such that $(x_{j}(\omega))_{j=1}^{\infty}$ is dense in K∩C(ω) on the event {K∩C≠∅}. Then

$$\{{ \omega\in\varOmega}:{ F(\omega) \cap K \neq\emptyset}\} = \bigcup_{M=1}^{\infty} \bigcap_{k=1}^{\infty} \bigcup_{j = 1}^{\infty} \bigl\{ {\omega\in\varOmega}:{\varphi\big(\omega, kx_j(\omega )\big) \ge-M }\bigr\} \in\mathcal{H}. $$

Closedness of A is clear. Now, for a fixed $K \subset\mathbb{R}^{d}$ which is supposed to be compact, we choose a sequence of $\mathcal {H}$-measurable random variables $(y_{j})_{j=1}^{\infty}$ with values in K such that $(y_{j}(\omega))_{j=1}^{\infty}$ is dense in K∩C(ω) on the event {K∩C≠∅}. Further, let $(c_{n})_{n=1}^{\infty}$ be a sequence of $\mathcal{H}$-measurable random variables such that $(c_{n}(\omega))_{n=1}^{\infty}$ is dense in C(ω). Then

$$\begin{aligned} &\{{ \omega\in\varOmega}:{ A (\omega) \cap K \neq\emptyset}\} \\ &\quad= \bigcap_{m=1}^{\infty} \bigcup_{j = 1}^{\infty} \bigcap_{n=1}^{\infty} \biggl\{ {\omega\in\varOmega}\,:\,{ \varphi\big(\omega, y_j(\omega )\big) \ge\varphi \big(\omega, c_n(\omega) \big) - \frac{1}{m} }\biggr\} \in\mathcal{H}. \end{aligned}$$

Now, by straightforward measurable selection arguments (see Theorem III.6 in [2] and use the fact that in $\mathbb{R}^{d}$, every open set is a union of compact sets), α and β can be defined in an $\mathcal{H}$-measurable way. □

Remark A.2

We can find an equivalent measure P′ such that $\mathbf {E}_{\mathbf{P}^{\prime}} [U(\ell(v)) \,\vert\, \mathcal{F}_{t-1}]$ and $\mathbf{E}_{\mathbf{P}^{\prime}} [U^{\prime}(\ell(v)) \,\vert\, \mathcal {F}_{t-1}]$ are finite for every $v \in\mathbf{L}^{0} (\mathbf{R}^{2}, \mathcal{F}_{t-1})$. Further, we can find a version of $\mathbf {E}_{\mathbf{P}^{\prime}} [U(\ell(a)) \,\vert\, \mathcal{F}_{t-1}]$ for every $a \in\mathbb{R}^{2}$ such that for every scenario ω∈Ω, the mapping $a \mapsto\mathbf{E}_{\mathbf{P}^{\prime}} [U(\ell (a)) \,\vert\, \mathcal{F}_{t-1}](\omega)$ is finite and continuous.

To see this, we choose a function $g : \mathbb{R}_{+} \rightarrow (0,\infty)$ which is decreasing and continuous such that

$$\vert U(a \cdot x) \vert g(\vert x \vert) \le \frac{1}{g(\vert a \vert)} + c \quad\forall a, x \in\mathbb{R}^d, $$

where c is a fixed constant. If U is bounded from above by 0, g can be defined by g(t):=(max{1,|U(−t ²)|})⁻¹ (see [18]). Denote by P′ the equivalent probability measure with density

and denote by k the regular conditional P-distribution for $( \underline{V}_{t}, \overline{V}_{t} )$ given $\mathcal{F}_{t-1}$. Then for any $v \in\mathbf{L}^{0} (\mathbb{R}^{2}, \mathcal{F}_{t-1} )$, we have by disintegration

with . Using the estimate from above plus dominated convergence, the claim follows easily.  □