Shadow prices, fractional Brownian motion, and portfolio optimisation under transaction costs

Abstract

The present paper accomplishes a major step towards a reconciliation of two conflicting approaches in mathematical finance: on the one hand, the mainstream approach based on the notion of no arbitrage (Black, Merton & Scholes), and on the other hand, the consideration of non-semimartingale price processes, the archetype of which being fractional Brownian motion (Mandelbrot). Imposing (arbitrarily small) proportional transaction costs and considering logarithmic utility optimisers, we are able to show the existence of a semimartingale, frictionless shadow price process for an exponential fractional Brownian financial market.

Appendix: Fluctuations of fractional Brownian motion

In this appendix, we establish a control on the number of fluctuations of fractional Brownian motion (Proposition A.1). It allows us to show the finiteness of the indirect utility for the proof of Theorem 2.4, and may also be interesting in its own right. Although the results stated here only deal with the case of fractional Brownian motion, our techniques can actually also be applied to a broader class of Gaussian processes.

Let $B^{H} = (B^{H}_{t})_{t \geq0}$ be a standard fractional Brownian motion with Hurst parameter $H \in(0, 1]$. Fix $\delta> 0$ and define the $\delta$-fluctuation times of $B^{H}$, denoted by $(\tau_{j} (\omega ))_{j \in \mathbb {N}}$, resp. the number of $\delta$-fluctuations of $B^{H}$ up to time $T \in(0, \infty)$, denoted by $F^{(\delta)}_{T} (\omega)$, exactly as we defined these concepts for $X$ in (4.3) and (4.4).

The main result of this appendix is the following proposition:

Proposition A.1

With the notation from above, there exist finite positive constants $C = C (H)$, $C' = C' (H)$ such that

$$ \mathbb {P}{\brack {F^{(\delta)}_{T} \geq n}} \leq\exp \left(-C^{-1} \delta^{2} T^{-2 H} n (n^{2 H \wedge1} - C' \log n)\right)\quad\textit{for all }n \geq2 . $$

(A.1)

The interest of Proposition A.1 for this article lies in the following corollary:

Corollary A.2

With the above notation, for any $\delta> 0$, the random variable $F^{(\delta)}_{T}$ has exponential moments of all orders, that is,

$$ \mathbb{\operatorname{E}}{\brack {\exp(a F^{(\delta)}_{T})}} < \infty\quad\textit{for all }a\in \mathbb {R}. $$

(A.2)

Moreover, if $H \geq1/2$, this random variable even has a Gaussian moment, that is, there exists $a>0$ such that ⁴

$$ \mathbb{\operatorname{E}}{\brack {\exp\big(a (F^{(\delta)}_{T})^{2}\big)} }< \infty. $$

(A.3)

Proof of Corollary A.2

For $f (x) = \exp(ax)$ and $f (x) = \exp(a x^{2})$, we have

$$\mathbb{\operatorname{E}}{\brack {f (F^{(\delta)}_{T})}} = f (0) + \int_{0}^{\infty}f' (x) \mathbb {P}{\brack {F^{(\delta)}_{T} \geq x}} \mathord {\mathrm {d} \mathord {x}} $$

by Fubini’s theorem. Combining this with the estimate (A.1) gives (A.2) and (A.3). □

Proof of Proposition A.1

Throughout the proof, we denote by $C, C' > 0$ constants only depending on $H$, but whose precise value may vary from appearance to appearance.

Let $n, m \in \mathbb {N}$ be such that $m > n \geq2$. We divide $[0, T]$ into $m$ subintervals $I_{k} := [\frac{k}{m}T, \frac{k + 1}{m}T]$ for $k = 0, \ldots, m - 1$. Denote their midpoints by $t_{k} := \frac{k + 1/2}{m} T$. Then we can estimate the probability of the set

$$A_{1} := \bigcup_{k = 0}^{m - 1} \bigg\{ \exists t \in I_{k} : \left\lvert B^{H}_{t} - B^{H}_{t_{k}}\right\rvert > \frac{1}{4} \delta\bigg\} $$

by

$$\begin{aligned} \mathbb {P}{\brack {A_{1}}} & \leq m \mathbb {P}{\brack {\exists t \in I_{k} : \left\lvert B^{H}_{t} - B^{H}_{t_{k}}\right\rvert > \frac{1}{4} \delta}} \\ &= m \mathbb {P}{\brack {\sup_{\left\lvert t\right\rvert \leq1} \left\lvert B^{H}_{t}\right\rvert > (T / 2 m)^{-H} \frac{1}{4} \delta}} \\ &\leq C' m \exp \left(-C^{-1} \left((m / T)^{H} \delta\right)^{2}\right) , \end{aligned}$$

(A.4)

where the equality comes from translation and scale invariance of fractional Brownian motion⁵ and the last inequality from Fernique’s theorem [13, Lemma 2.2.5].

On the complement $A_{1}^{{c}}$ of $A_{1}$, we then have that

$$ \sup_{t \in I_{k}} \left\lvert B^{H}_{t} - B^{H}_{t_{k}}\right\rvert \leq\frac{1}{4} \delta\quad \text{for all $k = 0, \ldots, m - 1$} . $$

(A.5)

Suppose now that $F^{(\delta)}_{T}(\omega)\geq n$. Then there must be at least $(n + 1)$ “random indices” $0 = K_{0} (\omega) < K_{1} (\omega) < \cdots< K_{n}(\omega) < m$ with $\tau_{j}(\omega)\in I_{K_{j}(\omega)}$ for $j = 0, \ldots, n$. Because of (A.5) and $\left\lvert B^{H}_{\tau_{j}} - B^{H}_{\tau_{j + 1}}\right\rvert = \delta$, we then must have $\left\lvert B^{H}_{t_{K_{j} }} - B^{H}_{t_{K_{j + 1}}}\right\rvert \geq\frac{1}{2} \delta$ for $j = 0, \ldots, n - 1$ on $\{F^{(\delta)}_{T} \geq n\} \cap A_{1}^{{c}}$.

In order to estimate $\mathbb {P}[\{F^{(\delta)}_{T}\geq n\}\cap A_{1}^{{c}}]$, it therefore only remains to bound the probability of the event

$$A_{2} := \bigcap_{j=0}^{n - 1}\bigg\{ |B^{H}_{t_{K_{j}}}-B^{H}_{t_{K_{j+1}}}|\geq\frac{1}{2}\delta\bigg\} . $$

But the set $A_{2}$ depends on the realisation of the “random indices” $(K_{j} (\omega))_{0 \leq j \leq n}$. To get rid of this dependence, we simply estimate the probability of the event

$$A_{3} := \bigcap_{j=0}^{n - 1} \left\lbrace \left\lvert B^{H}_{t_{k_{j}}} - B^{H}_{t_{k_{j + 1}}}\right\rvert \geq\frac{1}{2} \delta\right\rbrace $$

for all $\binom{m - 1}{n}$ possible realisations $0 = k_{0} < k_{1} < \cdots< k_{n} < m$ of our “random indices”. For this, fix an arbitrary realisation of indices $(k_{j})_{0 \leq j \leq n}$ and set

$$\Delta_{j} := B^{H}_{t_{k_{j + 1}}} - B^{H}_{t_{k_{j}}} \quad\text{for $j = 0, \ldots, m - 1$.} $$

Then

$$A_{3} = \bigcap_{j = 0}^{n - 1} \left\lbrace |\Delta_{j}| \geq\frac{1}{2} \delta\right\rbrace = \bigcap_{j = 0}^{n - 1} \left\lbrace \mathrm{\operatorname{sgn}}(\Delta_{j}) \Delta_{j} \geq \frac{1}{2} \delta\right\rbrace \subseteq \left\lbrace \sum_{j = 0}^{n - 1} \mathrm{\operatorname{sgn}}(\Delta_{j}) \Delta_{j} \geq\frac{1}{2} n \delta\right\rbrace , $$

so that

$$A_{3} \subseteq\bigcup\bigg\{ \Big\{ \sum_{j = 0}^{n - 1} \epsilon_{j} \Delta_{j} \geq\frac{1}{2} n \delta\Big\} : (\epsilon_{j})_{0 \leq j < n} \in\{-1, +1\}^{n}\bigg\} . $$

For fixed $(\epsilon_{j})_{0 \leq j < n} \in\{-1, +1\}^{n}$, we have that $\sum_{j = 0}^{n - 1} \epsilon_{j} \Delta_{j}$ is a centred normally distributed random variable with variance

$$ \operatorname{Var} \left[\sum_{j = 0}^{n - 1} \epsilon_{j} \Delta_{j}\right] = \sum_{j, j' = 0}^{n - 1} \epsilon_{j} \epsilon_{j'} \operatorname{Cov}(\Delta_{j}, \Delta_{j'}) \leq\sum_{j, j' = 0}^{n - 1} \left\lvert \operatorname{Cov}(\Delta_{j}, \Delta_{j'})\right\rvert . $$

(A.6)

To estimate (A.6), we distinguish the case $H \geq1/2$ from the case $H < 1/2$. If $H \geq1/2$, the covariance $\operatorname{Cov}(\Delta_{j}, \Delta_{j'})$ is always nonnegative so that

$$\sum_{j, j' = 0}^{n - 1} \left\lvert \operatorname{Cov}(\Delta_{j}, \Delta_{j'})\right\rvert = \operatorname{Var} \left[{\sum_{j = 0}^{n - 1} \Delta_{i}}\right] = \operatorname{Var}[B^{H}_{t_{k_{n}}} - B^{H}_{t_{k_{0}}}] = \left\lvert t_{k_{0}} - t_{k_{n}}\right\rvert ^{2H} \leq T^{2 H}. $$

If $H < 1/2$, the covariance $\operatorname{Cov}(\Delta_{j}, \Delta_{j'})$ is nonpositive as soon as $j \neq j'$, so that

$$\begin{aligned} \begin{aligned} \sum_{j' = 0}^{n - 1} \left\lvert \operatorname{Cov}(\Delta_{j}, \Delta_{j'})\right\rvert &= \operatorname{Var}[\Delta_{j}] - \operatorname{Cov} \left(\Delta_{j}, \sum_{j' < j} \Delta _{j'}\right) - \operatorname{Cov} \left(\Delta_{j}, \sum_{j' > j} \Delta_{j'}\right) \\ &= \operatorname{Var}[\Delta_{j}] - \operatorname{Cov}(\Delta_{j}, B^{H}_{t_{k_{0}}}-B^{H}_{t_{k_{j}}}) - \operatorname{Cov}(\Delta_{j}, B^{H}_{t_{k_{j + 1}}} - B^{H}_{t_{k_{n}}}). \end{aligned} \end{aligned}$$

But for $0 \leq t \leq u \leq v \leq T$, it follows from the definition of fractional Brownian motion that

$$-\operatorname{Cov}(B^{H}_{u} - B^{H}_{t}, B^{H}_{v} - B^{H}_{u}) = \tfrac{1}{2} (\left\lvert t - u\right\rvert ^{2 H} + \left\lvert u - v\right\rvert ^{2 H} - \left\lvert t - v\right\rvert ^{2 H}) \leq\tfrac{1}{2} \left\lvert t - u\right\rvert ^{2 H}. $$

Therefore, we have that

$$\begin{aligned} \sum_{j' = 0}^{n - 1} \left\lvert \operatorname{Cov}(\Delta_{j}, \Delta_{j'})\right\rvert &\leq \left\lvert t_{k_{j}} - t_{k_{j + 1}}\right\rvert ^{2 H} + \frac{1}{2} \left\lvert t_{k_{j}} - t_{k_{j + 1}}\right\rvert ^{2 H} + \frac{1}{2} \left\lvert t_{k_{j}} - t_{k_{j + 1}}\right\rvert ^{2 H} \\ &= 2\left\lvert t_{k_{j}} - t_{k_{j + 1}}\right\rvert ^{2 H} \end{aligned}$$

and thus

$$ \operatorname{Var}\bigg[\sum_{j = 0}^{n - 1} \epsilon_{j} \Delta_{j}\bigg] \leq2 \sum _{j = 0}^{n - 1} \left\lvert t_{k_{j}} - t_{k_{j + 1}}\right\rvert ^{2 H}. $$

(A.7)

But since $H < 1/2$, the function $x \mapsto x^{2H}$ is concave, so that the right-hand side of (A.7) is bounded above by

$$2 n \left(\frac{\sum_{j = 0}^{n - 1} \left\lvert t_{k_{j}} - t_{k_{j + 1}}\right\rvert }{n}\right)^{2 H} = 2 n (\left\lvert t_{k_{0}} - t_{k_{n}}\right\rvert / n)^{2 H} \leq2 n (T / n)^{2 H} = 2 n^{1 - 2 H} T^{2 H} . $$

Hence in both cases, we can estimate

$$\operatorname{Var} \left[\sum_{j = 0}^{n - 1} \epsilon_{j} \Delta_{j}\right] \leq2 T^{2 H} n^{(1 - 2 H)_{+}}. $$

So, using the classical bound that $\mathbb {P}[Z \geq x] \leq e^{-x^{2} / 2}$ for any standard normally distributed random variable $Z\sim\mathcal {N}(0,1)$, we have that

$$\mathbb {P}{\brack {\sum_{j = 0}^{n - 1} \epsilon_{j} \Delta_{j} \geq\frac {1}{2} n \delta}} \leq\exp\bigg(-\frac{1}{16} T^{-2 H} \delta^{2} n^{1 + (2 H \wedge1)}\bigg) $$

for all $2^{n}$ possible choices of $(\epsilon_{j})_{0 \leq j < n} \in\{ -1, +1\}^{n}$, and therefore

$$\mathbb {P}{\brack {A_{3}}} \leq2^{n} \exp\bigg(-\frac{1}{16} T^{-2 H} \delta^{2} n^{1 + (2 H \wedge1)}\bigg) . $$

Combining that estimate with (A.4) and using that $\binom{m - 1}{n} \leq m^{n}$, we finally get that

$$\begin{aligned} \mathbb {P}[F^{(\delta)}_{T} \geq n] &\leq \mathbb {P}{\brack {A_{1}}}+\mathbb {P}[\{F^{(\delta )}_{T} \geq n\}\cap A^{c}_{1}]\\ &\leq C' m \exp(-C^{-1} T^{-2 H} \delta^{2} m^{2 H})\\ &\phantom{=:}+ 2^{n} m^{n} \exp\bigg(-\frac{1}{16} T^{-2 H} \delta^{2} n^{1 + (2 H \wedge1)}\bigg). \end{aligned}$$

Now it only remains to choose $m$ to be $\lceil n^{1/H} \rceil$ to obtain

$$\mathbb {P}[F^{(\delta)}_{T} \geq n]\leq\exp \left(-C^{-1} \delta^{2} T^{-2 H} n (n^{2 H \wedge1} - C' \log n)\right) , $$

which completes the proof. □