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Outperformance portfolio optimization via the equivalence of pure and randomized hypothesis testing

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Abstract

We study the portfolio optimization problem of maximizing the outperformance probability over a random benchmark through dynamic trading with a fixed initial capital. Under a general incomplete market framework, this stochastic control problem can be formulated as a composite pure hypothesis testing problem. We analyze the connection between this pure testing problem and its randomized counterpart, and from the latter we derive a dual representation for the maximal outperformance probability. Moreover, in a complete market setting, we provide a closed-form solution to the problem of beating a leveraged exchange traded fund. For a general benchmark under an incomplete stochastic factor model, we provide the Hamilton–Jacobi–Bellman PDE characterization for the maximal outperformance probability.

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Acknowledgements

The authors would like to thank the Editor and two anonymous referees for their insightful remarks, as well as Jun Sekine, Birgit Rudloff, and James Martin for their helpful discussions.

Tim Leung’s work is partially supported by NSF grant DMS-0908295. Qingshuo Song’s work is partially supported by SRG grant 7002818 and GRF grant CityU 103310 of Hong Kong.

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Appendix

Appendix

1.1 A.1 The role of \(\operatorname {co}(\mathcal{H})\) in V(x)

In this example, we show that the representation of V(x) in (2.8) does not hold if \(\operatorname {co}(\mathcal{H})\) is replaced by the smaller set \(\mathcal{H}\).

Example A.1

Let Ω=[0,1], and let \(\mathbb{P}\) be Lebesgue measure, i.e., \(\mathbb{P}[(a,b)] = b-a\) for ab. Let \(\mathcal{G} = \{G \equiv1\}\) and \(\mathcal{H} = \{H_{1}, H_{2}\}\) with

$$H_1(\omega) = I_{\{1/2 \le\omega\le1\}} + 1, \qquad H_2(\omega) = I_{\{0 \le\omega\le1/2\}} + 1, \quad\omega\in \varOmega. $$

For the randomized hypothesis testing problem (2.3) with x=1, it is easy to see, e.g., from (2.8), that

$$V(1) = \inf_{a \ge0} \Big\{xa + \inf_{\mathcal{G} \times \operatorname {co}(\mathcal{H})} \mathbb{E} [(G- a H)^+]\Big\} \Big|_{x=1} = \frac{2}{3}, $$

along with the optimizers

$$\hat{G} = 1, \qquad\hat{H} = \frac{1}{2} (H_1 + H_2), \qquad\hat {a} = 2/3. $$

In this simple example, the uniqueness follows immediately.

Now, if one switches from \(\operatorname {co}(\mathcal{H})\) to \(\mathcal{H}\) in (2.8), then a strictly larger value will result; in fact,

$$\inf_{a \ge0} \Big\{xa + \inf_{\mathcal{G} \times\mathcal{H}} \mathbb{E} [(G- a H)^+]\Big\} \Big|_{x=1} = \frac{3}{4}>\frac{2}{3}=V(1). $$

1.2 A.2 On the positivity of \(\hat{a}\)

Theorem 2.3 shows that the minimizer \(\hat{a}\) takes values in [0,∞) rather than (0,∞) as claimed in Proposition 3.1 and Lemma 4.3 in Cvitanić and Karatzas [7]. To illustrate this issue, we first give an example where \(\hat{a}\) takes the value zero. Then we provide a sufficient condition for \(\hat{a}>0\).

Example A.2

Let \(N_{g} := \bigcap_{G\in\mathcal{G}} \{G=0\}\) and x>0.

  1. (i)

    If \(\mathbb{P}[N_{g}]=1\), then \(\mathbb{E} [(G- aH)^{+}] = 0\) for all G,H,a. Thus, \(\hat{a} = 0\) is the unique minimizer of \(\{xa + \inf_{\mathcal{G} \times\mathcal{H}} \mathbb{E} [(G- a H)^{+}]\}\).

  2. (ii)

    If \(0<\mathbb{P}[N_{g}] <1\) and \(x > \sup_{H\in\mathcal{H}} \mathbb{E} [(H I_{N_{g}^{c}})]\), then there also exists a counterexample such that \(\hat{a} = 0 \) minimizes \(\{xa + \inf_{\mathcal{G} \times\mathcal{H}} \mathbb{E} [(G- a H)^{+}]\}\). Let us consider the following scenario. The sample space is Ω=[0,1], \(\mathbb{P}\) is Lebesgue measure on [0,1], \(\mathcal{G} = \{G\}\) with G=2I [1/2,1], and \(\mathcal{H} = \{H\}\) with H≡1. Then one can check that N g ={G=0}=[0,1/2), which in turn implies that \(G = I_{N_{g}^{c}} / \mathbb{P}(N_{g}^{c})\), and

    $$\begin{aligned} xa + \inf_{\mathcal{G} \times\mathcal{H}} \mathbb{E} [(G- a H)^+] &= xa + \mathbb{E}[(G- zH)^+] \\ &= \left\{ \begin{array}{l@{\quad}l} xa & \text{if}\ a \geq\frac{1}{P[N_g^c]},\\ 1+ a (x - \mathbb{P}[N_g^c] ) & \text{if}\ 0\leq a < \frac{1}{\mathbb{P}[N_g^c]}. \end{array} \right. \end{aligned}$$
    (A.1)

    Since \(x > \sup_{H\in\mathcal{H}}\mathbb{E} [HI_{N_{g}^{c}}] = \mathbb{P}[N_{g}^{c}]\), \(\hat{a} = 0\) is the unique minimizer of (A.1).

Proposition A.3

If

$$ 0<x<\sup_{\mathcal{H}} \mathbb{E} \Big[ H I_{ \underset{G\in\mathcal{G}}{\cap} \{G>0\} }\Big], $$
(A.2)

then there exists \((\hat{G}, \hat{H}, \hat{a}, \hat{X}) \in \mathcal{G} \times\overline{\operatorname {co}(\mathcal{H})} \times(0,\infty) \times \mathcal{X}_{x}\) satisfying (2.5)(2.7). In particular,

$$\hat{a} = \mathop {\mathrm {arg\,min}}_{a \geq0}\Big\{xa + \inf_{\mathcal{G} \times \overline{\operatorname {co}(\mathcal{H})}} \mathbb{E} [(G- a H)^+] \Big\}>0. $$

Proof

Define the function \(f_{x}(a) := xa + \inf_{\mathcal{G} \times \overline{\operatorname {co}(\mathcal{H})}} \mathbb{E} [(G- a H)^{+}]\), which is Lipschitz-continuous (see Lemma 4.1 of [7]). Since \(f_{x}(0) = \inf_{\mathcal{G}} \mathbb{E}[G]\) is in [0,∞) and lim a→∞ f x (a)=∞, there exists a finite \(\hat{a} \ge0\) that minimizes f x (a).

Now suppose that \(\hat{a} = 0\) is a minimizer of f x (a). Then it follows that f x (a)≥f x (0), ∀a>0, which leads to

$$\begin{aligned} xa &\ge\displaystyle\inf_{\mathcal{G}} \mathbb{E}[G] - \inf_{\mathcal{G} \times\overline{\operatorname {co}(\mathcal{H})}} \mathbb{E} [(G- aH)^+] \\ \displaystyle & \ge\mathbb{E}[\tilde{G}] - \displaystyle\inf_{\overline{\operatorname {co}(\mathcal{H})}} \mathbb{E} [(\tilde{G}- aH)^+] \\ &\ge\displaystyle a \sup_{\overline{\operatorname {co}(\mathcal{H})}} \mathbb{E} \big[H I_{\{ \tilde{G} \ge aH\}}\big] \ge\displaystyle a \sup_{\mathcal{H}} \mathbb{E} \big[H I_{\{\tilde{G} \ge aH\} }\big]. \end{aligned}$$
(A.3)

In (A.3), \(\tilde{G}\) minimizes \(\mathbb{E}[G]\) over \(\mathcal{G}\), and its existence follows from convexity and closedness of \(\mathcal{G}\). Taking the limit a↘0 yields a contradiction to (A.2) because

$$x\ge\sup_{\mathcal{H}} \mathbb{E}\big[H I_{\{\tilde{G} >0\}}\big ] \ge \sup_{\mathcal{H}} \mathbb{E}\big[ H I_{\cap_\mathcal{G} \{G>0\} }\big]. $$

Hence, we conclude that \(\hat{a}>0\). □

1.3 A.3 Counterexample for Remark 2.5

Let Ω={ω 1,ω 2}, P[{ω 1}]=P[{ω 2}]=1/2. Then any random variable in \(\mathcal{G}, \mathcal{H}\) or in \(\mathcal{X}_{x}\), \(\mathcal{I}_{x}\) can be represented as a point in \({\mathbb{R}}^{2}\). Let \(\mathcal{H}\) be the line segment connecting (2,4) and (6,2), and \(\mathcal{G} = \{(2,2)\}\). Given x≥0, \(\mathcal{X}_{x}\) is the convex quadrangle with four vertices (0,0),(x/3,0),(x/5,2x/5), (0,x/2) intersected with {(x 1,x 2) | 0≤x 1,x 2≤1}. For all \(H =(h_{1},h_{2})\in\mathcal{H}\) and X=(x 1,x 2), the constraint \(\mathbb{E}[HX] \leq x\) implies that \(\frac{h_{1}}{2}x_{1} + \frac{h_{2}}{2}x_{2} \leq x\). The set \(\{(x_{1},x_{2}): \frac{h_{1}}{2} x_{1} + \frac{h_{2}}{2}x_{2} \le x\}\) is the lower half-plane bounded by the line h 1 x 1+h 2 x 2=2x, which passes through (x/5,2x/5) since h 1+2h 2=5. Hence, we have

$$V(x)=\sup_{(x_1,x_2)\in\mathcal{X}_x} (x_1+x_2), \quad\text{and} \quad V_1(x)=\sup_{(x_1,x_2)\in\mathcal{I}_x} (x_1+x_2), $$

where \(\mathcal{I}_{x}=\mathcal{X}_{x}\cap\{(0,0),(0,1),(1,0),(1,1)\}\). In summary, the values are given by the following table:

x

V(x)

V 1(x)

0≤x<2

\(\frac{3}{5}x\)

0

\(2\leq x < \frac{5}{2}\)

\(\frac{3}{5}x\)

1

\(\frac{5}{2}\leq x < 4\)

\(\frac{x}{3}+\frac{2}{3}\)

1

x≥4

2

2

By inspecting the values of V 1(x) we see that its smallest concave majorant must take the value \(\frac{x}{2}\) in [0,4]. Therefore, V(x) is not the smallest concave majorant of V 1(x).

1.4 A.4 Counterexample for Remark 2.11

With reference to Theorem 2.3, we show via an example that one cannot remove the independence requirement in (C2) of Theorem 2.10 when \(\mathcal{G}\) and \(\mathcal{H}\) are not both singletons.

Example A.4

Let Ω={0,1}×[0,1], \(\mathcal{F}_{T} = \mathcal{B}(\varOmega)\). Let μ be Lebesgue measure on [0,1]. Define \(\mathbb{P}\) by

$$\mathbb{P}[\{0\} \times A] = \mathbb{P}[\{1\} \times A] = \frac{1}{2} \mu(A) \quad\forall A\in\mathcal{B}([0,1]). $$

Let \(H_{0}: \{0,1\} \to\mathbb{R}\) be given by H 0(0)=1/2 and H 0(1)=3/2, and \(f:[0,1]\to \mathbb{R}\) be an arbitrarily fixed probability density function. Define the set

$$\mathcal{H} = \{H: \varOmega\to\mathbb{R}: H(\alpha, a) = H_0(\alpha) f(a),\ (\alpha, a) \in\varOmega\} $$

and the singleton \(\mathcal{G} = \{G \equiv1\}\). Let U be a uniform random variable on \((\varOmega, \mathcal{F}_{T}, \mathbb{P})\), so that \(\mathbb{P}[U\le a] = a\) for a∈[0,1].

The pure hypothesis testing problem is

$$V_1 = \sup_{A\in\mathcal{F}_T} \mathbb{E}[I_A] \quad\text{subject to}\quad \sup_{H\in\mathcal{H}} \mathbb{E}[ HI_A] \le1/2. $$

Direct computation gives the success set \(\hat{A} = \{0\}\) and the value of the pure hypothesis test V 1=1/2. On the other hand, the randomized hypothesis testing problem is

$$V = \sup_{X\in\mathcal{X}} \mathbb{E}[X] \quad \text{subject to}\quad \sup_{H\in\mathcal{H}} \mathbb{E}[ H X] \le1/2. $$

We find that \(\hat{H}(\alpha, a) = H_{0}(\alpha)\) and \(\hat{X} = I_{ \{\alpha= 0\}} + 1/3 I_{\{\alpha= 1\}}\) solve this randomized hypothesis test with the optimal value V=2/3.

This shows that the values of the pure and randomized hypothesis tests are different. If one were to construct an indicator version of the randomized test as in (2.13), namely

$$\bar{X} := I_{\{\alpha= 0\}} + I_{\{\alpha= 1\}} I_{\{U<1/3\}}, $$

then although this test \(\bar{X} \) still satisfies \(\mathbb{E}[ \hat{H} \bar{X}] = 1/2\), it does not solve either the pure or the randomized hypothesis test. Indeed, for \(\tilde{H} (\alpha, a) = 3 I_{a<1/3} H_{0}(\alpha) \in \mathcal{H}\), we observe the violation \(\mathbb{E}[ \tilde{H} \bar{X}] = 1> 1/2\).

1.5 A.5 A property of nondegenerate martingales

On the probability space \((\varOmega, \mathcal{F}, \mathbb{P})\) with filtration \((\mathcal{F}_{t})_{0\le t\le1}\), we denote by W a standard Brownian motion. Let Y be a \((\mathbb{P}, \mathcal{F}_{t})\)-martingale defined by

$$Y_t = \int_0^t \sigma_r \,d W_r, \quad t\in[0, 1], $$

where (σ t ) is a bounded \((\mathcal{F}_{t})\)-adapted process.

Proposition A.5

Assume that c<σ t <C for some positive constants c and C. Then

$$\mathbb{P}[Y_1 = b] = 0 $$

for all constants b.

To prove this proposition, we use the following two facts. We first define the function \(f: \mathbb{R}^{+} \times\mathbb{R}^{+} \times\mathbb{R} \to[0,1]\) by

$$f(x,y,u) = \mathbb{P}[W_t = u \hbox{ for some } t\in(x,y)]. $$
  1. 1.

    By direct computation we obtain

    $$\sup_{u\in\mathbb{R}} f(x,y,u) = f(x,y,0) < 1. $$
  2. 2.

    By a scaling argument we have

    $$f(\lambda x, \lambda y, u) = f\bigg(x,y, \frac{u}{\sqrt{\lambda}}\bigg) \quad\forall\lambda>0. $$

Now we are ready to present the

Proof of Proposition A.5

Since Y is a continuous process,

$$\{Y_1 = b\} \in\sigma(\{\mathcal{F}_t: t<1\})=: \mathcal{F}_{ 1-}. $$

By Lévy’s zero–one law we have

$$I_{\{Y_1 = b\}} = \lim_{t\uparrow1} \mathbb{P}[Y_1 = b |\mathcal{F}_t\} \quad\text{a.s.} $$

Therefore, it is enough to show that there exists a∈(0,1) such that

$$\mathbb{P}[Y_1 = b|\mathcal{F}_t] < a < 1 \quad \forall t\in(0,1). $$

Note that the martingale (Y s |Y t =u:s>t) has the same distribution as a time-changed Brownian motion starting from state u. Combining this with the estimate \(c^{2} (1-t) \le\int_{t}^{1} \sigma_{r}^{2} dr \le C^{2} (1-t)\), we have for some standard Brownian motion B that

$$\begin{aligned} \mathbb{P}[Y_1 = b | Y_t = u] &= \mathbb{P}\big[B_r = b - u \hbox{ for some } r \in \big( c^2(1-t), C^2(1-t)\big)\big] \\ &= f\bigg(c^2, C^2, \frac{b - u}{\sqrt{1 - t}}\bigg) \le f(c^2, C^2, 0). \end{aligned}$$

Since f(c 2,C 2,0) is independent of t and strictly less than 1, we can simply take a=f(c 2,C 2,0). □

One may wonder whether the condition on σ in Proposition A.5 can be relaxed to σ t >0 a.s. for all t. The answer is negative, as shown by the counterexample in [22].

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Leung, T., Song, Q. & Yang, J. Outperformance portfolio optimization via the equivalence of pure and randomized hypothesis testing. Finance Stoch 17, 839–870 (2013). https://doi.org/10.1007/s00780-013-0213-8

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