Portfolio optimisation under non-linear drawdown constraints in a semimartingale financial model

Abstract

A drawdown constraint forces the current wealth to remain above a given function of its maximum to date. We consider the portfolio optimisation problem of maximising the long-term growth rate of the expected utility of wealth subject to a drawdown constraint, as in the original setup of Grossman and Zhou (Math. Finance 3:241–276, 1993). We work in an abstract semimartingale financial market model with a general class of utility functions and drawdown constraints. We solve the problem by showing that it is in fact equivalent to an unconstrained problem with a suitably modified utility function. Both the value function and the optimal investment policy for the drawdown problem are given explicitly in terms of their counterparts in the unconstrained problem.

Appendix

We state and prove here lemmas used in the proofs in the main body of the paper. Note, however, that the first two lemmas may be of independent interest. Lemma A.1 shows that when computing CER _U , it is enough to consider wealth processes which dominate a given fraction of the numeraire. Lemma A.2 studies the convergence $\mathrm {CER}_{U_{n}}\to \mathrm {CER}_{U}$ as U _n →U.

Recall the general setup introduced in Sect. 2 and the objectives CER and given in (2.2), (2.3) and (5.1), respectively.

Lemma A.1

Let U be a continuous nondecreasing function with a well-defined locally bounded right derivative $U'_{+}$. Assume U is either positive or negative and satisfies

$$ \limsup_{x\to\infty} \frac{xU'_+(x)}{|U(x)|} < \infty. $$

(A.1)

Then for any 0<y<v ₀,

Further, if N is nondecreasing, or in general if $N_{t}\ge\underline {N}$ for all t≥0 and some constant $\underline{N}>0$, then also

Proof

For $V \in \mathcal {A}(v_{0})$ and some 0<ε<1, consider the process $\tilde{V} \in \mathcal {A}(v_{0})$ given by⁴ $\tilde{V}_{t} = \varepsilon v_{0} + (1-\varepsilon) V_{t} \geq\varepsilon v_{0}$, t≥0. As U satisfies (A.1) and $\tilde{V}_{t} \geq \varepsilon v_{0}$, we are able to use Lemma A.3 below to deduce that for some $\gamma\in\mathbb{R} \setminus\{0\}$,

$$ (1-\varepsilon)^{\gamma}U\left(V_t\right)\leq(1-\varepsilon)^{\gamma }U\Big(\frac{1}{1-\varepsilon}\tilde{V}_t\Big) \leq U (\tilde{V}_t ), $$

where we used $\frac{1}{1-\varepsilon}\tilde{V} \geq V$. Taking expectations, applying $\frac{1}{t} \log$ and taking the limit as t→∞, we deduce that

$$ \mathcal {R}_U(V)\leq \mathcal {R}_U(\tilde{V}). $$

Thus, taking ε=y/v ₀ we obtain

$$ \sup_{V \in \mathcal {A}(v_0)} \mathcal {R}_U (V) \leq \sup_{V \in \mathcal {A}(v_0), V \geq y} \mathcal {R}_U (V). $$

The reverse inequality is trivial and the first equality in the statement follows.

Now consider $V \in \mathcal {A}(v_{0})$ such that V≥y for some y>0. Note that we have $V^{1}:=\frac{1}{v_{0}}V\in \mathcal {A}(1)$. If v ₀>1, Lemma A.3 and monotonicity of U yield, for some $\gamma\in \mathbb {R}\setminus\{0\}$,

$$ U(V^1_t)\leq U(V_t) = U(v_0 V^1_t) \leq v_0^{\gamma} U(V^1_t),\quad t\geq0. $$

If 0<v ₀<1, we obtain similarly

$$ v_0^\gamma U(V^1_t)=v_0^\gamma U\bigg(\frac{1}{v_0}V_t \bigg)\leq U(V_t)\leq U(V^1_t). $$

It follows that

$$ \sup_{V \in \mathcal {A}(v_0),\ V \geq y} \mathcal {R}_U (V) = \sup_{V^1 \in \mathcal {A}(1),\ V^1 \geq y/v_0} \mathcal {R}_U (V^1). $$

The equality CER _U (1)=CER _U (v ₀) now follows from the first equality in the statement which we established above.

For the second pair of equations in the statement of the lemma, we have $\tilde{V}_{t}N_{t} = \varepsilon v_{0} N_{t} + (1-\varepsilon ) V_{t}N_{t} \geq\varepsilon v_{0} N_{t} \geq\underline{N}\varepsilon v_{0}>0$, and the arguments are then entirely analogous. □

Lemma A.2

Let U _n ,U be nondecreasing functions of the same sign, continuous with a well-defined locally bounded right derivative, and satisfying Assumption 4.1 and (A.1). Assume further that for some c,c ₁>0 and some 0<ν<1,

$$ \forall \delta>0\ \exists n_\delta\ \forall n\geq n_\delta\quad c_1 U(x)\leq U_n(x)\leq U(cx^{1+\delta}),\quad x\geq\nu. $$

If U _n ,U are negative, we have for any v ₀>0 that

(A.2)

If U _n ,U are positive, the above holds assuming that CER _G (1)<∞, where we define G(x):=U(x)^1+δ for some δ>0. Consequently, we then have CER _U (1)<∞.

If N is bounded away from zero, $N_{t}\geq\underline{N}$, t≥0 for some $\underline{N}>0$, then the above results hold with CER replaced by .

Proof

We prove the statements for CER and simultaneously. They follow, respectively, by taking ξ=V and ξ=VN, $V\in \mathcal {A}(1)$, in what follows. When considering the latter, the assumption that N is bounded away from zero is in place. Observe that by Lemma A.1, it is sufficient to consider ξ≥ν.

Assume that for n and K large enough and any ξ≥ν, we have

(A.3)

Take δ>0 and large K,T,n so that the assumptions yield

where we used Lemma A.3 below to obtain the last inequality. Recall that we define logx=−log(−x) for x<0. Taking expectations, applying $\frac{1}{T}\log$ and taking the limit as T→∞ in the above, we conclude thanks to (A.3) that

$$\mathcal {R}_U(\xi)\leq \mathcal {R}_{U_n}(\xi)\leq \mathcal {R}_U(\xi)+|\gamma|\delta\log K. $$

This is true for n large enough and any ξ, and hence also when we take the supremum over ξ. We deduce (A.2) by letting δ→0.

It remains to argue (A.3). We prove this separately for positive and negative U. Consider first U _n ,U≥0. Assumption 4.1 implies that there exist $\tilde{c} >0$ and ε>0 such that $\nu\leq x < \tilde{c}U(x)^{1/\varepsilon}$. For any δ′>0, using Lemma A.3, we obtain

From the proof of Lemma A.3 below, it is clear that U and γ have the same sign, and we conclude that for some δ′≤δ, we have

$$ U\big(x^{1+\delta^\prime}\big)^{1+\delta^\prime} \leq c_2 U\big (x^{1+\delta}\big)=c_2 G(x), \quad x \geq\nu, $$

(A.4)

for some c ₂>0.

Using Chebyshev’s inequality we obtain

$$ \mathbb {P}\big[\xi_T > K^T\big] \leq\frac{\mathbb {E}[ U_n(\xi_T)]}{U_n(K^T)} \leq\frac {\mathbb {E}[ U_n(\xi_T)]}{\tilde{c}^{-\varepsilon} c_1 K^{\varepsilon T}}. $$

In the last inequality, we used again the fact that U _n (x)≥c ₁ U(x) for x≥ν and that $U(x) \geq\tilde{c}^{-\varepsilon} x^{\varepsilon}$. Take n>n _δ′ with δ′ as in (A.4). Combining the above and using twice Hölder’s inequality with p=1+δ′, 1/p+1/q=1, we obtain

Let C _G denote CER _G (1) or , depending on whether we consider ξ=V or ξ=VN. Let γ′ be the constant resulting from Lemma A.3 applied with x ₀=v ^1+δ′. We can then continue the above chain of inequalities with

where to get the second inequality we used (A.4) and the fact that for any κ>0, for T large enough, we have $\mathbb {E}[G(\xi_{T})]\leq\exp((C_{G}(1)+\kappa)T)$. Above, c ₃ is a positive constant which can be made explicit from the above computation. For K large enough, the above is decreasing exponentially in T. Combining the two displays above, we conclude that for any κ>0, n>n _δ′, K large enough and all T large enough, we have and hence

where we wrote $\kappa=\frac{\kappa}{ \mathbb {E}[U_{n}(\xi_{T})]}\mathbb {E}[U_{n}(\xi _{T})]\leq\frac{\kappa}{U_{n}(\nu)}\mathbb {E}[U_{n}(\xi_{T})]$ and used the assumption U _n ≥c ₁ U. The first equality in (A.3) now follows by taking expectations, applying $\frac{1}{T}\log$ and letting T→∞. Analogous but simplified arguments yield the second equality in (A.3).

It remains to show (A.3) when U _n ,U<0. We detail the arguments for U _n and the first equality in (A.3). Obviously , so (A.3) holds if $\zeta _{n}:=\mathcal {R}_{U_{n}}(\xi)=\infty$. Assume now that ζ _n <∞ and note also that ζ _n ≥0 since ξ≥ν. Using Assumption 4.1 on U, we see that there exists ε>0 such that $0>U(x)\geq-\tilde{c} x^{-\varepsilon}$, x≥ν. This yields

It follows that

where $c_{4}=c_{1}\tilde{c}$, we took κ>0 arbitrary and T large enough. Taking finally K>exp((ζ _n +κ)/ε), applying $\frac{1}{T}\log$ and letting T→∞, we see that (A.3) holds. □

The following lemma is a slight extension of the first part of Lemma 6.3 in Kramkov and Schachermayer [24].

Lemma A.3

Let $U: (0,\infty) \rightarrow\mathbb {R}$ be a continuous nondecreasing function, either strictly positive or strictly negative, with a well-defined and locally bounded right derivative and which satisfies (A.1). Then for any x ₀>0, there exists $\gamma\in\mathbb{R}\setminus\{0\}$ such that

$$ U(x) \leq U(\lambda x) \leq\lambda^ {\gamma} U(x) \quad \mbox{\textit{for all}}\ \lambda> 1, x\geq x_0. $$

Proof

Let x ₀>0. From (A.1), the fact that U is monotone and of constant sign, and since $U'_{+}$ is locally bounded, there exists $\gamma\in\mathbb{R}\setminus\{0\}$ such that

$$ \frac{xU'_+(x)}{\gamma U(x)} < 1, \quad x \geq x_0, $$

where γ has the same sign as U.

Fix x≥x ₀ and define functions F(λ):=U(λx) and G(λ):=λ ^γ U(x) for λ>1. Then F(1)=G(1) and $F'_{+}(1) = xU'_{+}(x) < \gamma U(x) = G'_{+}(1)$. Hence F(λ)<G(λ) for λ∈(1,1+ε) for some ε>0. Assume that F(λ)>G(λ) for some λ∈(1,∞); then from continuity of F and G, there exists a point λ ^∗>1 such that F(λ ^∗)=G(λ ^∗) and $F'_{+}(\lambda^{*}) \geq G'_{+}(\lambda^{*})$. But

$$ F'_+(\lambda^*) = x U'_+(\lambda^* x) < \frac{\gamma}{\lambda ^*}U(\lambda^* x) = \frac{\gamma}{\lambda^*} F(\lambda^*) = \frac{\gamma }{\lambda^*} G(\lambda^*) = G'_+(\lambda^*), $$

which gives a contradiction. □

Lemma A.4

Suppose U is a utility function which satisfies the first condition of Assumption 5.1. Then for any x ₀>0,ε>0, there exist c ₋,c ₊>0 such that

$$c_- H^{({\gamma(1-\varepsilon)})}(x)\leq U(x)\leq c_+ H^{({\gamma (1+\varepsilon)})}(x),\quad x\geq x_0. $$

Proof

Let us consider U>0, the case of U<0 being entirely analogous. The assumption on U means that for any ϵ>0, there exists y ₀>0 such that

$$ \frac{yU'_+(y)}{U(y)} \in\big(\gamma(1- \varepsilon), \gamma(1+ \varepsilon)\big) \quad\text{for}\ y \in[y_0, \infty). $$

For x≥y ₀ we express U(x) as

$$ U(x)= U(y_0)\exp\left\{ \int_{y_0}^{x} \frac{yU'_+(y)}{U(y)}\frac{\mathrm {d}y}{y} \right\}, $$

which establishes the claim for x≥y ₀ with c _±=U(y ₀)/H ^{(−γ(1±ε))}(y ₀). It follows that the claim holds for x≥x ₀ for any x ₀>0 with

 □