Hedge and mutual funds’ fees and the separation of private investments

Abstract

A fund manager invests both the fund’s assets and own private wealth in separate but potentially correlated risky assets, aiming to maximize expected utility from private wealth in the long run. If relative risk aversion and investment opportunities are constant, we find that the fund’s portfolio depends only on the fund’s investment opportunities, and the private portfolio only on private opportunities. This conclusion is valid both for a hedge fund manager, who is paid performance fees with a high-water mark provision, and for a mutual fund manager, who is paid management fees proportional to the fund’s assets. The manager invests earned fees in the safe asset, allocating remaining private wealth in a constant-proportion portfolio, while the fund is managed as another constant-proportion portfolio. The optimal welfare is the maximum between the optimal welfares of each investment opportunity, with no diversification gain. In particular, the manager does not use private investments to hedge future income from fees.

Appendix A: Verification theorems

This appendix contains the proofs of Theorems 2.1 and 2.2.

1.1 A.1 Proof of Theorem 2.1

We start by defining the following processes, which represent cumulative log-returns, before fees, on the fund and wealth, respectively:

$$\begin{aligned} R^{X}_{t} =& \int_{0}^{t}\bigg(\Big(\mu ^{X}\pi ^{X}_{s}-\frac{1}{2}(\sigma ^{X}\pi ^{X}_{s})^{2}\Big)ds + \sigma ^{X}\pi ^{X}_{s} dW_{s}^{X}\bigg),\\ R^{F}_{t} =& \int_{0}^{t}\bigg(\Big(\mu ^{F}\pi ^{F}_{s}-\frac{1}{2}(\sigma ^{F}\pi ^{F}_{s})^{2}\Big)ds + \sigma ^{F}\pi ^{F}_{s} dW_{s}^{F}\bigg),\\ R^{X}_{t,T} =& R^{X}_{T}-R^{X}_{t},\\ R^{F}_{t,T} =& R^{F}_{T}-R^{F}_{t}. \end{aligned}$$

R ^X and R ^F depend on π ^X and π ^F, respectively, and should be denoted as $R^{X,\pi ^{X}}$ and $R^{F,\pi ^{F}}$. We drop the superscripts π ^X and π ^F for ease of notation, unless it causes ambiguity.

Equations (2.1), (2.2), Proposition 7 and Lemma 8 in [12] imply that

$$\begin{aligned} X_{t} =& X_{0}e^{R^{X}_{t} - \alpha\bar{R}^{X}_{t}}, \\ \bar{X}_{t} =& X_{0}e^{(1-\alpha)\bar{R}^{X}_{t}}, \\ F_{t} =&F_{0}e^{R^{F}_{t}} + \frac{\alpha}{1-\alpha}\int_{0}^{t}e^{R^{F}_{s,t}}d\bar{X}_{s} . \end{aligned}$$

(A.1)

The discussion begins with two simple lemmas that are used often in the proof of the theorems.

Lemma A.1

Let $\left(X_{t}\right)_{t\geq0}$ and $\left(Y_{t}\right)_{t\geq0}$ be two continuous stochastic processes, and define $\bar{X}_{t} = \displaystyle\max_{0\leq s\leq t}X_{s}$ and $\underline{Y}_{t} = \displaystyle\min_{0\leq s\leq t}Y_{s}$. Then $\bar{X}_{t} + \underline{Y}_{t} \leq(\overline{X+Y})_{t}$.

Proof

Since $(\overline{X+Y})_{t} \geq X_{s} + Y_{s} \geq X_{s} + \underline{Y}_{t}$ for every 0≤s≤t, it follows that $(\overline {X+Y})_{t}\geq\bar{X}_{t} + \underline{Y}_{t}$. □

Lemma A.2

Let $\left(\mathcal{G}_{t}\right)_{0\leq t\leq T}$ be a continuous filtration, $\mathcal{F} \subseteq\mathcal{G}_{0}$ a σ-algebra, and denote by $\mathbb{E}_{\mathcal{F}}$ and $\mathbb{E}_{\mathcal{G}_{t}}$ the conditional expectation with respect to $\mathcal{F}$ and $\mathcal {G}_{t}$, respectively. If $\left(A_{t}\right)_{t\geq0}$ is a continuous and increasing process adapted to $(\mathcal{G}_{t})$ for 0≤t≤T, and $\left(X_{t}\right)_{t\geq0}$ is a positive, continuous stochastic process such that $\mathbb{E}_{\mathcal{G}_{t}}[X_{t,T}] \leq C$, for every 0≤t≤T and some constant C, where $X_{t,T} = \frac{X_{T}}{X_{t}}$, then $\mathbb{E}_{\mathcal{F}}[\int_{0}^{T}X_{t,T}dA_{t}] \leq C\mathbb {E}_{\mathcal{F}}[A_{T}-A_{0}]$.

Proof

Since A is an increasing process, for a partition $0 = t^{n}_{0} < t^{n}_{1} < \cdots< t^{n}_{n} = T$ of [0,T] with $t^{n}_{k}= \frac{k}{n}T$ for 1≤k≤n,

$$ \int_{0}^{T}X_{t,T}dA_{t} = \lim_{n\rightarrow\infty}\sum _{k=1}^{n}X_{t^{n}_{k},T}\big(A_{t^{n}_{k}} - A_{t^{n}_{k-1}}\big). $$

Thus,

$$ \mathbb{E}_{\mathcal{F}}\left[\int_{0}^{T}X_{t,T}dA_{t}\right] = \mathbb {E}_{\mathcal{F}}\bigg[\lim_{n\rightarrow\infty}\sum _{k=1}^{n}X_{t^{n}_{k},T}\big(A_{t^{n}_{k}} - A_{t^{n}_{k-1}}\big)\bigg], $$

and by Fatou’s lemma and the tower property of conditional expectation, the right-hand side is less than or equal to

$$\begin{aligned} &\liminf_{n\rightarrow\infty}\mathbb{E}_{\mathcal{F}}\bigg[\sum _{k=1}^{n}X_{t^{n}_{k},T}\big(A_{t^{n}_{k}} - A_{t^{n}_{k-1}}\big)\bigg] \\ &\quad = \liminf_{n\rightarrow\infty}\mathbb{E}_{\mathcal{F}}\bigg[\sum _{k=1}^{n}\mathbb{E}_{\mathcal{G}_{t^{n}_{k}}}\big[X_{t^{n}_{k},T}\big]\big(A_{t^{n}_{k}} - A_{t^{n}_{k-1}}\big)\bigg]. \end{aligned}$$

(A.2)

Since $\mathbb{E}_{\mathcal{G}_{t}}[X_{t,T}] \leq C$ for every 0≤t≤T, (A.2) is less than or equal to

$$\begin{aligned} &C\liminf_{n\rightarrow\infty}\mathbb{E}_{\mathcal{F}}\bigg[\sum _{k=1}^{n}\big(A_{t^{n}_{k}} - A_{t^{n}_{k-1}}\big)\bigg]\\ &\quad = C\liminf_{n\rightarrow\infty}\mathbb{E}_{\mathcal{F}}\left[A_{T} - A_{0}\right] = C\mathbb{E}_{\mathcal{F}}\left[A_{T} - A_{0}\right]. \end{aligned}$$

 □

The proof of Theorem 2.1 is divided into two steps. First, any investment policies π ^X and π ^F satisfy the estimate

$$\operatorname {ESR}_{\gamma}(\pi^{X},\pi^{F}) \leq\max\left(\frac{(1-\alpha)(\nu ^{X})^2}{2\gamma^{*}},\frac{(\nu ^{F})^2}{2\gamma}\right), $$

which is proved in Lemma A.3. Second, this upper bound is achieved by the candidate optimal policies in (2.6) and (2.7), as proved in Lemma A.8.

Lemma A.3

For a hedge fund manager compensated by high-water mark performance fees with rate 0<α<1, the $\operatorname {ESR}$ induced by any investment strategies π ^X and π ^F satisfies

$$ \operatorname {ESR}_{\gamma}(\pi ^{F},\pi ^{F}) \leq\lambda_1 = \max\left(\frac{(1-\alpha)(\nu ^{X})^{2}}{2\gamma^{*}},\frac{(\nu ^{F})^{2}}{2\gamma}\right), \quad \textit{for\ any}\ 0 < \gamma\leq1. $$

Proof

We prove this lemma for logarithmic utility and power utility, respectively.

Case 1. Logarithmic utility. For convenience of notation, define

$$\tilde{X}_{t}= \begin{cases} 0 & \mbox{for }t < 0,\\ \textstyle{\frac{1-\alpha}{\alpha}}F_{0} + \bar{X}_{t}-X_{0} & \mbox{for } t\geq0. \end{cases} $$

Then $\tilde{X}$ is an increasing process, which has a jump at t=0 and then grows with $\bar{X}$. From (A.1), $F_{t} = \frac{\alpha}{1-\alpha}\int _{0}^{T}e^{R^{F}_{t,T}}d\tilde{X}_{t}$, and (2.5) can be rewritten as

$$\begin{aligned} &\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln \left(\frac{\alpha}{1-\alpha}\int_{0}^{T}e^{R^{F}_{t,T}}d\tilde {X}_{t}\right)\right] \\ &\quad = \limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln \int_{0}^{T}e^{R^{F}_{t,T}}d\tilde{X}_{t}\right] \\ &\quad = \lambda_1 +\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln\int _{0}^{T}e^{-\lambda_1 T + R^{F}_{t,T}}d\tilde{X}_{t}\right]. \end{aligned}$$

(A.3)

Since d〈W ^X,W ^F〉 _t =ρdt, we can rewrite $W^{X}_{t} = \rho W^{F}_{t} + \sqrt{1-\rho^{2}}W^{\perp}_{t}$, where W ^⊥ is a Brownian motion independent of W ^F. Denote $\mathbb{E}_{W^{\perp}_{T}}$ as the expectation conditional on $(W^{\perp}_{s})_{0\leq s\leq T}$ (the whole trajectory of W ^⊥ until T). Letting

$$ L_T = \int_{0}^{T}e^{R^{F}_{t,T} - \int_{t}^{T}(\frac{1}{2}(\nu ^{F})^{2}ds +\nu ^{F}dW^{F}_{s})}e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline {\rho W^{F}_{t}}}d\tilde{X}_{t} , $$

Lemma A.4 below implies that

$$\begin{aligned} \limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln\int _{0}^{T}e^{-\lambda_1 T+R^{F}_{t,T}}d\tilde{X}_{t}\right] \leq&\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln L_T\right] \\ =& \limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\big[\mathbb {E}_{W^{\perp}_{T}}\left[\ln L_T\right]\big]\\ \leq&\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\big[\ln\mathbb {E}_{W^{\perp}_{T}}\left[L_T\right]\big], \end{aligned}$$

where the first equality follows from the tower property of conditional expectation, and the second inequality from Jensen’s inequality. Then Lemma A.5 below implies that $M_{t} = e^{R^{F}_{0,t} - \int_{0}^{t}(\frac{1}{2}(\nu ^{F})^{2}ds + \nu ^{F}dW^{F}_{s})}$ is a supermartingale with respect to the filtration generated by $(W^{F}_{s})_{0\leq s\leq t}$ and $(W^{\perp}_{s})_{0\leq s\leq T}$ (the past of W ^F and W ^⊥, plus the future of W ^⊥). Thus,

$$ \mathbb{E}_{W^{\perp}_{T},W^{F}_{t}}\left[e^{R^{F}_{t,T} - \int _{t}^{T}(\frac{1}{2}(\nu ^{F})^{2}ds +\nu ^{F}dW^{F}_{s})}\right] \leq1,\quad 0\leq t \leq T. $$

(A.4)

In addition, since $A_{t} = \int_{0}^{t}e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}}d\tilde{X}_{t}$ is an increasing process, (A.4) and Lemma A.2 imply that

$$\begin{aligned} &\mathbb{E}_{W^{\perp}_{T}}\left[L_T\right] \leq\mathbb{E}_{W^{\perp }_{T}}\left[\int_{0}^{T}e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}}d\tilde{X}_{t}\right]. \end{aligned}$$

(A.5)

Then, from (A.3) and (A.5), it follows that

$$\begin{aligned} &\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln F_{T}\right ]\\ &\quad \leq\lambda_1 +\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\bigg[\ln\mathbb {E}_{W^{\perp}_{T}}\Big[\int_{0}^{T}e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}}d\tilde{X}_{t}\Big]\bigg]. \end{aligned}$$

Then Lemma A.6 below proves that

$$ \limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\bigg[\ln\mathbb {E}_{W^{\perp}_{T}}\Big[\int_{0}^{T}e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}}d\tilde{X}_{t}\Big]\bigg] \leq0, $$

whence $\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln F_{T}\right]\leq\lambda_{1}$.

Case 2. Power utility. Define $\tilde{F}_{t} = F_{t} + \bar{X}_{t}$. Then $\tilde{F}_{t} \geq F_{t}$ and $\tilde{F}_{t} \geq\bar{X}_{t}$ for every t≥0. Thus, the ESR of F is less than or equal to the ESR of $\tilde{F}$. Lemma A.7 below shows that this upper bound is also less than or equal to λ ₁. □

Lemma A.4

For λ ₁ defined in Lemma A.3,

$$\begin{aligned} &\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln\int _{0}^{T}e^{-\lambda_1 T}e^{R^{F}_{t,T}}d\tilde{X}_{t}\right]\\ &\quad \leq\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln\int _{0}^{T}e^{R^{F}_{t,T} - \int_{t}^{T}(\frac{1}{2}(\nu ^{F})^{2}ds +\nu ^{F}dW^{F}_{s})}e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}}d\tilde{X}_{t}\right]. \end{aligned}$$

Proof

Define a stochastic process by $N^{T}_{s} = W^{F}_{T} - W^{F}_{T-s}$, for 0≤s≤T, which has the same distribution as $W^{F}_{s}$. It follows that

$$\begin{aligned} &\lim_{T \rightarrow\infty}\frac{1}{T}\mathbb{E}\Big[-\nu ^{F}\bar{N}^{T}_{T}-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{T}}\Big]\\ &\quad = \lim_{T \rightarrow\infty}\frac{1}{T}\mathbb{E}\big[-\nu ^{F}\bar{N}^{T}_{T}\big] + \lim_{T \rightarrow\infty}\frac{1}{T}\mathbb{E}\Big[-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{T}}\Big]\\ &\quad =\lim_{T \rightarrow\infty}\frac{1}{T}\mathbb{E}\big[-\nu ^{F}\bar{W}^{F}_{T}\big] + \lim_{T \rightarrow\infty}\frac{1}{T}\mathbb{E}\big[-(1-\alpha)\nu ^{X}|\rho| \bar{W}^{F}_{T}\big]=0, \end{aligned}$$

where the last equality uses the fact that, for a, b≠0 and a Brownian motion W,

$$ \lim_{T \rightarrow\infty}\frac{1}{b T}\mathbb{E}[a\bar{W}_{T}] =\lim_{T \rightarrow\infty}\frac{1}{b T}\int_{0}^{\infty}\frac{a x}{\sqrt{2\pi T}}e^{-\frac{x^{2}}{2T}}dx = 0. $$

(A.6)

Thus,

$$\begin{aligned} &\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln\int _{0}^{T}e^{-\lambda_1 T+ R^{F}_{t,T}}d\tilde{X}_{t}\right] \\ &\quad =\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln\int _{0}^{T}e^{-\lambda_1 T + R^{F}_{t,T}}d\tilde{X}_{t}\right] \\ &\qquad{} +\lim_{T \rightarrow\infty}\frac{1}{T}\mathbb{E}\Big[\!-\nu ^{F}\bar{N}^{T}_{T}-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{T}}\Big] \\ &\quad \leq\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln\int _{0}^{T}e^{-\lambda_1 T-\nu ^{F}\bar{N}^{T}_{T}-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{T}} + R^{F}_{t,T}}d\tilde{X}_{t}\right]. \end{aligned}$$

(A.7)

Now note that we have $\bar{N}^{T}_{T} \geq W^{F}_{T} - W^{F}_{t}$, $\lambda_{1} T \geq\frac{1}{2}(\nu ^{F})^{2}(T-t) + \frac{1-\alpha}{2}(\nu ^{X})^{2}t$ and $(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{T}} \geq(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}$, for any 0≤t≤T. Thus,

$$\begin{aligned} &{}-\lambda_1 T-\nu ^{F}\bar{N}^{T}_{T}-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{T}} + R^{F}_{t,T} \\ &\quad \leq-\frac{(\nu ^{F})^{2}(T-t)}{2} - \frac{(1-\alpha)(\nu ^{X})^{2}t}{2} - \nu ^{F}(W^{F}_{T} - W^{F}_{t}) \\ &\qquad{}-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}} + R^{F}_{t,T} \\ &\quad =R^{F}_{t,T} - \int_{t}^{T}\left(\frac{1}{2}(\nu ^{F})^{2}ds +\nu ^{F}dW^{F}_{s}\right)-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}. \end{aligned}$$

Plugging this inequality into (A.7) yields

$$\begin{aligned} &\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln\int _{0}^{T}e^{-\lambda_1 T}e^{R^{F}_{t,T}}d\tilde{X}_{t}\right] \\ &\quad \leq\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln\int _{0}^{T}e^{R^{F}_{t,T} - \int_{t}^{T}(\frac{1}{2}(\nu ^{F})^{2}ds +\nu ^{F}dW^{F}_{s})-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline {\rho W^{F}_{t}}}d\tilde{X}_{t}\right]. \end{aligned}$$

 □

Lemma A.5

Let $\pi= \left(\pi_{t}\right)_{t\geq0}$ be adapted to $(\mathcal {F}_{t})_{t\geq0}$, the filtration generated by $(W^{F}_{s})_{0\leq s\leq t}$ and $(W^{X}_{s})_{0\leq s\leq t}$. Define $(\mathcal{G}_{t})_{t \geq0}$ as the filtration generated by $(W^{F}_{s})_{s\leq t}$ and $(W^{X}_{s})_{s\leq T}$. Then $M_{t} = e^{\int_{0}^{t}(\pi_{s} dW^{F}_{s} - \frac{1}{2}\pi _{s}^{2}ds)}$ is a supermartingale with respect to $(\mathcal {G}_{t})_{t\geq0}$.

Proof

Suppose π is a simple process, i.e., $\pi_{t} = \sum^{n}_{i=1}\tilde{\pi}_{i}\textbf {1}_{(t_{i-1},t_{i}]}$ for a partition of [0,T], 0=t ₀<t ₁<t ₂<⋯<t _n =T, and $\tilde{\pi}_{i}$ is $\mathcal {F}_{t_{i-1}}$-measurable, for i=1,…,n. Then for any 0≤s<t≤T,

$$ \mathbb{E}_{W^{X}_{T},W^{F}_{s}}\left[e^{\int_{0}^{t}(\pi_{u} dW^{F}_{u} - \frac{1}{2}\pi_{u}^{2}du)}\right] = e^{\int_{0}^{s}(\pi_{u} dW^{F}_{u} - \frac{1}{2}\pi_{u}^{2}du)}\mathbb {E}_{W^{X}_{T},W^{F}_{s}}\left[e^{\int_{s}^{t}(\pi_{u} dW^{F}_{u} - \frac{1}{2}\pi_{u}^{2}du)}\right]. $$

Since there exist 1≤k _s ≤k _t ≤n such that $t_{k_{s}-1} \leq s \leq t_{k_{s}}$ and $t_{k_{t}-1} \leq t \leq t_{k_{t}}$, s, t and all the division points in between form a partition of [s,t], denoted by s=u ₀<u ₁<u ₂<⋯<u _m =t, with π _t equal to a constant $\bar{\pi}_{j}$ (= $\tilde{\pi}_{i}$ for some k _s −1≤i≤k _t ) on (u _j−1,u _j ], for j=1,…,m. Then, since W ^F has independent increments,

$$\begin{aligned} \mathbb{E}_{W^{X}_{T},W^{F}_{s}}\Big[e^{\int_{s}^{t}(\pi_{u} dW^{F}_{u} - \frac{1}{2}\pi_{u}^{2}du)}\Big] =& \mathbb{E}_{W^{X}_{T},W^{F}_{s}}\bigg[e^{\sum_{j=1}^{m}\bar {\pi}_{j}(W^{F}_{u_{j}}-W^{F}_{u_{j-1}}) - \frac{1}{2}\bar{\pi }_{j}^{2}(u_{j}-u_{j-1})}\bigg]\\ =& \prod_{j=1}^{m}\mathbb{E}_{W^{X}_{T},W^{F}_{s}}\Big[e^{\bar{\pi }_{j}(W^{F}_{u_{j}}-W^{F}_{u_{j-1}}) - \frac{1}{2}\bar{\pi }_{j}^{2}(u_{j}-u_{j-1})}\Big] \leq1, \end{aligned}$$

and thus $\mathbb{E}_{W^{X}_{T},W^{F}_{s}}[e^{\int_{0}^{t}(\pi_{u} dW^{F}_{u} - \frac{1}{2}\pi_{u}^{2}du)}]\leq e^{\int_{0}^{s}(\pi_{u} dW^{F}_{u} - \frac{1}{2}\pi_{u}^{2}du)}$.

For a general π, from the definition of the stochastic integral, there exists a sequence of simple processes $(\pi^{n})_{n=1}^{\infty}$ such that

$$ \int_{0}^{t}\left(\pi^{n}_{s} dW^{F}_{s} - \frac{1}{2}(\pi _{s}^{n})^{2}ds\right) \stackrel{n\uparrow\infty}{\longrightarrow} \int_{0}^{t}\left(\pi_{s} dW^{F}_{s} - \frac{1}{2}\left(\pi_{s}\right )^{2}ds\right)\quad \text{a.s.} $$

Hence, for 0≤s≤t≤T,

$$\begin{aligned} \mathbb{E}_{W^{X}_{T},W^{F}_{s}}\left[e^{\int_{0}^{t}(\pi_{u} dW^{F}_{u} - \frac{1}{2}\pi_{u}^{2}du)}\right] =\mathbb {E}_{W^{X}_{T},W^{F}_{s}}\left[\liminf_{n\rightarrow\infty}e^{\int _{0}^{t}(\pi^{n}_{u} dW^{F}_{u} - \frac{1}{2}(\pi^{n}_{u})^{2}du)}\right]. \end{aligned}$$

By Fatou’s lemma, this is less than or equal to

$$\begin{aligned} \liminf_{n\rightarrow\infty}\mathbb{E}_{W^{X}_{T},W^{F}_{s}}\left [e^{\int_{0}^{t}(\pi^{n}_{u} dW^{F}_{u} - \frac{1}{2}(\pi _{u}^{n})^{2}du)}\right] \leq&\liminf_{n\rightarrow\infty} e^{\int_{0}^{s}(\pi^n_{u} dW^{F}_{u} - \frac{1}{2}(\pi^n_{u})^{2}du)}\\ =& e^{\int_{0}^{s}(\pi_{u} dW^{F}_{u} - \frac{1}{2}\pi_{u}^{2}du)}, \end{aligned}$$

which confirms that M _t is a supermartingale with respect to $(\mathcal{G}_{t})_{t\geq0}$. □

Lemma A.6

$\displaystyle\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\bigg[\ln \mathbb{E}_{W^{\perp}_{T}}\Big[\int_{0}^{T}e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}}d\tilde{X}_{t}\Big]\bigg] \leq0$.

Proof

By integration by parts,

$$\begin{aligned} &\int_{0}^{T}e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline {\rho W^{F}_{t}}}d\tilde{X}_{t} \\ &\quad =\frac{(1-\alpha)F_{0}}{\alpha} + \int_{0}^{T}e^{-\frac{1-\alpha }{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}}d\bar{X}_{t} \\ &\quad =\frac{(1-\alpha)F_{0}}{\alpha} + e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}T-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{T}}}\bar{X}_{T} - X_{0} \\ &\qquad{} +\frac{1-\alpha}{2}(\nu ^{X})^{2}\int_{0}^{T}e^{-\frac {1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}}\bar{X}_{t}dt \\ &\qquad{} + (1-\alpha)\nu ^{X}\int_{0}^{T}e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}}\bar{X}_{t}d(\overline {\rho W^{F}_t}). \end{aligned}$$

(A.8)

Since $\bar{X}_{t} = X_{0}e^{(1-\alpha)\bar{R}^{X}_{t}}$, Lemma A.1 implies that

$$ e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}}\bar{X}_{t} \leq X_0e^{(1-\alpha)(\overline{R^{X}_{\cdot} - \frac{(\nu ^{X})^{2}}{2}\cdot- \nu ^{X}\rho W^{F}_{\cdot}})_{t}}. $$

Thus, from (A.8),

$$\begin{aligned} &\int_{0}^{T}e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\bar{W}^{F}_{t}}d\tilde{X}_{t} \\ &\quad \leq\frac{(1-\alpha)F_{0}}{\alpha} + X_0e^{(1-\alpha)(\overline {R^{X}_{\cdot} - \frac{1}{2}(\nu ^{X})^{2}\cdot- \nu ^{X}\rho W^{F}_{\cdot }})_{T}} \\ &\qquad{} + \frac{1-\alpha}{2}(\nu ^{X})^{2}X_0\int_{0}^{T}e^{(1-\alpha )(\overline{R_{\cdot} - \frac{1}{2}(\nu ^{X})^{2}\cdot- \nu ^{X}\rho W^{F}_{\cdot}})_{t}}dt \\ &\qquad{} + (1-\alpha)\nu ^{X}X_0\int_{0}^{T}e^{(1-\alpha)(\overline {R^{X}_{\cdot} - \frac{1}{2}(\nu ^{X})^{2}\cdot- \nu ^{X}\rho W^{F}_{\cdot }})_{t}}d(\overline{\rho W^{F}_t}). \end{aligned}$$

(A.9)

Since $e^{(1-\alpha)(\overline{R^{X}_{\cdot} - \frac{1}{2}(\nu ^{X})^{2}\cdot- \nu ^{X}\rho W^{F}_{\cdot}})_{t}} \leq e^{(1-\alpha)(\overline {R^{X}_{\cdot} - \frac{1}{2}(\nu ^{X})^{2}\cdot- \nu ^{X}\rho W^{F}_{\cdot }})_{T}}$ for every t≤T,

$$\begin{aligned} &\frac{1-\alpha}{2}(\nu ^{X})^{2}X_0\int_{0}^{T}e^{(1-\alpha)(\overline {R_{\cdot} - \frac{1}{2}(\nu ^{X})^{2}\cdot- \nu ^{X}\rho W^{F}_{\cdot }})_{t}}dt \\ &\qquad{} + (1-\alpha)\nu ^{X}X_0\int_{0}^{T}e^{(1-\alpha)(\overline {R^{X}_{\cdot} - \frac{1}{2}(\nu ^{X})^{2}\cdot- \nu ^{X}\rho W^{F}_{\cdot }})_{t}}d(\overline{\rho W^{F}_t}) \\ &\quad \leq\frac{1-\alpha}{2}(\nu ^{X})^{2}X_0e^{(1-\alpha)(\overline{R^{X}_{\cdot } - \frac{1}{2}(\nu ^{X})^{2}\cdot- \nu ^{X}\rho W^{F}_{\cdot}})_{T}}T \\ &\qquad{} + (1-\alpha)\nu ^{X}X_0 e^{(1-\alpha)(\overline{R^{X}_{\cdot} - \frac{1}{2}(\nu ^{X})^{2}\cdot- \nu ^{X}\rho W^{F}_{\cdot}})_{T}}\overline{\rho W^{F}_{T}} \\ &\quad =K_TL_T, \end{aligned}$$

(A.10)

where

$$\begin{aligned} K_{T} =& \left(1 + \frac{1-\alpha}{2}(\nu ^{X})^{2}T + (1-\alpha)\nu ^{X}|\rho |\bar{W}^{F}_{T}\right)X_0,\\ L_{T} =& e^{(1-\alpha)(\overline{R^{X}_{\cdot} - \frac{1}{2}(\nu ^{X})^{2}\cdot- (1-\alpha)\rho \nu ^{X}W^{F}_{\cdot}})_{T}}. \end{aligned}$$

Thus, from (A.9) and (A.10),

$$\begin{aligned} &\mathbb{E}_{W^{\perp}_{T}}\left[\int_{0}^{T}e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}}d\tilde{X}_{t}\right] \leq \mathbb{E}_{W^{\perp}_{T}}\left[\frac{(1-\alpha)F_{0}}{\alpha} + K_TL_T\right] \\ &\quad =\frac{(1-\alpha)F_{0}}{\alpha} + \mathbb{E}_{W^{\perp}_{T}}\left [K_TL_T\right] \leq\frac{(1-\alpha)F_{0}}{\alpha} \\ &\qquad {} + \mathbb{E}_{W^{\perp }_{T}}[L^{\delta}_{T}]^{\frac{1}{\delta}}\mathbb{E}_{W^{\perp}_{T}}\Big[K_{T}^{\frac{\delta}{\delta-1}}\Big]^{\frac{\delta-1}{\delta}}, \end{aligned}$$

(A.11)

for any δ>1, by Hölder’s inequality. Since δ>1 and therefore $\frac{\delta}{\delta-1}>1$,

$$\begin{aligned} &\mathbb{E}_{W^{\perp}_{T}}\Big[K_{T}^{\frac{\delta}{\delta-1}}\Big]^{\frac{\delta-1}{\delta}} \\ &\quad \leq\left(1 + \frac{1-\alpha}{2}(\nu ^{X})^{2}T + (1-\alpha)\nu ^{X}|\rho |\mathbb{E}_{W^{\perp}_{T}}[(\bar{W}^{F}_{T})^{\frac{\delta}{\delta -1}}]^{\frac{\delta-1}{\delta}}\right)X_0 \\ &\quad = \bigg(1 + \frac{1-\alpha}{2}(\nu ^{X})^{2}T + \sqrt{2}(1-\alpha)\nu ^{X}|\rho |\bigg(\frac{\varGamma(\frac{1+\frac{\delta}{\delta-1}}{2})}{\sqrt{\pi }}\bigg)^{\frac{\delta-1}{\delta}}\sqrt{T}\bigg)X_0, \end{aligned}$$

(A.12)

following from Minkowski’s inequality $\mathbb{E}[(f + g)^{p}]^{\frac{1}{p}} \leq\mathbb{E}[f^{p}]^{\frac {1}{p}} + \mathbb{E}[f^{p}]^{\frac{1}{p}}$. Then (A.11) and (A.12) imply that for any δ>1,

$$ \mathbb{E}_{W^{\perp}_{T}}\left[\int_{0}^{T}e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}}d\tilde{X}_{t}\right]\leq \frac{(1-\alpha)F_{0}}{\alpha} + C_{T}\mathbb{E}_{W^{\perp }_{T}}[L_T^{\delta}]^{\frac{1}{\delta}}, $$

where $C_{T} = (1 + \frac{1-\alpha}{2}(\nu ^{X})^{2}T + \sqrt{2}(1-\alpha)\nu ^{X}|\rho|(\frac{\varGamma(\frac{1+\frac{\delta}{\delta-1}}{2})}{\sqrt{\pi }})^{\frac{\delta-1}{\delta}}\sqrt{T})X_{0}$. Thus

$$\begin{aligned} &\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\bigg[\ln\mathbb {E}_{W^{\perp}_{T}}\Big[\int_{0}^{T}e^{-\frac{1-\alpha}{2}(\nu ^{X})^{2}t-(1-\alpha)\nu ^{X}\overline{\rho W^{F}_{t}}}d\tilde{X}_{t}\Big]\bigg] \\ &\quad \leq\limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln\left (\frac{(1-\alpha)F_{0}}{\alpha} + C_{T}\mathbb{E}_{W^{\perp }_{T}}[L^{\delta}_T]^{\frac{1}{\delta}}\right)\right] \\ &\quad \leq\limsup_{T\rightarrow\infty}\frac{1}{T}\ln C_{T} + \limsup _{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln\left(\frac{(1-\alpha )F_{0}}{\alpha C_T} + \mathbb{E}_{W^{\perp}_{T}}[L^{\delta}_T]^{\frac {1}{\delta}}\right)\right]. \end{aligned}$$

(A.13)

Note that the limit in the first term is 0. Furthermore, since $\frac {(1-\alpha)F_{0}}{\alpha C_{T}}\rightarrow0$ as T↑∞ and L _T ≥1, for T large enough, $\frac{(1-\alpha)F_{0}}{\alpha C_{T}} < 1\leq \mathbb{E}_{W^{\perp}_{T}}[L^{\delta}_{T}]^{\frac{1}{\delta}}$. Thus, (A.13) is less than or equal to

$$ \limsup_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\big[\ln\big(2 \mathbb {E}_{W^{\perp}_{T}}[L^{\delta}_T]^{\frac{1}{\delta}}\big)\big] =\limsup_{T\rightarrow\infty}\frac{1}{\delta T}\mathbb{E}\big[\ln\mathbb {E}_{W^{\perp}_{T}}[ L^{\delta}_T]\big]. $$

(A.14)

Since (A.6) implies that $\liminf_{T\rightarrow\infty}\frac {1}{\delta T}\mathbb{E}[-(1-\alpha)\sqrt{1-\rho^{2}}\delta \nu ^{X}\bar{W}^{\perp}_{T}] = 0$, (A.14) equals

$$\begin{aligned} &\limsup_{T\rightarrow\infty}\frac{1}{\delta T}\mathbb{E}\left[\ln \mathbb{E}_{W^{\perp}_{T}}\Big[ e^{(1-\alpha)\delta(\overline {R^{X}_{\cdot} - \frac{1}{2}(\nu ^{X})^{2}\cdot- \rho \nu ^{X}W^{F}_{\cdot }})_{T}}\Big]\right] \\ &\qquad{} +\liminf_{T\rightarrow\infty}\frac{1}{\delta T}\mathbb {E}\big[-(1-\alpha)\sqrt{1-\rho^{2}}\delta \nu ^{X}\bar{W}^{\perp}_{T}\big] \\ &\quad \leq\limsup_{T\rightarrow\infty}\frac{1}{\delta T}\mathbb{E}\left[\ln \mathbb{E}_{W^{\perp}_{T}}\Big[ e^{(1-\alpha)\delta(\overline {R^{X}_{\cdot} - \frac{1}{2}(\nu ^{X})^{2}\cdot- \rho \nu ^{X}W^{F}_{\cdot }})_{T}-(1-\alpha)\sqrt{1-\rho^{2}}\delta \nu ^{X}\bar{W}^{\perp}_{T}}\Big]\right]. \end{aligned}$$

(A.15)

Then by Lemma A.1, the sum of the running maximum and the running minimum is less than or equal to the running maximum of the sum, and (A.15) is less than or equal to

$$\begin{aligned} &\limsup_{T\rightarrow\infty}\frac{1}{\delta T}\mathbb{E}\left[\ln \mathbb{E}_{W^{\perp}_{T}}\Big[ e^{(1-\alpha)\delta(\overline {R^{X}_{\cdot} - \frac{1}{2}(\nu ^{X})^{2}\cdot- \rho \nu ^{X}W^{F}_{\cdot }-\sqrt{1-\rho^{2}}\nu ^{X}W^{\perp}_{\cdot}})_{T}}\Big]\right] \\ &\quad =\limsup_{T\rightarrow\infty}\frac{1}{\delta T}\mathbb{E}\left[\ln \mathbb{E}_{W^{\perp}_{T}}\Big[ e^{(1-\alpha)\delta(\overline {R^{X}_{\cdot} - \frac{1}{2}(\nu ^{X})^{2}\cdot- \nu ^{X}W^{X}_{\cdot }})_{T}}\Big]\right] \\ &\quad \leq\limsup_{T\rightarrow\infty}\frac{1}{\delta T}\ln\mathbb{E}\left [e^{(1-\alpha)\delta(\overline{R^{X}_{\cdot} - \frac{1}{2}(\nu ^{X})^{2}\cdot- \nu ^{X}W^{X}_{\cdot}})_{T}}\right], \end{aligned}$$

(A.16)

where (A.16) follows from Jensen’s inequality and the tower property of conditional expectation.

Now $M_{t} = e^{R^{X}_{t} - \frac{1}{2}(\nu ^{X})^{2}t - \nu ^{X}W^{X}_{t}}$ is a local martingale with respect to the filtration generated by $(W^{F}_{s})_{0\leq s \leq t}$ and $(W^{\perp}_{s})_{0\leq s \leq t}$. Then, since $\bar{M}_{t} \leq\bar{M}_{\infty}$, which in turn is dominated by a random variable X, and X ⁻¹ is uniformly distributed on [0,1] (cf. (54) in [12]), for $1<\delta< \frac{1}{1-\alpha}$, (A.16) is less than or equal to

$$\begin{aligned} \limsup_{T\rightarrow\infty}\frac{1}{\delta T}\ln\mathbb{E}\big[(\bar{M}_{\infty})^{(1-\alpha)\delta}\big] \leq&\limsup_{T\rightarrow\infty}\frac{1}{\delta T}\ln\int _{0}^{1}x^{-(1-\alpha)\delta}dx\\ =&\limsup_{T\rightarrow\infty}\frac{1}{\delta T}\ln\frac{1}{1-(1-\alpha )\delta}=0, \end{aligned}$$

which concludes the proof. □

For the rest of this paper, let p=1−γ.

Lemma A.7

$\displaystyle\limsup_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb {E}[\tilde{F}^{p}_{T}]^{\frac{1}{p}}\leq\lambda_1$ for any 0<p<1.

Proof

Let $\tilde{\pi}^{F}_{t} = \frac{(\tilde{F}_{t} - \bar{X}_{t})}{\tilde {F}_{t}}\pi ^{F}_{t}$. Investing $\pi ^{F}_{t}$ of $\tilde{F}_{t} - \bar{X}_{t}$ in the risky asset is equivalent to investing $\tilde{\pi}^{F}_{t}$ of $\tilde{F}_{t}$, and thus $\tilde{\pi}^{F}$ can be regarded as an investment strategy for $\tilde{F}$. Then

$$\begin{aligned} d\tilde{F}^{p}_{t} =& p\tilde{F}_{t}^{p-1}(\tilde{F}_{t} - \bar{X}_{t})(\pi ^{F}_{t}\mu ^{F}dt +\pi ^{F}_{t}\sigma ^{F}dW^{F})\\ &{}+ \frac{p(p-1)}{2}\tilde{F}^{p-2}_{t} (\tilde{F}_{t} - \bar{X}_{t})^{2}(\pi ^{F}_{t}\sigma ^{F})^{2}dt + \frac{1}{1-\alpha}p\tilde {F}_{t}^{p-1}d\bar{X}_{t}\\ =& p\tilde{F}_{t}^{p}\left(\Big(\tilde{\pi}^{F}_{t}\mu ^{F}+ \frac {(p-1)}{2}(\tilde{\pi}^{F}_{t}\sigma ^{F})^{2}\Big)dt + \tilde{\pi }^{F}_{t}\sigma ^{F}dW^{F}\right) \\ &{}+ \frac{1}{1-\alpha}p\tilde{F}_{t}^{p-1}d\bar{X}_{t}. \end{aligned}$$

Solving this equation gives

$$\begin{aligned} \tilde{F}^{p}_{T} = A_{T} + B_{T} , \end{aligned}$$

where $A_{T} = F^{p}_{0}e^{pR^{F,\tilde{\pi}^{F}}_{T}}$ and $B_{T} = \frac {1}{1-\alpha}\int_{0}^{T}pe^{pR^{F,\tilde{\pi}^{F}}_{t,T}}\tilde {F}^{p-1}_{t}d\bar{X}_{t}$. Thus,

$$\begin{aligned} \limsup_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}[\tilde {F}^{p}_{T}]^{\frac{1}{p}} =\limsup_{T\rightarrow\infty}\frac{1}{T}\ln \mathbb{E}\left[A_T + B_T\right]^{\frac{1}{p}}. \end{aligned}$$

Since 0<p<1, Lemma 1.2.15 in [4] implies that for any positive processes $\left(f_{t}\right)_{t\geq0}$ and $\left(g_{t}\right)_{t\geq0}$,

$$ \limsup_{T\rightarrow\infty}\frac{1}{T}\ln\left(f_{T} + g_{T}\right )^{\frac{1}{p}} = \max\left(\limsup_{T\rightarrow\infty}\frac{1}{T}\ln f_{T}^{\frac{1}{p}}, \limsup_{T\rightarrow\infty}\frac{1}{T}\ln g_{T}^{\frac{1}{p}}\right). $$

It follows that

$$\begin{aligned} &\limsup_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\left[A_T + B_T\right]^{\frac{1}{p}} \\ &\quad = \max\left(\limsup_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\left [A_T\right]^{\frac{1}{p}}, \limsup_{T\rightarrow\infty}\frac{1}{T}\ln \mathbb{E}\left[B_T\right]^{\frac{1}{p}}\right). \end{aligned}$$

(A.17)

Note that since 0<p<1, by Hölder’s inequality,

$$ \mathbb{E}[A_T]^{\frac{1}{p}}\mathbb{E}\Big[e^{q(-\nu ^{F}W^{F}_{T} - \frac {(\nu ^{F})^{2}}{2}T)}\Big]^{\frac{1}{q}} \leq\mathbb{E}\Big[F_{0}e^{R^{F,\hat{\pi}^{F}}_{T}-\nu ^{F}W^{F}_{T} - \frac{(\nu ^{F})^{2}}{2}T}\Big]\leq F_{0}, $$

where $q = \frac{p}{p-1}$. Dividing by the second factor on the left-hand side, it follows that

$$ \mathbb{E}\left[A_T\right]^{\frac{1}{p}} \leq F_0\mathbb{E}\Big[e^{q(-\nu ^{F}W^{F}_{T} - \frac{(\nu ^{F})^{2}}{2}T)}\Big]^{-\frac{1}{q}} = F_0e^{\frac{(\nu ^{F})^{2}}{2(1-p)}T}. $$

(A.18)

Thus,

$$ \limsup_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\left[A_T\right ]^{\frac{1}{p}} \leq\limsup_{T\rightarrow\infty}\frac{1}{T}\ln F_0e^{\frac{(\nu ^{F})^{2}}{2(1-p)}T}= \frac{(\nu ^{F})^{2}}{2(1-p)}. $$

(A.19)

For the second term in (A.17), since p<1 and $\tilde{F}_{t} \geq \bar{X}_{t}$, we get $\tilde{F}^{p-1}_{t} \leq\bar{X}_{t}^{p-1}$ and

$$\begin{aligned} \mathbb{E}\big[e^{-p\lambda_1T}B_T\big] \leq&\frac{1}{1-\alpha}\mathbb {E}\left[\int_{0}^{T}e^{pR^{F,\tilde{\pi}^{F}}_{t,T}-p\lambda _1(T-t)}e^{-p\lambda_1 t}p\bar{X}_{t}^{p-1}d\bar{X}_{t}\right]\\ =&\frac{1}{1-\alpha}\mathbb{E}\left[\int_{0}^{T}e^{pR^{F,\tilde{\pi }^{F}}_{t,T}-p\lambda_1(T-t)}e^{-p\lambda_1 t}d\bar{X}_{t}^{p}\right]. \end{aligned}$$

(A.18) implies that $\mathbb{E}_{t}[e^{pR^{F,\tilde{\pi }^{F}}_{t,T}-p\lambda_{1}(T-t)}]\leq1$. Then, since $\int _{0}^{T}e^{-p\lambda_{1} t}d\bar{X}_{t}^{p}$ is an increasing process, Lemma A.2 implies that

$$ \mathbb{E}\left[\int_{0}^{T}e^{pR^{F,\tilde{\pi}^{F}}_{t,T}-p\lambda _1(T-t)}e^{-p\lambda_1 t}d\bar{X}_{t}^{p}\right]\leq \mathbb{E}\left [\int_{0}^{T}e^{-p\lambda_1 t}d\bar{X}_{t}^{p}\right], $$

and hence

$$\begin{aligned} \limsup_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\left[B_T\right ]^{\frac{1}{p}} =& \lambda_1 + \limsup_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\big[e^{-p\lambda_1T}B_T\big]^{\frac{1}{p}} \\ \leq&\lambda_1 + \limsup_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb {E}\left[\int_{0}^{T}e^{-p\lambda_1 t}d\bar{X}_{t}^{p}\right]^{\frac {1}{p}}. \end{aligned}$$

(A.20)

Now integration by parts implies that

$$\begin{aligned} \int_{0}^{T}e^{-p\lambda_1 t}d\bar{X}_{t}^{p} =& e^{-p\lambda_1 T}\bar{X}_{T}^{p} - X_0^p + p\lambda_1\int_{0}^{T}e^{-p\lambda_1 t}\bar{X}_{t}^{p}dt \\ \leq& e^{-p\lambda_1 T}\bar{X}_T^{p}+ p\lambda_1\int_{0}^{T}e^{-p\lambda _1 t}\bar{X}_t^{p}dt. \end{aligned}$$

(A.21)

Lemma A.1 implies that $e^{-p\lambda_{1} t}\bar{X}_{t}^{p} \leq X^{p}_{0}e^{(1-\alpha)p(\overline{R^{X}_{\cdot} - \frac{\lambda_{1}}{1-\alpha }\cdot})_{t}}$ for every 0≤t≤T. Thus (A.21) is less than or equal to

$$\begin{aligned} &X_0^pe^{(1-\alpha)p(\overline{R^{X}_{\cdot} - \frac{\lambda_1}{1-\alpha }\cdot})_{T}}+ X^p_0 p\lambda_1\int_{0}^{T}e^{(1-\alpha)p(\overline {R^{X}_{\cdot} - \frac{\lambda_1}{1-\alpha}\cdot})_{t}}dt\\ &\quad \leq X_0^pe^{(1-\alpha)p(\overline{R^{X}_{\cdot} - \frac{\lambda _1}{1-\alpha}\cdot})_{T}} + X_0^pp\lambda_1 Te^{(1-\alpha)p(\overline {R^{X}_{\cdot} - \frac{\lambda_1}{1-\alpha}\cdot})_{T}}\\ &\quad =X_0^p\left(1+ p\lambda_1 T\right)e^{(1-\alpha)p(\overline{R^{X}_{\cdot } - \frac{\lambda_1}{1-\alpha}\cdot})_{T}}. \end{aligned}$$

Now, Lemma 9 in [12] with $\varphi- r = \frac {\lambda_{1}}{1-\alpha}$ implies that

$$\begin{aligned} &\limsup_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\Big[X_0^p(1+p\lambda_1 T)e^{p(\overline{R^{X}_{\cdot}-\frac{\lambda _1}{1-\alpha}\cdot})_{T}}\Big]^{\frac{1}{p}} \\ &\quad =\limsup_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\Big[e^{p(\overline{R^{X}_{\cdot}-\frac{\lambda_1}{1-\alpha}\cdot})_{T}}\Big]^{\frac{1}{p}}\leq0. \end{aligned}$$

(A.22)

Thus (A.20) and (A.22) imply that

$$\begin{aligned} \limsup_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\left[B_T\right ]^{\frac{1}{p}} \leq&\lambda_1 + \limsup_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb {E}\left[\int_{0}^{T}e^{-p\lambda_1 t}d\bar{X}_{t}^{p}\right]^{\frac {1}{p}} \\ \leq&\lambda_1 + \limsup_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb {E}\Big[(1+p\lambda_1 T)e^{p(\overline{R^{X}_{\cdot}-\frac{\lambda _1}{1-\alpha}\cdot})_{T}}\Big]^{\frac{1}{p}} \\ \leq&\lambda_1. \end{aligned}$$

(A.23)

Then (A.17), (A.19) and (A.23) imply that

$$ \limsup_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}[\tilde {F}^{p}_{T}]^{\frac{1}{p}} \leq\max\left(\frac{(\nu ^{F})^{2}}{2(1-p)}, \lambda_1\right) = \lambda_1. $$

 □

To prove Theorem 2.1, it now remains to show that the upper bound in Lemma A.3 is achieved by the ESR induced by the strategies in (2.6) and (2.7), and hence that they are optimal. Plugging them into the dynamics of X and F, the corresponding fund’s value and wealth processes follow

$$\begin{aligned} d\hat{X}_{t} =& \hat{X}_{t}\left(\frac{(\nu ^{X})^{2}}{1-(1-\alpha)p} dt + \frac{\nu ^{X}}{1-(1-\alpha)p} dW^{X}_{t}\right) - \frac{\alpha}{1-\alpha }d\bar{\hat{X}}_{t},\\ d\hat{F}_{t} =& \left(\hat{F}_{t}-\frac{\alpha}{1-\alpha}(\bar{\hat{X}}_{t}-X_{0})\right )\left(\frac{(\nu ^{F})^{2}}{1-p} dt + \frac{\nu ^{F}}{1-p} dW^{F}_{t}\right) +\frac{\alpha}{1-\alpha}d\bar {\hat{X}}_{t}. \end{aligned}$$

Solving the above equations implies that

$$\begin{aligned} \bar{\hat{X}}_{T} =& X_{0}e^{(1-\alpha)\bar{\hat{R}}^{X}_{T}}, \\ \hat{F}_{T} =& F_{0}e^{\hat{R}^{F}_{T}}+ \frac{\alpha}{1-\alpha}(\bar{\hat{X}}_{T}-X_{0}), \end{aligned}$$

where $\hat{R}^{X}_{T} = \frac{\left(1-2(1-\alpha)p\right)(\nu ^{X})^{2}}{2\left(1-(1-\alpha)p\right)^{2}}T + \frac{\nu ^{X}}{1-(1-\alpha)p}W^{X}_{T}$, $\hat{R}^{F}_{T} = \frac {(1-2p)(\nu ^{F})^{2}}{2(1-p)^{2}}T + \frac{\nu ^{F}}{1-p} W^{F}_{T}$.

Lemma A.8

For $\hat {\pi }^{X}$ and $\hat {\pi }^{F}$ in (2.6) and (2.7), we have $\operatorname {ESR}_{\gamma}(\hat{\pi}^{X},\hat{\pi}^{F}) = \lambda_{1}$ for any 0<γ≤1.

Proof

Let $G_{t} = F_{0}e^{\hat{R}^{F}_{T}}$ and $H_{t} = \frac{\alpha}{1-\alpha}(\bar{\hat{X}}_{t}-X_{0})$; then $\hat{F}_{T} = G_{T} + H_{T}$. From Lemma A.3, it suffices to prove that $\operatorname {ESR}_{\gamma}(\hat{\pi}^{X},\hat{\pi}^{F}) \geq\lambda_{1}$.

Case 1. Logarithmic utility (p=0). Since H is a positive process,

$$ \lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}[\ln\hat{F}_{T}]=\lim _{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln(G_{T} + H_{T})\right ]\geq\lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln G_{T}\right ]=\frac{(\nu ^{F})^{2}}{2}. $$

Likewise, since G is a positive process, Lemma A.9 below implies that

$$\begin{aligned} \lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}[\ln\hat{F}_{T}] \geq&\lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln H_{T}\right ]=\lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\big[\ln\big(\bar{\hat {X}}_{T}-X_{0}\big)\big]\\ =& \lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\Big[\ln\Big(e^{(1-\alpha)\bar{\hat{R}}^{X}_{T}}-1\Big)\Big] = \lim_{T\rightarrow \infty}\frac{1}{T}\mathbb{E}\big[(1-\alpha)\bar{\hat{R}}^{X}_{T}\big]\\ =& \lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\bigg[(1-\alpha)\bigg(\overline{\frac{(\nu ^{X})^{2}}{2}\cdot+ \nu ^{X}W^{X}_{\cdot}}\bigg)_{T}\bigg]. \end{aligned}$$

Since $(\overline{\frac{(\nu ^{X})^{2}}{2}\cdot+ \nu ^{X}W^{X}_{\cdot}})_{T} \geq\frac{(\nu ^{X})^{2}T}{2} + \nu ^{X}\underline{W}^{X}_{T} = \frac{(\nu ^{X})^{2}T}{2} - \nu ^{X}\overline{(-W^{X})}_{T}$, which follows from Lemma A.1,

$$\begin{aligned} &\lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\bigg[(1-\alpha)\bigg(\overline{\frac{(\nu ^{X})^{2}}{2}\cdot+ \nu ^{X}W^{X}_{\cdot}}\bigg)_{T}\bigg]\\ &\quad \geq\frac{(1-\alpha)(\nu ^{X})^{2} }{2} - \lim_{T\rightarrow\infty}\frac {1}{T}\mathbb{E}\big[(1-\alpha) \nu ^{X}\overline{-W}^{X}_{T}\big] = \frac{(1-\alpha)(\nu ^{X})^{2}}{2}, \end{aligned}$$

where the last equality follows from (A.6). Thus,

$$\lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}[\ln\hat{F}_{T}]\geq\max \bigg(\frac{(\nu ^{F})^{2}}{2},\frac{(1-\alpha)(\nu ^{X})^{2}}{2}\bigg)=\lambda_1. $$

Case 2. Power utility (0<p<1). Since H is a positive process,

$$\begin{aligned} \lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}[\hat {F}^{p}_{T}]^{\frac{1}{p}} =&\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}[(G_{T} + H_{T})^{p}]^{\frac{1}{p}}\\ \geq&\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}[G^{p}_{T}]^{\frac {1}{p}}=\frac{(\nu ^{F})^{2}}{2(1-p)}. \end{aligned}$$

Likewise, since G is a positive process, Lemma A.9 below implies that

$$\begin{aligned} \lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}[\hat {F}^{p}_{T}]^{\frac{1}{p}} \geq&\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}[H^{p}_{T}]^{\frac {1}{p}}\\ =&\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\big[\big(\bar{\hat {X}}_{T}-X_{0}\big)^{p}\big]^{\frac{1}{p}}\\ =&\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\Big[\Big(e^{(1-\alpha)\bar{\hat{R}}^{X}_{T}}-1\Big)^{p}\Big]^{\frac{1}{p}}\\ =&\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\big[e^{(1-\alpha )p\bar{\hat{R}}^{X}_{T}}\big]^{\frac{1}{p}}=\frac{(1-\alpha)(\nu ^{X})^{2}}{2\left(1-(1-\alpha)p\right)}, \end{aligned}$$

where the last equality follows from Lemma 11 in [12]. Thus,

$$ \displaystyle\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}[\hat {F}^{p}_{T}]^{\frac{1}{p}}\geq\max\left(\frac{(\nu ^{F})^{2}}{2(1-p)},\frac {(1-\alpha)(\nu ^{X})^{2}}{2\left(1-(1-\alpha)p\right)}\right)=\lambda_1. $$

 □

Lemma A.9

For $\hat{R}^{X}_{T} = \frac{(1-2(1-\alpha)p)(\nu ^{X})^{2}}{2\left (1-(1-\alpha)p\right)^{2}}T + \frac{\nu ^{X}}{1-(1-\alpha)p} W^{X}_{T}$, when p=0,

$$ \lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\Big[\ln\Big(e^{(1-\alpha )\bar{\hat{R}}^{X}_{T}}-1\Big)\Big] = \lim_{T\rightarrow\infty}\frac {1}{T}\mathbb{E}\big[(1-\alpha)\bar{\hat{R}}^{X}_{T}\big], $$

and when 0<p<1,

$$ \lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\Big[\Big(e^{(1-\alpha )\bar{\hat{R}}^{X}_{T}}-1\Big)^{p}\Big]^{\frac{1}{p}} = \lim _{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\Big[e^{(1-\alpha)p\bar {\hat{R}}^{X}_{T}}\Big]^{\frac{1}{p}}. $$

Proof

Since $e^{(1-\alpha)\bar{\hat{R}}^{X}_{T}}-1\leq e^{(1-\alpha)\bar{\hat {R}}^{X}_{T}}$, the “≤” part of the two assertions is straightforward. It suffices to prove the “≥” part.

When p=0, since $e^{(1-\alpha)\bar{\hat{R}}^{X}_{T}} = e^{(1-\alpha )\bar{\hat{R}}^{X}_{T}} -1 +1$, the concavity of the logarithm implies that

$$\begin{aligned} &(1-\alpha)\bar{\hat{R}}^{X}_{T} = \ln e^{(1-\alpha)\bar{\hat {R}}^{X}_{T}} \leq\ln\big(e^{(1-\alpha)\bar{\hat{R}}^{X}_{T}}-1\big) + \frac{1}{e^{(1-\alpha)\bar{\hat{R}}^{X}_{T}}-1}. \end{aligned}$$

(A.24)

Since ν ^X>0, $e^{(1-\alpha)\hat{R}^{X}_{T}} = e^{\frac{(1-\alpha)(\nu ^{X})^{2}}{2}T + (1-\alpha)\nu ^{X}W^{X}_{T}} \geq2e^{(1-\alpha)\nu ^{X}W^{X}_{T}}$ for sufficiently large T, and hence

$$ e^{(1-\alpha)\bar{\hat{R}}^{X}_{T}} \geq2e^{(1-\alpha)\nu ^{X}\bar{W}^{X}_{T}} \geq1 + e^{(1-\alpha)\nu ^{X}\bar{W}^{X}_{T}}. $$

(A.25)

Thus,

$$ \lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\bigg[\frac {1}{e^{(1-\alpha)\bar{\hat{R}}^{X}_{T}}-1}\bigg] \leq\lim _{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\big[e^{-(1-\alpha)\nu ^{X}\bar{W}^{X}_{T}}\big]=0. $$

(A.26)

Then, (A.24) and (A.26) imply that

$$\begin{aligned} &\lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\big[(1-\alpha)\bar{\hat {R}}^{X}_{T} \big] \\ &\quad \leq\lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\Big[\ln\Big(e^{(1-\alpha)\bar{\hat{R}}^{X}_{T}} -1\Big)\Big] +\lim_{T\rightarrow \infty}\frac{1}{T}\mathbb{E}\bigg[\frac{1}{e^{(1-\alpha)\bar{\hat {R}}^{X}_{T}}-1}\bigg]\\ &\quad \leq\lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\Big[\ln\Big(e^{(1-\alpha)\bar{\hat{R}}^{X}_{T}} -1\Big)\Big]. \end{aligned}$$

When 0<p<1, from the concavity of the function f(x)=x ^p,

$$ e^{(1-\alpha)p\bar{\hat{R}}^{X}_{T}} \leq\big(e^{(1-\alpha)\bar{\hat {R}}^{X}_{T}}-1\big)^{p} + p\big(e^{(1-\alpha)\bar{\hat {R}}^{X}_{T}}-1\big)^{p-1}. $$

Thus, from Lemma 1.2.15 in [4],

$$\begin{aligned} &\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\big[(1-\alpha)p\bar {\hat{R}}^{X}_{T} \big]^{\frac{1}{p}} \\ &\quad \leq\lim_{T\rightarrow\infty }\frac{1}{T}\ln\mathbb{E}[A_T + B_T]^{\frac{1}{p}} \\ &\quad = \max\bigg(\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb {E}[A_T]^{\frac{1}{p}}, \lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb {E}\Big[\Big(e^{(1-\alpha)\bar{\hat{R}}^{X}_{T}}-1\Big)^{p}\Big]^{\frac {1}{p}}\bigg), \end{aligned}$$

(A.27)

where $A_{T} = p(e^{(1-\alpha)\bar{\hat{R}}^{X}_{T}}-1)^{p-1}$. Since p−1<0, from (A.25),

$$\begin{aligned} \lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}[A_T]^{\frac{1}{p}} \leq\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\big[pe^{(1-\alpha )(p-1)\nu ^{X}\bar{W}^{X}_{T}}\big]^{\frac{1}{p}}\leq0. \end{aligned}$$

(A.28)

Thus, (A.27) and (A.28) imply that

$$\begin{aligned} &\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\Big[e^{(1-\alpha )p\bar{\hat{R}}^{X}_{T}} \Big]^{\frac{1}{p}}\\ &\quad \leq\max\bigg(\lim _{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\Big[\Big(e^{(1-\alpha)\bar {\hat{R}}^{X}_{T}}-1\Big)^{p}\Big]^{\frac{1}{p}}, 0\bigg)\\ &\quad \leq\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\Big[\Big(e^{(1-\alpha)\bar{\hat{R}}^{X}_{T}} -1\Big)^{p}\Big]^{\frac{1}{p}}, \end{aligned}$$

because from (A.25),

$$\begin{aligned} \lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\Big[\Big(e^{(1-\alpha )\bar{\hat{R}}^{X}_{T}} -1\Big)^{p}\Big]^{\frac{1}{p}} \geq&\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\big[e^{(1-\alpha )p\nu ^{X}\bar{W}^{X}_{T}}\big]^{\frac{1}{p}} \\ =& \frac{(1-\alpha)^2(\nu ^{X})^{2}p}{2}>0. \end{aligned}$$

 □

1.2 A.2 Proof of Theorem 2.2

Solving equations (2.3) and (2.4) gives

$$\begin{aligned} X_t =& X_0e^{R^X_t - \varphi t},\\ F_t =& F_0e^{R^F_t} + \int_0^t \varphi e^{R^F_{s,t}}X_sds. \end{aligned}$$

Theorem 2.2 is proved by arguments similar to those for Theorem 2.1: first prove that for general strategies, the ESR of wealth is bounded above by (2.11), and then show that the ESR induced by the candidate strategies $\hat {\pi }^{X}$ and $\hat {\pi }^{F}$ in (2.9) and (2.10) achieves this upper bound.

Lemma A.10

For a mutual fund manager compensated by proportional fees with rate φ>0, the $\operatorname {ESR}$ induced by any investment strategies π ^X and π ^F satisfies

$$ \operatorname {ESR}_{\gamma}(\pi ^{F},\pi ^{F}) \leq\lambda_2 = \max\left(\frac{(\nu ^{X})^{2}}{2\gamma}-\varphi,\frac{(\nu ^{F})^2}{2\gamma}\right), \quad \textit{for\ any}\ 0< \gamma\leq1. $$

Proof

We prove this lemma for logarithmic utility and power utility, respectively.

Case 1. Logarithmic utility. We have

$$ \lim_{T\rightarrow\infty} \frac{1}{T} \mathbb{E}\left[\ln F_{T}\right] = \lambda_2 + \lim_{T\rightarrow\infty} \frac{1}{T} \mathbb{E}\left [\ln\left(F_0e^{R^F_T-\lambda_2 T} + \varphi\int _0^Te^{R^F_{t,T}-\lambda_2 T}X_tdt\right)\right]. $$

Then (A.6) implies that, with $N^{T}_{t}$ defined in the proof of Lemma A.4,

$$\begin{aligned} &\lim_{T\rightarrow\infty} \frac{1}{T} \mathbb{E}\left[\ln\left (F_0e^{R^F_T-\lambda_2 T} + \varphi\int_0^Te^{R^F_{t,T}-\lambda_2 T}X_tdt\right)\right] \\ &\quad =\lim_{T\rightarrow\infty} \frac{1}{T} \mathbb{E}\left[\ln\left (F_0e^{R^F_T-\lambda_2 T} + \varphi\int_0^Te^{R^F_{t,T}-\lambda_2 T}X_tdt\right)\right] \\ &\qquad{} + \lim_{T \rightarrow\infty}\frac{1}{T}\mathbb{E}\Big[-\nu ^{F}\bar{N}^{T}_{T}-\nu ^{X}\overline{\rho W^{F}_{T}}\Big] \\ &\quad \leq\lim_{T\rightarrow\infty} \frac{1}{T} \mathbb{E}\left[\ln A_T\right] , \end{aligned}$$

(A.29)

where the last inequality follows from the definition of λ ₂, the fact that $\bar{N}^{T}_{T} \geq W_{t,T}$ for any 0≤t≤T and

$$\begin{aligned} A_T =& F_0e^{R^F_T- \frac{(\nu ^{F})^2T}{2} - \nu ^{F}W^F_T} \\ &{} + \varphi \int_0^T\!\!\!e^{R^F_{t,T}- \frac{(\nu ^{F})^2(T-t)}{2}-\nu ^{F}W^F_{t,T}}e^{-(\frac{(\nu ^{X})^2}{2}-\varphi)t - \rho \nu ^{X}W^F_t}X_tdt. \end{aligned}$$

Then with W ^⊥ being a Brownian motion independent of W ^F and such that we have $W^{X}_{t} = \rho W^{F}_{t} + \sqrt{1-\rho^{2}}W^{\perp}_{t}$, Jensen’s inequality implies that (A.29) is less than or equal to $\lim_{T\rightarrow\infty} \frac{1}{T} \mathbb{E}[\ln\mathbb {E}_{W^{\perp}_{T}}[A_{T}]]$.

From Lemma A.5, $M_{t} = e^{R^{F}_{t}- \frac{(\nu ^{F})^{2}t}{2} - \nu ^{F}W^{F}_{t}}$ is a supermartingale with respect to the filtration generated by $(W^{\perp}_{s})_{0\leq s\leq T}$ and $(W^{F}_{s})_{0\leq s\leq t}$. Then, Lemma A.2 implies that

$$\begin{aligned} &\lim_{T\rightarrow\infty} \frac{1}{T} \mathbb{E}\big[\ln\mathbb {E}_{W^{\perp}_T}[A_T]\big] \\ &\quad \leq\lim_{T\rightarrow\infty} \frac{1}{T} \mathbb{E}\left[\ln\mathbb {E}_{W^{\perp}_T}\Big[F_0+ \varphi\int_0^Te^{-(\frac{(\nu ^{X})^2}{2}-\varphi )t - \rho \nu ^{X}W^F_t}X_tdt\Big]\right] \\ &\quad =\lim_{T\rightarrow\infty} \frac{1}{T} \mathbb{E}\left[\ln\mathbb {E}_{W^{\perp}_T}\Big[F_0+ \varphi\int_0^Te^{-(\frac{(\nu ^{X})^2}{2}-\varphi )t - \rho \nu ^{X}W^F_t}X_tdt\Big]\right] \\ &\qquad{}+ \lim_{T\rightarrow\infty} \frac{1}{T} \mathbb{E}\big[\ln e^{-\sqrt{1-\rho^2}\nu ^{X}\bar{W}^{\perp}_T}\big], \end{aligned}$$

(A.30)

where the last equality follows from (A.6). Then, since $\bar{W}^{\perp}_{T} \geq W^{\perp}_{t}$ for every 0≤t≤T and $X_{t} = X_{0}e^{R^{X}_{t}-\varphi t}$, (A.30) is less than or equal to

$$\begin{aligned} &\lim_{T\rightarrow\infty} \frac{1}{T} \mathbb{E}\left[\ln\mathbb {E}_{W^{\perp}_T}\Big[F_0 e^{-\sqrt{1-\rho^2}\nu ^{X}\bar{W}^{\perp}_T}+ \varphi X_0\int_0^Te^{R^X_t-\frac{(\nu ^{X})^2 t}{2} - \nu ^{X}W^X_t}dt\Big]\right]\\ &\quad \leq\lim_{T\rightarrow\infty} \frac{1}{T}\ln\mathbb{E}\left[F_0 e^{-\sqrt{1-\rho^2}\nu ^{X}\bar{W}^{\perp}_T}+ \varphi X_0\int _0^Te^{R^X_t-\frac{(\nu ^{X})^2 t}{2} - \nu ^{X}W^X_t}dt\right], \end{aligned}$$

where the last inequality follows from Jensen’s inequality and the tower property of conditional expectation. Then since $G_{t} = e^{R^{X}_{t}-\frac{(\nu ^{X})^{2} t}{2} - \nu ^{X}W^{X}_{t}}$ is a supermartingale, by Fubini’s theorem, the above equals

$$\begin{aligned} &\lim_{T\rightarrow\infty} \frac{1}{T}\ln\bigg(\mathbb{E}\big[F_0 e^{-\sqrt{1-\rho^2}\nu ^{X}\bar{W}^{\perp}_T}\big]+ \varphi X_0\int _0^T\mathbb{E}\Big[e^{R^X_t-\frac{(\nu ^{X})^2 t}{2} - \nu ^{X}W^X_t}\Big]dt\bigg)\\ &\quad \leq\lim_{T\rightarrow\infty} \frac{1}{T}\ln\left(F_0 + \varphi X_0 \int_0^T1dt\right)= \lim_{T\rightarrow\infty} \frac{1}{T}\ln\left(F_0 + \varphi X_0T\right)\leq0, \end{aligned}$$

and this implies that $\lim_{T\rightarrow\infty} \frac{1}{T} \mathbb {E}\left[\ln F_{T}\right]\leq\lambda_{2}$.

Case 2. Power utility. Define $\tilde{F}_{t} = F_{t} + \bar{X}_{t}$, which implies that $\tilde{F}_{t} \geq\bar{X}_{t}$, $\tilde{F}_{t} \geq F_{t}$. Thus the ESR of F is less than or equal to the ESR of $\tilde{F}$, which will be proved in the following to be also bounded above by λ ₂. Notice that this is similar to the technique used to deal with the power utility case in the proof of Lemma A.3: we add a positive and increasing process to wealth without increasing the ESR. Here we choose $\bar{X}$, because the property $\bar{X}_{t}\geq X_{t}$ helps to derive (A.32) below, though the mutual fund manager is not compensated by high-water mark performance fees.

Let $\tilde{\pi}^{F}_{t} = \frac{\tilde{F}_{t} - \bar{X}_{t}}{\tilde{F}_{t}}\pi ^{F}_{t}$; then for 0<p<1,

$$\begin{aligned} d\tilde{F}^p_t =& p\tilde{F}^{p-1}_t(\tilde{F}_t - \bar{X}_t)(\pi ^{F}_t\mu ^{F}dt+\pi ^{F}_t\sigma ^{F}dW^F_t) \\ &{}+ \frac{p(p-1)}{2}\tilde{F}^{p-2}_t(\tilde{F}_t - \bar{X}_t)^2(\pi ^{F}_t\sigma ^{F})^2dt + \varphi p\tilde{F}^{p-1}_t X_tdt + p\tilde {F}^{p-1} d\bar{X}_t\\ =&p\tilde{F}^p\left(\Big(\tilde{\pi}^F_t\mu ^{F}+ \frac{p-1}{2}(\tilde{\pi }^F_t\sigma ^{F})^2\Big)dt + \tilde{\pi}^F_t\sigma ^{F}dW^F_t\right)\\ &{}+ \varphi p\tilde{F}^{p-1}_t X_tdt + p\tilde{F}^{p-1} d\bar{X}_t. \end{aligned}$$

By solving this equation, $\tilde{F}^{p}$ can be represented as a sum of three positive processes,

$$ \tilde{F}^p_T = F_0^pe^{pR^{F,\tilde{\pi}^F}_T} + \varphi p\int _0^Te^{pR^{F,\tilde{\pi}^F}_{t,T}}\tilde{F}^{p-1}_t X_tdt + p\int _0^Te^{pR^{F,\tilde{\pi}^F}_{t,T}}\tilde{F}^{p-1}d\bar{X}_t. $$

(A.31)

Then from Lemma 1.2.15 in [4], it suffices to prove that the ESR of each of the three terms in (A.31) is less than or equal to λ ₂. From (A.19),

$$\lim_{T\rightarrow\infty} \frac{1}{T} \ln\mathbb{E}\Big[F_0^pe^{pR^{F,\tilde{\pi}^F}_T} \Big]^{\frac{1}{p}} \leq\frac{(\nu ^{F})^2}{2(1-p)}\leq\lambda_2. $$

For the second term in (A.31), since $\tilde{F}_{t}\geq\bar{X}_{t} \geq X_{t}$ and p−1<0,

$$\begin{aligned} &\lim_{T\rightarrow\infty} \frac{1}{T} \ln\mathbb{E}\left[\varphi p\int _0^Te^{pR^{F,\tilde{\pi}^F}_{t,T}}\tilde{F}^{p-1}_t X_tdt\right]^{\frac {1}{p}} \\ &\quad \leq\lim_{T\rightarrow\infty} \frac{1}{T} \ln\mathbb{E}\left[\int _0^Te^{pR^{F,\tilde{\pi}^F}_{t,T}} X^{p}_tdt\right]^{\frac{1}{p}} \\ &\quad =\lambda_2 + \lim_{T\rightarrow\infty} \frac{1}{T} \ln\mathbb{E}\left [\int_0^Te^{pR^{F,\tilde{\pi}^F}_{t,T}-p\lambda_2 (T-t)} X^{p}_te^{-p\lambda_2 t}dt\right]^{\frac{1}{p}} \\ &\quad = \lambda_2 + \lim_{T\rightarrow\infty} \frac{1}{pT} \ln\int _0^T\mathbb{E}\left[e^{pR^{F,\tilde{\pi}^F}_{t,T}-p\lambda_2 (T-t)} X^{p}_te^{-p\lambda_2 t}\right] dt, \end{aligned}$$

(A.32)

where the last equality follows from Fubini’s theorem. Then, since (A.18) implies that $\mathbb{E}_{t}[e^{pR^{F,\tilde{\pi}^{F}}_{t,T}}] \leq e^{\frac{p(\nu ^{F})^{2}}{2(1-p)}(T-t)}\leq e^{p\lambda_{2}(T-t)}$, from the tower property of conditional expectation,

$$\begin{aligned} &\lim_{T\rightarrow\infty} \frac{1}{pT} \ln\int_0^T\mathbb{E}\left [e^{pR^{F,\tilde{\pi}^F}_{t,T}-p\lambda_2 (T-t)} X^{p}_te^{-p\lambda t}\right] dt\\ &\quad =\lim_{T\rightarrow\infty} \frac{1}{pT} \ln\int_0^T\mathbb{E}\left [\mathbb{E}_t\Big[e^{pR^{F,\tilde{\pi}^F}_{t,T}-p\lambda_2 (T-t)}\Big] X^{p}_te^{-p\lambda_2 t}\right] dt\\ &\quad \leq\lim_{T\rightarrow\infty} \frac{1}{pT} \ln\int_0^T\mathbb {E}[X^{p}_te^{-p\lambda_2 t}] dt. \end{aligned}$$

Since $\mathbb{E}[X^{p}_{t}] = \mathbb{E}[X^{p}_{0}e^{pR^{X}_{t} - p\varphi t}] \leq X^{p}_{0}e^{p(\frac{(\nu ^{X})^{2}}{2(1-p)}-\varphi)t}$, which follows from an argument similar to (A.18), and $\lambda_{2}\geq\frac{(\nu ^{X})^{2}}{2(1-p)}-\varphi$,

$$ \lim_{T\rightarrow\infty} \frac{1}{pT} \ln\int_0^T\mathbb {E}[X^{p}_te^{-p\lambda_2 t}]dt \leq\lim_{T\rightarrow\infty} \frac {1}{pT}\ln(X_0^pT) = 0, $$

which implies that $\lim_{T\rightarrow\infty} \frac{1}{T} \ln \mathbb{E}[\varphi p\int_{0}^{T}e^{pR^{F,\tilde{\pi}^{F}}_{t,T}}\tilde {F}^{p-1}_{t} X_{t}dt]^{\frac{1}{p}}\leq\lambda_{2}$.

Finally, since $\tilde{F}_{t}\geq\bar{X}_{t}$ and p−1<0, following arguments similar to those in the proof of Lemma A.7 yields

$$\begin{aligned} \lim_{T\rightarrow\infty} \frac{1}{T} \ln\mathbb{E}\bigg[p\int_0^T\!\! e^{pR^{F,\tilde{\pi}^F}_{t,T}}\tilde{F}^{p-1}d\bar{X}_t\bigg]^{\frac{1}{p}} \leq&\lim_{T\rightarrow\infty} \frac{1}{pT} \ln\mathbb{E}\bigg[p\int _0^T\!\!e^{pR^{F,\tilde{\pi}^F}_{t,T}}\bar{X}_t^{p-1}d\bar{X}_t\bigg] \\ =&\lim_{T\rightarrow\infty} \frac{1}{pT} \ln\mathbb{E}\bigg[\int_0^T\! \!e^{pR^{F,\tilde{\pi}^F}_{t,T}}d\bar{X}_t^p\bigg]\\ \leq&\lambda_2 + \lim_{T\rightarrow\infty} \frac{1}{pT} \ln\mathbb {E}\bigg[\int_0^T\!\!e^{-p\lambda_2 t}d\bar{X}_t^p\bigg]. \end{aligned}$$

By integration by parts, $\int_{0}^{T}e^{-p\lambda_{2} t}d\bar{X}_{t}^{p} \leq (1+p\lambda_{2} T)X^{p}_{0}e^{p(\overline{R^{X}_{\cdot}-\varphi\cdot-\lambda _{2}\cdot})_{T}}$. Then, applying Lemma 9 in [12] to the case of α=0 and r=−λ ₂ implies that

$$\begin{aligned} \lim_{T\rightarrow\infty} \frac{1}{pT} \ln\mathbb{E}\left[\int _0^Te^{-p\lambda_2 t}d\bar{X}_t^p\right] \leq&\lim_{T\rightarrow\infty} \frac{1}{pT} \ln\mathbb{E}\Big[e^{p(\overline{R^X_{\cdot}-\varphi\cdot-\lambda_2\cdot})_T}\Big]\\ \leq& \frac{(\nu ^{X})^2}{2(1-p)} -\varphi-\lambda_2\leq0, \end{aligned}$$

which indicates that $\lim_{T\rightarrow\infty} \frac{1}{T} \ln\mathbb {E}[p\int_{0}^{T}e^{pR^{F,\tilde{\pi}^{F}}_{t,T}}\tilde{F}^{p-1}d\bar{X}_{t}]^{\frac{1}{p}}\leq\lambda_{2}$. □

Now it remains to prove that the ESR induced by the candidate strategies in (2.9) and (2.10) achieves (2.11).

Lemma A.11

For $\hat {\pi }^{X}$ and $\hat {\pi }^{F}$ in (2.9) and (2.10), we have $\operatorname {ESR}_{\gamma}(\hat{\pi}^{X},\hat{\pi}^{F}) = \lambda_{2}$ for any 0<γ≤1.

Proof

Plugging $\hat {\pi }^{X}$ and $\hat {\pi }^{F}$ into (2.3) and (2.4), with $\hat{R}^{X} = \frac{(1-2p)(\nu ^{X})^{2}T}{2(1-p)^{2}} + \frac{\nu ^{X}}{1-p}W^{X}_{T}$ and $\hat{R}^{F} = \frac{(1-2p)(\nu ^{F})^{2}T}{2(1-p)^{2}} + \frac {\nu ^{F}}{1-p}W^{F}_{T}$, the corresponding fund’s value and wealth satisfy

$$\begin{aligned} \hat{X}_T =& X_0e^{\hat{R}^X_T-\varphi T},\\ \hat{F}_T =& F_0e^{\hat{R}^F_T} + \varphi\int_0^T\hat{X}_tdt. \end{aligned}$$

Let $G_{t} = F_{0}e^{\hat{R}^{F}_{T}}$ and $H_{t} =\varphi\int_{0}^{T}\hat {X}_{t}dt$; then $\hat{F}_{T} = G_{T} + H_{T}$. From Lemma A.10, it suffices to prove that $\operatorname {ESR}_{\gamma}(\hat {\pi }^{X},\hat {\pi }^{F}) \geq\lambda_{2}$.

Case 1. Logarithmic utility. Since H is a positive process,

$$ \lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}[\ln\hat{F}_{T}]=\lim _{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln(G_{T} + H_{T})\right ]\geq\lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln G_{T}\right ]=\frac{(\nu ^{F})^{2}}{2}. $$

Likewise, since G is a positive process,

$$\begin{aligned} \lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}[\ln\hat{F}_{T}] \geq&\lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln H_{T}\right ] \\ =&\lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}\left[\ln\left(\varphi \int_0^T\hat{X}_tdt\right)\right] \\ =& \lim_{T\rightarrow\infty} \frac{1}{T} \mathbb{E}\left[\ln\int _0^Te^{(\frac{(\nu ^{X})^2}{2}-\varphi)t + \nu ^{X}W^X_t}dt \right] \\ \geq&\lim_{T\rightarrow\infty} \frac{1}{T} \mathbb{E}\left[\ln\int _0^Te^{(\frac{(\nu ^{X})^2}{2}-\varphi)t + \nu ^{X}\underline{W}^X_{t}}dt \right]. \end{aligned}$$

(A.33)

Then since $\underline{W}^{X}_{t}\geq\underline{W}^{X}_{T}$ for every 0≤t≤T, (A.33) is greater than or equal to

$$\begin{aligned} \lim_{T\rightarrow\infty} \frac{1}{T} \mathbb{E}\left[\nu ^{X}\underline{W}^X_{T} + \ln\int_0^Te^{(\frac{(\nu ^{X})^2}{2}-\varphi)t} dt \right] =& \lim_{T\rightarrow\infty} \frac{1}{T}\ln\int_0^Te^{(\frac{(\nu ^{X})^2}{2}-\varphi)t} dt\\ =&\lim_{T\rightarrow\infty} \frac{1}{T}\ln\frac{e^{(\frac{(\nu ^{X})^2}{2}-\varphi)T}-1}{\frac{(\nu ^{X})^2}{2}-\varphi}\\ =& \max\left(\frac{(\nu ^{X})^2}{2}-\varphi,0\right), \end{aligned}$$

where the first equality follows from (A.6). Thus,

$$\lim_{T\rightarrow\infty}\frac{1}{T}\mathbb{E}[\ln\hat{F}_{T}] \geq\max \bigg(\frac{(\nu ^{F})^{2}}{2}, \max\Big(\frac{(\nu ^{X})^2}{2}-\varphi,0\Big)\bigg) = \lambda_2. $$

Case 2. Power utility. Since H is a positive process,

$$ \lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}[\hat{F}^p_{T}]^{\frac {1}{p}}\geq\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb {E}[G^p_{T}]^{\frac{1}{p}}=\frac{(\nu ^{F})^{2}}{2(1-p)}. $$

Likewise, since G is a positive process,

$$\begin{aligned} \lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}[\hat{F}^p_{T}]^{\frac{1}{p}} \geq&\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}[ H^p_{T}]^{\frac {1}{p}}\\ =&\lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}\bigg[\varphi^p\bigg(\int_0^T\hat{X}_tdt\bigg)^p\bigg]^{\frac{1}{p}} \\ =&\lim_{T\rightarrow\infty} \frac{1}{pT} \ln\mathbb{E}\bigg[\bigg(\int _0^Te^{(\frac{(1-2p)(\nu ^{X})^2}{2(1-p)^2}-\varphi)t + \frac{\nu ^{X}}{1-p} W^X_t}dt \bigg)^p\bigg] \\ =& \lim_{T\rightarrow\infty} \frac{1}{pT} \ln\mathbb{E}\bigg[T^p\bigg(\int_0^T\frac{1}{T}e^{(\frac{(1-2p)(\nu ^{X})^2}{2(1-p)^2}-\varphi)t + \frac {\nu ^{X}}{1-p} W^X_t}dt \bigg)^p\bigg]. \end{aligned}$$

Since 0<p<1, Jensen’s inequality implies that

$$\begin{aligned} &\mathbb{E}\bigg[T^p\bigg(\int_0^T\frac{1}{T}e^{(\frac{(1-2p)(\nu ^{X})^2}{2(1-p)^2}-\varphi)t + \frac{\nu ^{X}}{1-p} W^X_t}dt \bigg)^p\bigg] \\ &\quad \geq\mathbb{E}\bigg[T^{p-1}\int_0^Te^{(\frac{p(1-2p)(\nu ^{X})^2}{2(1-p)^2}-p\varphi)t + \frac{p\nu ^{X}}{1-p} W^X_t}dt\bigg] \\ &\quad =T^{p-1}\int_0^T\mathbb{E}\bigg[e^{(\frac{p(1-2p)(\nu ^{X})^2}{2(1-p)^2}-p\varphi)t + \frac{p\nu ^{X}}{1-p} W^X_t}\bigg]dt = T^{p-1}\int_0^T e^{(\frac{p(\nu ^{X})^2}{2(1-p)}-p\varphi)t}dt, \end{aligned}$$

where the first equality follows from Fubini’s theorem. Thus,

$$\begin{aligned} \lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}[\hat{F}^p_{T}]^{\frac{1}{p}} \geq&\lim_{T\rightarrow\infty} \frac{1}{pT} \ln\mathbb{E}\bigg[T^p\bigg(\int_0^T\frac{1}{T}e^{(\frac{(1-2p)(\nu ^{X})^2}{2(1-p)^2}-\varphi )t + \frac{\nu ^{X}}{1-p} W^X_t}dt \bigg)^p\bigg]\\ \geq&\lim_{T\rightarrow\infty} \frac{1}{pT} \ln\left(T^{p-1}\int_0^T e^{(\frac{p(\nu ^{X})^2}{2(1-p)}-p\varphi)t}dt\right) \\ =& \max\left(\frac{(\nu ^{X})^2}{2(1-p)}-\varphi,0\right), \end{aligned}$$

which implies that

$$ \lim_{T\rightarrow\infty}\frac{1}{T}\ln\mathbb{E}[\hat{F}^p_{T}]^{\frac {1}{p}} \geq\max\left(\frac{(\nu ^{F})^{2}}{2(1-p)}, \max\Big(\frac{(\nu ^{X})^2}{2(1-p)}-\varphi,0\Big)\right) = \lambda_2. $$

 □