Efficient Approximations for Utility-Based Pricing

Abstract

In a context of illiquidity, the reservation price is a well-accepted alternative to the usual martingale approach which does not apply. However, this price is not available in closed form and requires numerical methods such as Monte Carlo or polynomial approximations to evaluate it. We show that these methods can be inaccurate and propose a deterministic decomposition of the reservation price using the Lambert function. This decomposition allows us to perform an improved Monte Carlo method, which we name Lambert Monte Carlo (LMC) and to give deterministic approximations of the reservation price and of the optimal strategies based on the Lambert function. We also give an answer to the problem of selecting a hedging asset that minimizes the reservation price and also the cash invested. Our theoretical results are illustrated by numerical simulations.

Appendix

1.1 Taylor Approximation of the Reservation Price

As announced in Section 3, we provide the Taylor expansion of the reservation price when the correlation is close to 1. Let $G = e^{\eta \sqrt{T}N-\frac{\eta ^2 T}{2}}$ and $g_c$ be the cumulant-generating function of G, i.e., $g_c(t)=\ln (\mathbb {E}(e^{tG}))$. Since G is a log-normal random variable, $g_c$ exists only for negative values of t. Furthermore, as G has moments of any orders, the function $g_c$ is $C^\infty$ on $(-\infty ,0)$, with limits at $0^-$, for all derivatives. So, the function $g_c$ admits a Taylor expansion up to an arbitrary order $n>0$ at $0^-$ and has the following decomposition:

$$\begin{aligned} \forall t\in (-\infty ,0) \quad g_c(t)=\sum _{k=1}^{n}{\chi _k\frac{t^k}{k!}}+t^n\epsilon (t)\quad \text{ with } \quad \underset{t\rightarrow 0^-}{\lim }\epsilon (t)=0. \end{aligned}$$

(40)

The coefficients $\chi _k$ are called the cumulants of the distribution of G and have the following expressions:

$$\begin{aligned} \chi _1= & {} 1 \quad \chi _2 = e^{\eta ^2T}-1\\ \chi _3= & {} e^{3\eta ^2T}-3e^{\eta ^2T}+2 \quad \chi _4 = e^{6\eta ^2T}-4e^{3\eta ^2T}-3e^{2\eta ^2T}+12e^{\eta ^2T}-6\\ \chi _5= & {} e^{10\eta ^2T}-5e^{6\eta ^2T}-10e^{4\eta ^2T}+20e^{3\eta ^2T}+30e^{2\eta ^2T}-60e^{\eta ^2T}+24. \end{aligned}$$

As a function of $\rho$, the reservation price can be written as follows (see, (7) with $h=id$):

$$\begin{aligned} \begin{aligned} p(\rho )&={ -\frac{e^{-rT}}{\gamma (1-\rho ^2)}g_c\left( -\lambda \gamma (1-\rho ^2) s_0e^{\left( \nu -\eta \rho s_R \right) T}\right) } \\ {}&= \sum _{k=0}^{n}c_k(1-\rho )^k+(1-\rho )^n \varepsilon (\rho ), \end{aligned} \end{aligned}$$

(41)

with $\lim _{\rho \rightarrow 1^-}\varepsilon (\rho )=0$ and where the coefficients $(c_k)_{k>0}$ can be expressed in terms of the sequence $(\chi _k)_{k>0}$, for any arbitrary order $n\ge 0$. Set $\alpha = \eta s_R T$, $\omega =1-\rho$, $\overline{p}=\lambda s_0 e^{(\nu -\eta s_R)T}$ and $\hat{p}=e^{-rT}\overline{p}$. Note that $\hat{p}$ corresponds to the expectation of $\lambda e^{-rT}S_T$ computed under the unique probability measure which turns the discounted value of the traded asset P into a martingale in the complete case, i.e., when $\rho =1$. Indeed,

$$\begin{aligned} \frac{d\widetilde{S}_t}{\widetilde{S}_t}= \left( \nu -r-\eta s_R\right) dt+\frac{\eta }{\sigma }\frac{d\widetilde{P}_t}{\widetilde{P}_t}. \end{aligned}$$

We also introduce $\varphi (\omega )= \overline{p}\gamma \omega (\omega -2) e^{\alpha \omega }$. The reservation price p takes the following form as a function of $\omega$

$$\begin{aligned} p(\omega )= -\frac{e^{-rT}}{\gamma \omega (2-\omega )}\ln \left( \mathbb {E}\left( e^{\varphi (\omega )G}\right) \right) . \end{aligned}$$

(42)

We first compute the Taylor expansion of order 5 at $0^+$ of $\ln (\mathbb {E}(e^{\varphi G}))$. For that we use (40) for $n=5$ and as $\lim _{\omega \rightarrow 0^+}\varphi (\omega )/ \omega =0$, we obtain that

$$\begin{aligned} \begin{aligned} \ln \left( \mathbb {E}\left( e^{\varphi (\omega )G}\right) \right) =&\ \chi _1 \varphi (\omega )+\frac{\chi _2}{2}{\varphi }^2(\omega )+\frac{\chi _3}{6}{\varphi }^3(\omega )\\ {}&+ \frac{\chi _4}{24}{\varphi }^4(\omega )+\frac{\chi _5}{120}{\varphi }^5(\omega ) + \omega ^5 \varepsilon (\omega ), \end{aligned} \end{aligned}$$

(43)

where $\varepsilon$ will denote in the sequel a function such that $\lim _{\omega \rightarrow 0^+} \varepsilon (\omega )=0$. Now, we compute the Taylor expansion of order 5 of the powers of $\varphi$.

$$\begin{aligned} \varphi (\omega )&= \overline{p}\gamma (\omega ^2-2\omega ) \left( 1+\alpha \omega + \frac{\alpha ^2}{2}\omega ^2+\frac{\alpha ^3}{6}\omega ^3+\frac{\alpha ^4}{24}\omega ^4+\omega ^4 \varepsilon (\omega )\right) \\&= \overline{p}\gamma \left( -2\omega + (1-2\alpha )\omega ^2+\alpha (1-\alpha )\omega ^3+ \alpha ^2\left( \frac{1}{2}-\frac{\alpha }{3}\right) \omega ^4\right. \\ {}&\quad + \left. \frac{\alpha ^3}{6}\left( 1-\frac{\alpha }{2}\right) \omega ^5 \right) +\omega ^5 \varepsilon (\omega ))\\ {\varphi }^2(\omega )&= (\overline{p}\gamma )^2 \left( 4\omega ^2+4(2\alpha -1)\omega ^3+ \left( 8\alpha ^2-8\alpha +1\right) \omega ^4\right. \\ {}&\quad + \left. \alpha \left( \frac{16}{3}\alpha ^2-8\alpha +2\right) \omega ^5 \right) +\omega ^5 \varepsilon (\omega )\\ {\varphi }^3(\omega )&= \begin{aligned} (\overline{p}\gamma )^3 \left( -8\omega ^3+12\left( 1-2\alpha \right) \omega ^4+6\left( -6\alpha ^2+6\alpha -1\right) \omega ^5 \right) +\omega ^5 \varepsilon (\omega ) \end{aligned}\\ {\varphi }^4(\omega )&= \begin{aligned} (\overline{p}\gamma )^4 \left( 16\omega ^4+32\left( 2\alpha -1\right) \omega ^5 \right) +\omega ^5 \varepsilon (\omega )\end{aligned}\\ {\varphi }^5(\omega )&= \begin{aligned} -32 (\overline{p}\gamma )^5\omega ^5+\omega ^5 \varepsilon (\omega ).\end{aligned} \end{aligned}$$

Plugging the previous expressions in (43), we find that $\ln (\mathbb {E}(e^{\varphi (\omega )G})) = \sum _{i=1}^5 a_i\omega ^i + \omega ^5 \varepsilon (\omega ),$ where,

$$\begin{aligned} a_1&= -2\chi _1 \overline{p}\gamma \quad a_2 = \chi _1 (1-2\alpha )\overline{p}\gamma +2\chi _2 (\overline{p}\gamma )^2\\ a_3&= \chi _1\alpha (1-\alpha )\overline{p}\gamma +2\chi _2 (2\alpha -1)(\overline{p}\gamma )^2- \frac{4}{3}\chi _3(\overline{p}\gamma )^3\\ a_4&= \frac{1}{6}\chi _1\alpha ^2(3-2\alpha )\overline{p}\gamma +\frac{1}{2}\chi _2 (8\alpha ^2-8\alpha +1)(\overline{p}\gamma )^2 + 2\chi _3 (1-2\alpha )(\overline{p}\gamma )^3+\frac{2}{3}\chi _4 (\overline{p}\gamma )^4\\ a_5&= \frac{1}{12}\chi _1\alpha ^3(2-\alpha )\overline{p}\gamma +\frac{1}{3}\chi _2 \alpha (8\alpha ^2-12\alpha +3)(\overline{p}\gamma )^2\\ {}&\quad + \chi _3 (-6\alpha ^2+6\alpha -1)(\overline{p}\gamma )^3+\frac{4}{3}\chi _4 (2\alpha -1)(\overline{p}\gamma )^4 -\frac{4}{15}\chi _5(\overline{p}\gamma )^5. \end{aligned}$$

We now proceed to the expansion of p in (42).

$$\begin{aligned} p(\omega )= & {} -\frac{e^{-rT}}{2\gamma }\frac{1}{\left( 1-\frac{\omega }{2}\right) \omega }\ln \left( \mathbb {E}\left( e^{\varphi (\omega )G}\right) \right) \\ {}= & {} -\frac{e^{-rT}}{2\gamma }\left( 1+\frac{\omega }{2}+\frac{\omega ^2}{4}+\frac{\omega ^3}{8}+\frac{\omega ^4}{16}+\omega ^4\varepsilon (\omega )\right) \frac{1}{\omega } \left( \sum _{i=1}^5 a_i \omega ^i+\omega ^5 \varepsilon (\omega ) \right) \\= & {} -\frac{e^{-rT}}{2\gamma } \left( a_1+\left( \frac{a_1}{2}+a_2\right) \omega ^1+ \left( \frac{a_1}{4}+\frac{a_2}{2}+a_3\right) \omega ^2 + \left( \frac{a_1}{8}+\frac{a_2}{4}+ \frac{a_3}{2}+a_4\right) \omega ^3\right. \\ {}{} & {} \quad + \left. \left( \frac{a_1}{16}+\frac{a_2}{8}+\frac{a_3}{4}+\frac{a_4}{2}+a_5\right) \omega ^4 \right) +\omega ^4 \varepsilon (\omega ). \end{aligned}$$

Recalling that $\omega =1-\rho$ and noting that for all $1 \le k\le 4$, $-2\gamma e^{r T} c_{k}=a_{k+1}-\gamma e^{r T}c_{k-1}$, we determine that

$$\begin{aligned} -2\gamma e^{rT}c_0&= -2 \overline{p}\gamma \\ -2\gamma e^{rT}c_1&= (1-2\alpha )\overline{p}\gamma +2\chi _2 (\overline{p}\gamma )^2 - \overline{p}\gamma = -2 \alpha \overline{p}\gamma +2\chi _2(\overline{p}\gamma )^2 \\ -2\gamma e^{rT}c_2&= \alpha (1-\alpha )\overline{p}\gamma +2\chi _2 (2\alpha -1)(\overline{p}\gamma )^2- \frac{4}{3}\chi _3(\overline{p}\gamma )^3 + \chi _2(\overline{p}\gamma )^2 - \alpha \overline{p}\gamma \\ {}&= - \alpha ^2 \overline{p}\gamma + \chi _2 (4\alpha -1)(\overline{p}\gamma )^2-\frac{4}{3}\chi _3(\overline{p}\gamma )^3\\ -2\gamma e^{rT}c_3&= \frac{1}{6}\alpha ^2(3-2\alpha )\overline{p}\gamma +\frac{1}{2}\chi _2 (8\alpha ^2-8\alpha +1)(\overline{p}\gamma )^2+ 2\chi _3 (1-2\alpha )(\overline{p}\gamma )^3\\ {}&\quad +\frac{2}{3}\chi _4(\overline{p}\gamma )^4 -\frac{1}{2}\alpha ^2\overline{p}\gamma + \frac{1}{2}\chi _2 (4\alpha -1)(\overline{p}\gamma )^2-\frac{2}{3}\chi _3(\overline{p}\gamma )^3 \\ {}&= -\frac{1}{3}\alpha ^3\overline{p}\gamma +2\chi _2 \alpha (2\alpha -1)(\overline{p}\gamma )^2 +\frac{4}{3}\chi _3 (1-3\alpha )(\overline{p}\gamma )^3+\frac{2}{3}\chi _4(\overline{p}\gamma )^4\\ -2\gamma e^{rT}c_4&= \frac{1}{12}\alpha ^3(2-\alpha )\overline{p}\gamma +\frac{1}{3}\chi _2 \alpha (8\alpha ^2-12\alpha +3)(\overline{p}\gamma )^2\\ {}&\quad + \chi _3 (-6\alpha ^2+6\alpha -1)(\overline{p}\gamma )^3+\frac{4}{3}\chi _4 (2\alpha -1)(\overline{p}\gamma )^4 -\frac{4}{15}\chi _5(\overline{p}\gamma )^5 -\frac{1}{6}\alpha ^3\overline{p}\gamma \\ {}&\quad +\chi _2 \alpha (2\alpha -1)(\overline{p}\gamma )^2 + \frac{2}{3}\chi _3 (1-3\alpha )(\overline{p}\gamma )^3+\frac{1}{3}\chi _4(\overline{p}\gamma )^4\\&= -\frac{1}{12}\alpha ^4\overline{p}\gamma + \frac{2}{3}\chi _2 \alpha ^2 (4\alpha -3)(\overline{p}\gamma )^2+\chi _3 \left( -6\alpha ^2+4\alpha -\frac{1}{3}\right) (\overline{p}\gamma )^3\\&\quad +\frac{1}{3}\chi _4(8\alpha -3)(\overline{p}\gamma )^4-\frac{4}{15}\chi _5 (\overline{p}\gamma )^5. \end{aligned}$$

So, we obtain that

$$\begin{aligned} c_0= & {} \hat{p}\quad c_1 = \alpha \hat{p}-\chi _2\gamma e^{rT}\hat{p}^2 \quad c_2 = \frac{1}{2}\alpha ^2\hat{p}-\frac{1}{2}\chi _2\gamma (4\alpha -1)e^{rT}\hat{p}^2+\frac{2}{3}\chi _3\gamma ^2e^{2rT}\hat{p}^3\\ c_3= & {} \frac{1}{6}\alpha ^3\hat{p}-\chi _2\gamma \alpha (2\alpha -1)e^{rT}\hat{p}^2+\frac{2}{3}\chi _3\gamma ^2(3\alpha -1)e^{2rT}\hat{p}^3-\frac{1}{3}\chi _4\gamma ^3 e^{3rT}\hat{p}^4\\ c_4= & {} \frac{1}{24}\alpha ^4\hat{p}-\frac{1}{3}\chi _2\gamma \alpha ^2(4\alpha -3)e^{rT}\hat{p}^2+\chi _3\gamma ^2\left( 3\alpha ^2-2\alpha +\frac{1}{6}\right) e^{2rT}\hat{p}^3\\ {}{} & {} -\frac{1}{6}\chi _4\gamma ^3(8\alpha -3)e^{3rT}\hat{p}^4+\frac{2}{15}\chi _5\gamma ^4e^{4rT}\hat{p}^5. \end{aligned}$$

1.2 Proofs of Section 4

1.2.1 Properties of the Lambert Function

In this section, we provides some useful properties of the Lambert function that will be used in the proofs of our results. By definition, W has the following properties:

$$\begin{aligned} \forall x>-\frac{1}{e} \quad W(x)e^{W(x)} = x \text{ and } \forall x>-1\quad W(xe^{x})=x. \end{aligned}$$

(44)

Lemma 2

The Lambert function W is strictly increasing $C^{\infty }$ and

$$\begin{aligned} \forall x>-\frac{1}{e},\; W'(x)=\frac{1}{e^{W(x)}+x}=\frac{W(x)}{x(1+W(x))}. \end{aligned}$$

(45)

Moreover, W admits a Taylor expansion around 0: $W(x)\!=\!x-x^{2}+x^2\epsilon (x) \text{ with } \underset{x \rightarrow 0}{\lim }\, \epsilon (x)\!=\!0$.

Finally, $W(x)\underset{x \rightarrow +\infty }{\sim }\ln (x)$.

Note that Lemma 2 and (44) imply that $W(0)=0$ and thus that $W(x)>0, \quad \forall x>0$.

Proof of Lemma 2

First, W is clearly $C^{\infty }$, being the inverse of $x\mapsto x e^x$. Now, we differentiate (44) and obtain that for $x>-\frac{1}{e}$,

$$\begin{aligned} \begin{aligned} (We^W)'(x)=1&\iff W'(x)e^{W(x)}+W'(x)W(x)e^{W(x)}=1 \\&\iff W'(x)e^{W(x)}\left( 1+W(x)\right) =1 \\&\iff W'(x)=\frac{1}{e^{W(x)}+x}=\frac{W(x)}{x(1+W(x))}, \end{aligned} \end{aligned}$$

(46)

where we have used for the last equivalence the fact that $\forall x>-\frac{1}{e},\;W(x)>-1$ and (44).

As $\forall x>-\frac{1}{e},\; W'(x)>0$, W is strictly increasing. Recalling that W is $C^{\infty }$, we determine that

$$\begin{aligned} W(x)=W(0)+W'(0)x+\frac{W''(0)}{2}x^2+x^2\epsilon (x)\quad \text{ with } \quad \underset{x \rightarrow 0}{\lim }\; \epsilon (x)=0. \end{aligned}$$

Using (44) and (45), $W(0)=0$ and $W'(0)=1$. Now differentiating (46), we determine that for $x>-\frac{1}{e}$

$$W''(x)=-\frac{1+W'(x)e^{W(x)}}{(x+e^{W(x)})^{2}}$$

and $W''(0)=-2$. Finally, as $\lim _{x\rightarrow +\infty } W(x)=+\infty$, (44) provides that

$$\begin{aligned} \lim _{x\rightarrow +\infty }\frac{\ln (x)}{W(x)}=1+\lim _{x\rightarrow +\infty }\frac{\ln (W(x))}{W(x)}=1. \end{aligned}$$

$\square$

1.2.2 Proof of Theorem 1

Theorem 1 provides a multiplicative decomposition of the Laplace transform of a function of a log-normal random variable. For that, we need to find the minimal value of $k_\beta$ (see (11)). This is the purpose of Lemma 3. Fix $\theta , v>0$. Let m be defined on $[0,\infty ]$ by

$$\begin{aligned} m(\beta ) = \left\{ \begin{array}{ll} -\frac{W(\theta v \eta ^2 T )}{\eta \sqrt{T}} &{} \text{ if }\;\; \beta \in \left[ \frac{1}{\theta v} \left( \frac{W(\theta v \eta ^2 T)}{\eta ^{2}T} + \frac{W^2(\theta v \eta ^2 T)}{2\eta ^{2} T}\right) ,+\infty \right] \\ 0 &{} \text{ elsewhere }. \end{array} \right. \end{aligned}$$

Lemma 3

Let $\beta \in [0,+\infty ]$ and $\theta , v>0$. Let u be any real number if $\beta =+\infty$ and $u=-v\beta$ otherwise. Then, $k_\beta$ reaches its minimum value in $m(\beta )$. Moreover,

$$\begin{aligned} k_\beta (m(\beta ))= & {} -\left( \theta v \beta -\frac{W(\theta v \eta ^2 T)}{\eta ^{2}T} - \frac{W^2(\theta v \eta ^2 T)}{2\eta ^{2} T}\right) _+ \end{aligned}$$

(47)

$$\begin{aligned} k_{+\infty }(m(+\infty ))= & {} \frac{W(\theta v \eta ^2 T)}{\eta ^{2}T} + \frac{W^2(\theta v \eta ^2 T)}{2\eta ^{2}T}+\theta u. \end{aligned}$$

(48)

Proof of Lemma 3

We first study $k: z\mapsto \theta \left( u+v e^{\eta \sqrt{T}z}\right) +\frac{z^2}{2}$. Then, we use (11) to link the minima of $k_\beta$ and of k according to the values of $\beta$. As $v>0$, for $z\ge 0$, we have $k'(z)>0$. Letting $z<0$, as W is strictly increasing and using (44), we obtain that

$$\begin{aligned} k'(z) \ge 0\Leftrightarrow & {} \theta v \eta ^2 T \ge - \eta \sqrt{T} z e^{-\eta \sqrt{T} z} \Leftrightarrow W( \theta v \eta ^2 T) \ge W(- \eta \sqrt{T} z e^{-\eta \sqrt{T} z} )\\ {}= & {} -\eta \sqrt{T}z \Leftrightarrow z \ge - \frac{W( \theta v \eta ^2 T)}{\eta \sqrt{T}}, \end{aligned}$$

which gives the variations of k. Moreover, using (44),

$$\begin{aligned} k \left( -\frac{W(\theta v \eta ^2 T )}{\eta \sqrt{T}} \right)= & {} \theta v e^{-W(\theta v \eta ^2 T) } + \frac{W^2(\theta v \eta ^2 T)}{2\eta ^{2}T}+\theta u\\ {}= & {} \frac{W(\theta v \eta ^2 T)}{\eta ^{2}T} + \frac{W^2(\theta v \eta ^2 T)}{2\eta ^{2} T}+\theta u. \end{aligned}$$

As $k_{\infty }=k$, we determine that $k_\beta$ reaches its minimum value in $m(+\infty )$ and that (48) holds true.

We now study $k_\beta$ if $\beta <\infty$. As $u=-v\beta$, $k_\beta$ is continuous.

Assume first that $\frac{W(\theta v \eta ^2 T)}{\eta ^{2}T} + \frac{W^2(\theta v \eta ^2 T)}{2\eta ^{2} T}\le \theta v\beta$. We show that $k_\beta$ reaches its minimal value at $-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}}$. As $v,\, \theta >0$, (44) implies that

$$\begin{aligned} e^{-W(\theta v \eta ^2 T)}=\frac{W(\theta v \eta ^2 T)}{\theta v \eta ^2 T}\le \beta . \end{aligned}$$

(49)

As (49) is equivalent to $-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}} \le \frac{\ln \beta }{\eta \sqrt{T}}$, (11) and (48) show that

$$\begin{aligned} k_\beta \left( -\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}}\right)= & {} k\left( -\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}}\right) = \frac{W(\theta v \eta ^2 T)}{\eta ^{2}T} + \frac{W^2(\theta v \eta ^2 T)}{2\eta ^{2} T}-\theta v \beta . \qquad \end{aligned}$$

(50)

As $k_\beta =k$ on $(-\infty ,\frac{\ln \beta }{\eta \sqrt{T}})$, it also shows that $k_\beta$ is decreasing on $(-\infty ,-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}})$ and increasing on $(-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}},\frac{\ln \beta }{\eta \sqrt{T}})$. Now, on $(\frac{\ln \beta }{\eta \sqrt{T}},+\infty )$ $k_\beta (z)=\frac{z^2}{2}$, we distinguish two cases. First, assume that $\beta \ge 1$. Then, $-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}} \le 0 \le \frac{\ln \beta }{\eta \sqrt{T}}$ and $k_\beta$ is increasing on $(\frac{\ln \beta }{\eta \sqrt{T}},+\infty )$. Thus, $k_\beta$ reaches its minimum value in $-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}}$. If now $\beta <1$, using (49) again, we find that $-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}} \le \frac{\ln \beta }{\eta \sqrt{T}}\le 0$. Therefore, $k_\beta$ is decreasing on $(\frac{\ln \beta }{\eta \sqrt{T}},0)$ and increasing on $(0,+\infty )$. Thus, 0 and $-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}}$ are two potential minima. However, using (50), we determine that $k_\beta (-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}}) \le 0 =k_\beta (0)$, and we conclude that $k_\beta$ reaches its minimal value at $-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}}$.

Suppose now that $\theta v\beta <\frac{W(\theta v \eta ^2 T)}{\eta ^{2}T} + \frac{W^2(\theta v \eta ^2 T)}{2\eta ^{2} T}$. This implies that $\beta < 1$. Indeed, using (44) and $1+x\le e^x$,

$$\begin{aligned} \beta < \frac{W(\theta v \eta ^2 T)}{\theta v\eta ^{2}T}\left( 1 + \frac{W(\theta v \eta ^2 T)}{2}\right) = e^{-W(\theta v \eta ^2 T)}\left( 1 + \frac{W(\theta v \eta ^2 T)}{2}\right) \le 1. \end{aligned}$$

We want to show that $k_\beta$ reaches its minimum value at 0. Then, as $\beta <1$, $k_\beta (0)=0$.

Assume first that $\beta \le \frac{W(\theta v \eta ^2 T)}{\theta v\eta ^{2}T}=e^{-W(\theta v \eta ^2 T)}$, then $\frac{\ln \beta }{\eta \sqrt{T}}\le -\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}}$. As $k_\beta =k$ on $(-\infty ,\frac{\ln \beta }{\eta \sqrt{T}})$, $k_\beta$ is decreasing on $(-\infty ,\frac{\ln \beta }{\eta \sqrt{T}})$. Moreover, on $(\frac{\ln \beta }{\eta \sqrt{T}},+\infty )$, $k_\beta (z)=\frac{z^2}{2}$ and $k_\beta$ is decreasing on $(\frac{\ln \beta }{\eta \sqrt{T}},0)$ and increasing on $(0,\infty )$. Thus, $k_\beta$ reaches its minimum value at 0.

Assuming now that $\beta > \frac{W(\theta v \eta ^2 T)}{\theta v\eta ^{2}T}=e^{-W(\theta v \eta ^2 T)}$, we determine that $-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}}<\frac{\ln \beta }{\eta \sqrt{T}}< 0$. Thus, as $k_\beta =k$ on $(-\infty ,\frac{\ln \beta }{\eta \sqrt{T}})$, $k_\beta$ is decreasing on $(-\infty ,-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}})$ and increasing on $(-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}}, \frac{\ln \beta }{\eta \sqrt{T}})$. Moreover, on $(\frac{\ln \beta }{\eta \sqrt{T}},\infty )$, $k_\beta (z)=\frac{z^2}{2}$, and $k_\beta$ is decreasing on $(\frac{\ln \beta }{\eta \sqrt{T}},0)$ and increasing on $(0,+\infty )$. Thus, $-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}}$ and 0 are two potential minima. Using (50), we determine that $k_\beta (-\frac{W(\theta v \eta ^2 T)}{\eta \sqrt{T}})>k_\beta (0)$ and we conclude that $k_\beta$ reaches its minimum in 0. $\square$

We are now in position to prove Theorem 1.

Proof of Theorem 1

We perform the Laplace method on (10) and thus, get out the maximum of $e^{-k_\beta }$ of the integral. This gives the deterministic part of the decomposition. Then, an affine change of variable is applied to the remaining integral in order to retrieve the required expression.

Let $\beta \in [0,+\infty ]$ and $\theta , v>0$. Let u be any real number if $\beta =+\infty$ and $u=-v\beta$ otherwise. Assume that (12) holds true. We set $\hat{\eta }=\eta \sqrt{T}$ and $\hat{w}=W(\theta v \hat{\eta }^2)$. Using (10), we have that

$$\begin{aligned} \begin{aligned} L_{f,\beta }(\theta )=&\ \exp (-k_\beta (m(\beta ))) \int _{\mathbb {R}} \exp \Big [ -\theta (f(e^{\hat{\eta } z})-u -v e^{\hat{\eta } z})1_{z\le \frac{\ln (\beta )}{\hat{\eta }}} \\ {}&- (k_{\beta }(z)-k_\beta (m(\beta ))) \Big ] \frac{dz}{\sqrt{2\pi }}. \end{aligned} \end{aligned}$$

(51)

Recall that when $\beta <+\infty$, we have chosen $u=-v\beta$. Thus, (12) and Lemma 3 imply that $m(\beta )=-\hat{w}/\hat{\eta }$. Using (47) and (48), we determine that

$$\begin{aligned} \exp \left[ -k_\beta (m(\beta ))\right]= & {} \exp {\left[ -\left( \theta u + \frac{\hat{w}}{\hat{\eta }^2}+\frac{\hat{w}^2}{2\hat{\eta }^2}\right) \right] }={L}_\beta (\theta ). \end{aligned}$$

(52)

Then, (51) and (52) with the change of variable $z=y-m(\beta )=y+\hat{w}/\hat{\eta }$ and the fact that $e^{-\hat{w}}=\hat{w}/(\theta v \hat{\eta }^2)$ (see (44)) imply that

$$\begin{aligned} \begin{aligned} L_{f,\beta }(\theta )&= {L}_\beta (\theta )\int _{\mathbb {R}}\exp \bigg [-\theta \left( f\left( \frac{\hat{w}}{\theta v \hat{\eta }^2}e^{\hat{\eta } y}\right) -u -v \frac{\hat{w}}{\theta v\hat{\eta }^2}e^{\hat{\eta } y}\right) 1_{y\le \frac{\ln \beta }{\hat{\eta }}+\frac{\hat{w}}{\hat{\eta }}}\\ {}&-\left( k_{\beta }\left( y-\frac{\hat{w}}{\hat{\eta }}\right) -k_\beta \left( -\frac{\hat{w}}{\hat{\eta }}\right) \right) \bigg ]\frac{dz}{\sqrt{2\pi }}. \end{aligned} \end{aligned}$$

(53)

Letting $y\in \mathbb {R}$, we find that

$$\begin{aligned} k_\beta \left( y-\frac{\hat{w}}{\hat{\eta }}\right)= & {} \theta (u+ ve^{\hat{\eta }y}e^{-\hat{w}})1_{y\le \frac{\ln \beta }{\hat{\eta }}+\frac{\hat{w}}{\hat{\eta }}}+\frac{(y-\frac{\hat{w}}{\hat{\eta }})^2}{2}\\ {}= & {} \theta u 1_{y\le \frac{\ln \beta }{\hat{\eta }}+\frac{\hat{w}}{\hat{\eta }}} +\frac{\hat{w}}{\hat{\eta }^2}e^{\hat{\eta }y}1_{y\le \frac{\ln \beta }{\hat{\eta }}+\frac{\hat{w}}{\hat{\eta }}}+\frac{\hat{w}^2}{2\hat{\eta }^2}-\frac{\hat{w}}{\hat{\eta }}y+\frac{y^2}{2}. \end{aligned}$$

We remark that $\frac{\ln \beta }{\hat{\eta }}+\frac{\hat{w}}{\hat{\eta }} \ge 0$. This is trivially true if $\beta =+\infty$. If $\beta <+\infty$, as $u=-v\beta$

$$\begin{aligned} \frac{\ln \beta }{\hat{\eta }}+ \frac{\hat{w}}{\hat{\eta }} \ge 0\Leftrightarrow & {} \beta e^{\hat{w}}= \beta \frac{\theta v \hat{\eta }^2}{\hat{w}} \ge 1 \Leftrightarrow \theta v \beta \ge \frac{\hat{w}}{\hat{\eta }^2}, \end{aligned}$$

which is true by (12). Thus, $k_\beta (-\frac{\hat{w}}{\hat{\eta }}) = \theta u + \frac{\hat{w}}{\hat{\eta }^2}+ \frac{\hat{w}^2}{2\hat{\eta }^2}$ and

$$\begin{aligned} k_\beta \left( y-\frac{\hat{w}}{\hat{\eta }}\right) -k_\beta \left( -\frac{\hat{w}}{\hat{\eta }}\right)= & {} \frac{\hat{w}}{\hat{\eta }^2}(e^{\hat{\eta } y}-1-\hat{\eta } y) -\left( \theta u+\frac{\hat{w}}{\hat{\eta }^2}e^{\hat{\eta } y}\right) 1_{y>\frac{\ln \beta }{\hat{\eta }}+\frac{\hat{w}}{\hat{\eta }}}+\frac{y^2}{2}.\nonumber \end{aligned}$$

Assume that $\beta <+\infty$. Then,

$$\begin{aligned} y>\frac{\ln \beta }{\hat{\eta }}+ \frac{\hat{w}}{\hat{\eta }}\Leftrightarrow & {} e^{\hat{\eta }y}> \beta e^{\hat{w}}= \beta \frac{\theta v \hat{\eta }^2}{\hat{w}} \Leftrightarrow \frac{\hat{w}}{\hat{\eta }^2}e^{\hat{\eta }y}> \theta v \beta . \end{aligned}$$

Thus, as $u=-v\beta$,

$$\begin{aligned} \left( \theta u+\frac{\hat{w}}{\hat{\eta }^2}e^{\hat{\eta }y}\right) 1_{y>\frac{\ln \beta }{\hat{\eta }}+\frac{\hat{w}}{\hat{\eta }}} = \left( \frac{\hat{w}}{\hat{\eta }^2}e^{\hat{\eta }y}-\theta v \beta \right) 1_{\frac{\hat{w}}{\hat{\eta }^2}e^{\hat{\eta }y}- \theta v \beta >0} =\left( \frac{\hat{w}}{\hat{\eta }^2}e^{\hat{\eta }y}-\theta v \beta \right) _+. \end{aligned}$$

The same obviously holds true when $\beta =+\infty$, with the convention $(-\infty )_+=0$. Thus, we get that

$$\begin{aligned} \exp \left[ -\left( k_\beta \left( y-\frac{\hat{w}}{\hat{\eta }}\right) -k_\beta \left( -\frac{\hat{w}}{\hat{\eta }}\right) \right) \right]= & {} \phi _{\beta }(y,\theta )e^{-\frac{y^2}{2}}. \end{aligned}$$

Therefore, (53) implies that

$$\begin{aligned} \begin{aligned} L_{f,\beta }(\theta )=&\;{L}_\beta (\theta ) \int _{\mathbb {R}}\exp \left[ -\theta \left( f\left( \frac{\hat{w}}{\theta v \hat{\eta }^2}e^{\hat{\eta }y}\right) -u-\frac{\hat{w}}{\theta \hat{\eta }^2}e^{\hat{\eta } y}\right) 1_{y\le \frac{\ln \beta }{\hat{\eta }}+\frac{\hat{w}}{\hat{\eta }}}\right] \times \\&\; \phi _\beta (y,\theta )e^{-\frac{y^2}{2}} \frac{dy}{\sqrt{2\pi }}={L}_\beta (\theta )I_{f,\beta }(\theta ). \end{aligned} \end{aligned}$$

$\square$

1.3 Proofs of Section 5

For ease of notation, we introduce the functions $\theta : \rho \in (-1,1)\mapsto \lambda \gamma (1-\rho ^2)$, $\widetilde{\theta }: \gamma \in (0,+\infty )\mapsto \lambda \gamma (1-\rho ^2)$ and

$$\begin{aligned} w \;:\;&\rho \in (-1,1)\mapsto W\left( s_0 \eta ^2 Te^{\left( \nu -\eta \rho s_R -\frac{\eta ^2}{2}\right) T} \theta (\rho )\right) \end{aligned}$$

(54)

$$\begin{aligned} \widetilde{w} \;:\;&\gamma \in (0,+\infty )\mapsto W\left( s_0 \eta ^2 Te^{\left( \nu -\eta \rho s_R -\frac{\eta ^2}{2}\right) T} \widetilde{\theta } (\gamma )\right) . \end{aligned}$$

(55)

Then, $A=a(\rho )$, $B=b(\rho )$, $D=d(\rho )$, $G=g(\rho )$ and $D=\widetilde{d}(\gamma )$, where

$$\begin{aligned} \begin{aligned} d\; :\;&\rho \in (-1,1) \mapsto \frac{ e^{-rT}}{\gamma \eta ^2 T(1-\rho ^2)} \left( {w}(\rho )+\frac{w^2 (\rho )}{2}\right) \\ a\; :\;&\rho \in \ (-1,1)\mapsto -\frac{ e^{-rT}}{\gamma (1-\rho ^2)}\ln \mathbb {E} \left( \exp \left[ - \frac{{w}(\rho )}{\eta ^2 T}\left( e^{\eta \sqrt{T}N}-1-\eta \sqrt{T}N\right) \right] \right) \quad \end{aligned} \end{aligned}$$

(56)

$$\begin{aligned} b\; :\;\rho \in \ (-1,1)\mapsto \frac{e^{-rT}}{\gamma (1-\rho ^2)}\frac{{w}(\rho )}{\eta ^2 T} \left( e^{\frac{\eta ^2}{2}T}-1\right) \end{aligned}$$

(57)

$$\begin{aligned} g\; :\; \rho \in (-1,1) \mapsto d(\rho )+b(\rho )= \frac{ e^{-rT}}{\gamma (1-\rho ^2)}\frac{w(\rho )}{\eta ^2 T}\left( e^{\frac{\eta ^2}{2} T}+\frac{w(\rho )}{2}\right) \end{aligned}$$

(58)

$$\begin{aligned} \widetilde{d}\; :\; \gamma \in (0,+\infty ) \mapsto \frac{ \lambda e^{-rT}}{\widetilde{\theta }(\gamma )\eta ^2 T} \left( \widetilde{w} (\gamma )+\frac{\widetilde{w}^2 (\gamma )}{2}\right) \end{aligned}$$

(59)

1.3.1 Proofs of Subsection 5.2

We start with the proof of Theorem 6. Proposition 1 gives some asymptotics of d and g, while Proposition 2 gives some greeks related to d.

Proof of Theorem 6

Assume for a moment that we have proved that for all $\rho \in (-1,1)$,

$$\begin{aligned} \left( b(\rho )-\frac{ e^{-r T}}{\gamma (1-\rho ^2)}\frac{{w}(\rho )^2}{2\eta ^4 T^2}e_2\right) _+\le a(\rho ) \le b(\rho ) \end{aligned}$$

(60)

Then, as $p=D+A$, Remark 5, (58) and (60) show that $D \le p \le d(\rho ) +b(\rho )=G$. Then, $V_D(x_0,\lambda )\le V(x_0,\lambda ) \le V_G(x_0,\lambda )$ follows from (22) and (8).

We prove now (60). Using the Jensen inequality for the convex function $x \mapsto e^{-x}$, we obtain that

$$\begin{aligned} \mathbb {E}\left( \exp \left[ -\frac{{w}(\rho )}{\eta ^2 T} \left( e^{\eta \sqrt{T} N}-1-\eta \sqrt{T} N\right) \right] \right)\ge & {} \exp \left[ -\frac{{w}(\rho )}{\eta ^2 T}\left( e^{\frac{\eta ^2}{2}T}-1\right) \right] . \end{aligned}$$

And so, using Remark 5, we determine that $0 \le a(\rho ) \le \frac{ \lambda e^{-rT}}{\theta (\rho )}\frac{{w}(\rho )}{\eta ^2 T} \left( e^{\frac{\eta ^2}{2}T}-1\right) =b(\rho )$. Then, as $e^{-x}\le 1-x+\frac{x^2}{2}$ for all $x\ge 0$, we have that

$$\begin{aligned} \mathbb {E}\left( \exp \left[ -\frac{{w}(\rho )}{\eta ^2 T} \left( e^{\eta \sqrt{T} N}-1-\eta \sqrt{T} N\right) \right] \right) \le 1-\frac{{w}(\rho )}{\eta ^2 T}\left( e^{\frac{\eta ^2 T}{2}}-1\right) +\frac{{w}(\rho )^2}{2\eta ^4 T^2}e_2, \end{aligned}$$

where $e_2=\mathbb {E}\left( \left( e^{\eta \sqrt{T}N}-1-\eta \sqrt{T}N\right) ^2\right) = e^{2\eta ^2 T}-2(1+\eta ^2 T)e^{\frac{\eta ^2 T}{2}}+\eta ^2 T +1$. Composing with the logarithm function and using the inequality $\ln (1+x)\le x$ yields

$$\begin{aligned} a(\rho ) \ge b(\rho )-\frac{\lambda e^{-r T}}{\theta (\rho )}\frac{{w}(\rho )^2}{2\eta ^4 T^2}e_2, \end{aligned}$$

(61)

which combined with $a(\rho )\ge 0$ yields in turn the lower bound for $a(\rho )$.

Now, we prove (24) and (25). As $D\le p=D+A \le D+B$, we find that

$$\begin{aligned} \begin{aligned}&1\le \frac{p}{D} \le 1+ \frac{e^{\frac{\eta ^2T}{2}}-1}{1+\frac{{w}(\rho )}{2}} \iff \frac{1+\frac{{w}(\rho )}{2}}{e^{\frac{\eta ^2T}{2}}+\frac{{w}(\rho )}{2}}\le \frac{D}{p}\le 1,\\&\frac{1+\frac{{w}(\rho )}{2}}{e^{\frac{\eta ^2T}{2}}+\frac{{w}(\rho )}{2}} = e^{-\frac{\eta ^2T}{2}}\frac{e^{\frac{\eta ^2T}{2}}+\frac{{w}(\rho )}{2}e^{\frac{\eta ^2T}{2}}}{e^{\frac{\eta ^2T}{2}}+\frac{{w}(\rho )}{2}}\ge e^{-\frac{\eta ^2T}{2}}, \end{aligned} \end{aligned}$$

as ${w}(\rho )\ge 0$. Moreover, using (58) and (61), $p=G+A-B \ge G-\frac{\lambda e^{-r T}}{\theta (\rho )}\frac{{w}(\rho )^2}{2\eta ^4 T^2}e_2$. As $p\ge D$, using (56), we obtain that

$$\begin{aligned} \begin{aligned} \frac{G}{p} \le 1+ \frac{\lambda e^{-r T}}{\theta (\rho )}\frac{{w}(\rho )^2}{2\eta ^4 T^2 p}e_2 \le&\;1+ \frac{\lambda e^{-r T}}{\theta (\rho )}\frac{{w}(\rho )^2}{2\eta ^4 T^2 D}e_2 \\&= 1+\frac{{w}(\rho )e_2}{2\eta ^2 T+\eta ^2 T{w}(\rho )}\le 1+\frac{e_2}{\eta ^2 T}. \end{aligned} \end{aligned}$$

$\square$

Proof of Proposition 1

The proof relies on the properties of the Lambert function given in Lemma 2. As ${w}(\rho ) > 0$, we find that $d>0$ and $g>0$. Lemma 2 imply that d and g are $C^1$ and that

$$\begin{aligned} \frac{w (\rho )}{\gamma (1-\rho ^2) \eta ^2T}\underset{\rho \rightarrow 1^-}{\sim } & {} s_0 e^{\left( \nu -\eta s_R -\frac{\eta ^2}{2}\right) T}. \end{aligned}$$

(62)

As $\lim _{\rho \mapsto 1^-}{w}( \rho )=0$, the first limits in (26) and (27) are proved. The case when $\rho$ goes to $-1^+$ is treated similarly. We deduce that d and g are bounded on $(-1,1)$.

Recalling (62), we obtain that

$$\begin{aligned} \lim _{\rho \rightarrow 1^-}b(\rho )= & {} \lambda e^{-rT} s_0 e^{\left( \nu -\eta s_R-\frac{\eta ^2}{2}\right) T}\left( e^{\frac{\eta ^2T}{2}}-1\right) \\ \lim _{\rho \rightarrow -1^+}b(\rho )= & {} \lambda e^{-rT} s_0 e^{\left( \nu +\eta s_R-\frac{\eta ^2}{2}\right) T}\left( e^{\frac{\eta ^2T}{2}}-1\right) . \end{aligned}$$

Since b is continuous on $(-1,1)$ (recall that W is continuous), b is bounded, and since $0\le a \le b$, a is also bounded.

Using Lemma 2, we determine that $\widetilde{w}(\gamma )\! \underset{\gamma \rightarrow 0^+}{\sim }\ \! s_0 \eta ^2 T e^{(\nu -\eta \rho s_R-\frac{\eta ^2}{2})T}\widetilde{\theta }(\gamma )$ and $\widetilde{w}(\gamma )\! \underset{\gamma \rightarrow +\infty }{\sim }\ \ln (\gamma )$. Thus,

$$\begin{aligned} \widetilde{d}(\gamma )\underset{\gamma \rightarrow 0^+}{\sim } & {} \lambda e^{-rT} s_0 e^{(\nu -\eta \rho s_R-\frac{\eta ^2}{2})T}\left( 1+\frac{\widetilde{w}(\gamma )}{2}\right) \underset{\gamma \rightarrow 0^+}{\sim }\ \lambda e^{-rT} s_0 e^{(\nu -\eta \rho s_R-\frac{\eta ^2}{2})T} \\ {}\underset{\gamma \rightarrow +\infty }{\sim } & {} \frac{ \lambda e^{-rT}\ln ^2( \gamma )}{2\widetilde{\theta }(\gamma )\eta ^2 T} \underset{\gamma \rightarrow +\infty }{\rightarrow }\ 0. \end{aligned}$$

$\square$

Proof of Proposition 2

The proof consists in the differentiation of d and $\widetilde{d}$ together with Lemma 2 in order to simplify the equalities involving the Lambert function. Let $\delta : \rho \in [-1,1]\mapsto \nu -\eta \rho s_R$. We easily determine that for all $\rho \in (-1,1)$,

$$\begin{aligned} d'(\rho )= & {} \frac{ \lambda e^{-rT} }{\eta ^2 T \theta (\rho )}\left( w' (\rho )(1+ {w (\rho )}) - \frac{\theta '(\rho )}{\theta (\rho )} w (\rho )\left( 1+ \frac{w (\rho )}{2}\right) \right) . \end{aligned}$$

As $s_0 \eta ^2 Te^{(\delta (\rho ) -\frac{\eta ^2}{2})T} \theta (\rho ) \ge 0$, (45) and (54) imply that

$$\begin{aligned} w'(\rho )&= s_0 \eta ^2 Te^{\left( \delta (\rho ) -\frac{\eta ^2}{2}\right) T} \theta (\rho ) W'\left( s_0 \eta ^2 Te^{\left( \delta (\rho ) -\frac{\eta ^2}{2}\right) T} \theta (\rho ) \right) \left( \frac{\theta '(\rho )}{\theta (\rho )} - \eta s_R T\right) \\&= \frac{\frac{\theta '(\rho )}{\theta (\rho )} -\eta s_R T }{1+ \frac{1}{W\left( s_0 \eta ^2 Te^{(\delta (\rho ) -\frac{\eta ^2}{2})T} \theta (\rho ) \right) }}=\frac{\frac{\theta '(\rho )}{\theta (\rho )} -\eta s_R T }{1+ \frac{1}{w(\rho )}}. \end{aligned}$$

It follows that

$$\begin{aligned} \begin{aligned} \Delta _\rho = d'(\rho )&= \frac{ \lambda e^{-rT} w(\rho )}{\eta ^2 T \theta (\rho )}\left( \frac{\theta '(\rho )}{\theta (\rho )} -\eta s_R T - \frac{\theta '(\rho )}{\theta (\rho )} \left( 1+ \frac{w (\rho )}{2}\right) \right) \\ {}&= -\frac{ \lambda e^{-rT} w(\rho )}{\eta ^2 T \theta (\rho )}\left( \eta s_R T - \frac{\rho }{1-\rho ^2}{w}(\rho )\right) . \end{aligned} \end{aligned}$$

(63)

Using (45), we see that $\widetilde{w}'(\gamma )=\frac{\widetilde{\theta }'(\gamma )}{\widetilde{\theta }(\gamma )}\frac{\widetilde{w}(\gamma )}{1+\widetilde{w}(\gamma )}$. The function $\widetilde{d}$ is $C^{1}$ (see Lemma 2) and we easily ascertain that

$$\begin{aligned} \begin{aligned} \Delta _\gamma =\widetilde{d}'(\gamma )&= \frac{\lambda e^{-rT}}{\eta ^2 T}\left( \frac{\widetilde{w}'(\gamma )}{\widetilde{\theta }(\gamma )}(1+\widetilde{w}(\gamma ))-\frac{\widetilde{\theta }'(\gamma )}{\widetilde{\theta }^2(\gamma )}\left( \widetilde{w}(\gamma )+\frac{\widetilde{w}^2(\gamma )}{2}\right) \right) \\&= -\frac{\lambda e^{-rT}}{2\widetilde{\theta }^2(\gamma )\eta ^2 T}\widetilde{w}^2(\gamma )\widetilde{\theta }'(\gamma ) = -\frac{ e^{-rT}}{2 \eta ^2 T\gamma ^2(1-\rho ^2)}\widetilde{w}^2(\gamma ). \end{aligned} \end{aligned}$$

$\square$

1.3.2 Proofs of Subsection 5.5

Proposition 3 gives the expression of the minimum of d in $\rho$ as well as the value of d at that minimum. The resulting minimum is then expanded in Lemma 1 using the Taylor expansion of the Lambert function given in Lemma 2. Proposition 4 highlights the link between the deterministic strategy given in (33) and $\rho ^*$. Proposition 5 gives the expression of the correlation between S and P for which an agent would not invest much in P in the case of a short maturity.

Proof of Proposition 3

As ${w}(\rho )> 0$ and $\theta (\rho )> 0$ for all $\rho \in (-1,1)$, (63) shows that the sign of $d'(\rho )$ is the sign of $\frac{\rho }{1-\rho ^2} {w}(\rho )-\eta s_R T$. Supposing first that $\mu \ge r$, then for $\rho \in (-1,0]$, $d'(\rho )\le 0$. If now $\rho \in (0,1)$

$$\begin{aligned} d'(\rho )\ge 0\iff & {} W\left( s_0\eta ^2 Te^{\left( \delta (\rho ) -\frac{\eta ^2}{2}\right) T} \theta (\rho ) \right) \ge \eta s_R T \frac{1-\rho ^2}{\rho }\\\iff & {} s_0 \eta ^2 Te^{\left( \nu -\eta \rho s_R-\frac{\eta ^2}{2}\right) T} \theta (\rho ) \ge \eta s_R T \frac{1-\rho ^2}{\rho } e^{\eta s_R \frac{1-\rho ^2}{\rho }T}\\\iff & {} \lambda \gamma s_0 \eta ^2 Te^{\left( \nu -\frac{\eta ^2}{2}\right) T} \ge \eta \frac{s_R}{\rho }T e^{\eta \frac{s_R}{ \rho }T }\\\iff & {} W\left( \lambda \gamma s_0 \eta ^2 Te^{\left( \nu -\frac{\eta ^2}{2}\right) T} \right) \ge \eta \frac{s_R}{\rho }T \\\iff & {} \rho \ge \rho ^*. \end{aligned}$$

For the second equivalence, we have composed by the reciprocal function of W and for the fourth one by W (see (44)).

Suppose now that $\mu \le r$. Then, for $\rho \in [0,1)$, $d'(\rho )\ge 0$. If now $\rho \in (-1,0)$, we prove as before that $d'(\rho )\ge 0$ if and only if $\rho \ge \rho ^*$. The results for the variation of d are as follow.

Setting $\kappa = \lambda \gamma s_0 \eta ^2 Te^{(\nu -\frac{\eta ^2}{2})T}>0$, then $\rho ^*=(\eta s_R T)/W(\kappa )$. Assume now that $-1<\rho ^*<1$,

$$\begin{aligned} \begin{aligned} w(\rho ^*)&= W\left( \kappa e^{-\eta \rho ^* s_R T}(1-\rho ^{* 2})\right) = W\left( e^{-\frac{\left( \eta s_R T\right) ^2}{W(\kappa )}} \kappa \frac{W^2(\kappa )-\left( \eta s_R T\right) ^2}{W^2(\kappa )}\right) \\&= W\left( e^{-\frac{\left( \eta s_R T\right) ^2}{W(\kappa )} + W(\kappa )} \left( -\frac{\left( \eta s_R T\right) ^2}{W(\kappa )} + W(\kappa )\right) \right) = -\frac{\left( \eta s_R T\right) ^2}{W(\kappa )} + W(\kappa )\\&= W(\kappa )(1-\rho ^{* 2})= {w}(0)(1-\rho ^{* 2}), \end{aligned} \end{aligned}$$

(64)

where we have used (44) for the third equality (as $\kappa >0$) and for the fourth one as $-1<\rho ^*<1$, and thus $-\frac{\left( \eta s_R T\right) ^2}{W(\kappa )} + W(\kappa )>0$. It follows that

$$\begin{aligned} d(\rho ^*)= & {} \frac{e^{-rT}}{\gamma \eta ^2 T (1-\rho ^{* 2})} w (\rho ^*)\left( 1 +\frac{w (\rho ^*)}{2}\right) \\= & {} \frac{ e^{-rT}}{ \gamma \eta ^2 T } W(\kappa )\left( 1 +\frac{1}{2}W(\kappa )\left( 1-\frac{\left( \eta s_R T\right) ^2}{W^2(\kappa )}\right) \right) \\= & {} \frac{ e^{-rT}}{ \gamma \eta ^2 T } \left( W(\kappa ) +\frac{W^2(\kappa )}{2} -\frac{1}{2}\left( \eta s_R T\right) ^2\right) = d(0)-\frac{e^{-rT}}{2\gamma } s_R^2 T. \end{aligned}$$

Using (57), (58) and (64), we obtain that

$$\begin{aligned} b(\rho ^*)&= \frac{e^{-rT}}{\gamma (1-\rho ^{* 2})}\frac{{w}(\rho ^*)}{\eta ^2 T} \left( e^{\frac{\eta ^2}{2}T}-1\right) =\frac{ e^{-rT}}{\gamma }\frac{{w}(0)}{\eta ^2 T} \left( e^{\frac{\eta ^2}{2}T}-1\right) =b(0)\\ g(\rho ^*)&= d(\rho ^*)+b(\rho ^*) = d(0)+b(0)-\frac{e^{-rT}}{2\gamma } s_R^2 T = g(0)-\frac{e^{-rT}}{2\gamma } s_R^2 T. \end{aligned}$$

So, using Remark 5, we determine that

$$\begin{aligned} p=d(\rho )+a(\rho ) \ge d(\rho )\ge \left\{ \begin{array}{ll} d(\rho ^*)= d(0)-\frac{e^{-rT}}{2\gamma }s_R^2 T &{} \text{ if } -1<\rho ^*<1 \\ \lim _{\rho \rightarrow 1^-}d(\rho ) &{} \text{ if }\; \rho ^* \ge 1\\ \lim _{\rho \rightarrow -1^+}d(\rho ) &{} \text{ if }\; \rho ^* \le -1. \end{array} \right. \end{aligned}$$

and (35) follows from (26). Then, (8) implies (36). $\square$

Proof of Proposition 4

Assume that $-1<\rho ^*<1$. Using (33) and (64), we get that

$$\begin{aligned} {\Pi }^{D,\lambda }_{\rho ^*}(0,s_0)= & {} e^{-r T}\left( \frac{s_R}{\gamma \sigma }-\frac{\rho ^*\; w(\rho ^*)}{\sigma \eta \gamma (1-\rho ^{* 2})T}\right) \\ {}= & {} e^{-r T}\left( \frac{s_R}{\gamma \sigma }-\frac{\rho ^*\; w(0)}{\sigma \eta \gamma T}\right) = e^{-rT}\left( \frac{s_R}{\gamma \sigma }-\frac{\eta s_R T\;}{\sigma \eta \gamma T}\right) = 0. \end{aligned}$$

$\square$

Proof of Proposition 5

Using (29), we have to prove that $\lim _{T\rightarrow 0^+}\frac{\partial p}{\partial s}(\rho ,s_0)=\lambda$. Differentiating in (7) with $h=id$, we get that

$$\begin{aligned} \frac{\partial p}{\partial s}(\rho ,s_0) = \lambda e^{-rT} \frac{\mathbb {E}\left( e^{\left( \nu -\eta \rho s_R-\frac{\eta ^2}{2}\right) T+\eta \sqrt{T}N} \exp \left[ -\lambda \gamma (1-\rho ^2)s_0 e^{\left( \nu -\eta \rho s_R-\frac{\eta ^2}{2}\right) T+\eta \sqrt{T}N}\right] \right) }{\mathbb {E}\left( \exp \left[ -\lambda \gamma (1-\rho ^2)s_0 e^{\left( \nu -\eta \rho s_R-\frac{\eta ^2}{2}\right) T+\eta \sqrt{T}N}\right] \right) }\underset{T\rightarrow 0^+}{\longrightarrow }\ \lambda , \end{aligned}$$

where we have used the dominated convergence theorem. The second assertion is trivial. $\square$

Proof of Lemma 1

We use the notation $\epsilon$ for any function vanishing when $T\rightarrow 0^+$. Using Lemma 2, we obtain that

$$\begin{aligned} W\left( \lambda \gamma \eta ^2 T s_0 e^{\left( \nu -\frac{\eta ^2}{2}\right) T}\right)&= \lambda \gamma \eta ^2 T s_0 e^{(\nu -\frac{\eta ^2}{2})T} \left( 1-\lambda \gamma \eta ^2 T s_0 e^{\left( \nu -\frac{\eta ^2}{2}\right) T}+T\epsilon (T)\right) \\&= \lambda \gamma \eta ^2 T s_0 \left( 1+\left( \nu -\frac{\eta ^2}{2}\right) T\right) \left( 1-\lambda \gamma \eta ^2 T s_0\right) +T\epsilon (T)\\&= \lambda \gamma \eta ^2 T s_0 \left( 1-\left( \lambda \gamma \eta ^2 s_0 -\left( \nu -\frac{\eta ^2}{2}\right) \right) T+T\epsilon (T)\right) . \end{aligned}$$

Using (34), we thus obtain that

$$\begin{aligned} \rho ^*= & {} s_R\frac{1}{\lambda \gamma \eta s_0}\left( 1+\left( \lambda \gamma \eta ^2 s_0 -\left( \nu -\frac{\eta ^2}{2}\right) \right) T+T\epsilon (T)\right) . \end{aligned}$$

Now, as $\nu > \frac{\eta ^2}{2}$, Lemma 2 implies again that

$$\begin{aligned} W\left( \lambda \gamma \eta ^2 T s_0 e^{\left( \nu -\frac{\eta ^2}{2}\right) T}\right) \underset{T\rightarrow +\infty }{\sim }\ {}&\left( \nu -\frac{\eta ^2}{2}\right) T \quad \text{ and } \quad \rho ^* \underset{T\rightarrow +\infty }{\sim }\ {}&s_R \frac{\eta }{\nu -\frac{\eta ^2}{2}}. \end{aligned}$$

$\square$