Affine forward variance models

Abstract

We introduce the class of affine forward variance (AFV) models of which both the conventional Heston model and the rough Heston model are special cases. We show that AFV models can be characterised by the affine form of their cumulant-generating function, which can be obtained as solution of a convolution Riccati equation. We further introduce the class of affine forward order flow intensity (AFI) models, which are structurally similar to AFV models, but driven by jump processes, and which include Hawkes-type models. We show that the cumulant-generating function of an AFI model satisfies a generalised convolution Riccati equation and that a high-frequency limit of AFI models converges in distribution to an AFV model.

Appendices

Appendix A: Some results on Volterra equations with convex nonlinearity

We show some results on Volterra equations with convex nonlinearity of the type appearing in Theorems 2.6 and 3.1. On the nonlinearity, we impose the following assumptions.

Assumption 1

The function $H: (-\infty , w_{\mathrm{max}}] \to \mathbb{R}$ is continuously differentiable and convex with a unique root $H(w_{*}) = 0$ in $(-\infty ,w_{\mathrm{max}}]$. Moreover, $H'(w_{*}) < 0$ and $H(w_{\mathrm{max}}) < 0$.

For a function $H$ satisfying Assumption A.1, we set

$$ w_{0} = \mathop {\mathrm {{argmin}}}_{w \in (-\infty ,w_{\mathrm{max}}]} H(w); $$

if the minimum is not unique (i.e., if $H$ has a flat part), then $w_{0}$ denotes the leftmost minimiser. Note that

• either $w_{0} = w_{\mathrm{max}}$, in which case $H$ is strictly decreasing on $(-\infty , w_{\mathrm{max}}]$,

• or $w_{0} < w_{\mathrm{max}}$, in which case $H$ is strictly decreasing on $(-\infty , w_{0})$ and increasing on $[w_{0},w_{\mathrm{max}}]$.

In any case, $w_{*} < w_{0} \le w_{\mathrm{max}}$ holds true. We also need the following definition.

Definition 2

Let $H$ be a function satisfying Assumption A.1. The decreasing envelope of $H$ is defined as

$$ \overline{H}(w) := \textstyle\begin{cases} H(w), \quad &w \le w_{0}, \\ H(w_{0}) ,\quad &w \in [w_{0},w_{\mathrm{max}}]. \end{cases} $$

Clearly, $\overline{H}$ also satisfies Assumption A.1, but is in addition decreasing and satisfies $\overline{H} \le H$. Both Assumption A.1 and Definition A.2 are illustrated in Fig. 1.

Fig. 1

Illustration of two convex functions $H_{1}$, $H_{2}$ satisfying Assumption A.1. While $H_{1}$ is monotone decreasing, $H_{2}$ is not, and its decreasing envelope $\overline{H}_{2}$ is also shown.

Lemma 3

Let$H: (-\infty , w_{\mathrm{max}}] \to \mathbb{R}$be a convex function that satisfies Assumption A.1; in particular, it has a root$H(w_{*}) = 0$. Then:

• (a) For any$a \in (w_{*},w_{\mathrm{max}}]$, the function

  $$ w \mapsto Q_{1}(w,a) = -\int _{w}^{a}\frac{d\zeta }{H(\zeta )} $$

  (A.1)

  maps$(w_{*},a]$onto$[0,\infty )$, is strictly decreasing and has an inverse$Q_{1}^{-1}(r,a)$which maps$[0,\infty )$onto$(w_{*},a]$.

• (b) For any$a \in (-\infty , w_{*})$, the function

  $$ w \mapsto Q_{2}(w,a) = \int _{a}^{w} \frac{d\zeta }{H(\zeta )} $$

  (A.2)

  maps$[a,w_{*})$onto$[0,\infty )$, is strictly increasing and has an inverse$Q_{2}^{-1}(r,a)$which maps$[0,\infty )$onto$[a,w_{*})$.

Remark 4

Analogously to (A.1), we denote by $\overline{Q}_{1}$ the function

$$ w \mapsto \overline{Q}_{1}(w,a) = -\int _{w}^{a}\frac{d\zeta }{ \overline{H}(\zeta )}, $$

(A.3)

where $\overline{H}$ is the decreasing envelope of $H$.

Proof of Lemma A.3

To show (a), note that the integrand $-1/H(\zeta )$ is strictly positive on $(w_{*},a)$. It follows that $Q_{1}(\cdot ,a)$ is strictly decreasing and maps $(w_{*},a]$ into $[0,\infty )$. It remains to show that the range of this map covers all of $[0,\infty )$. To this end, observe that by convexity, we have

$$ H(w) \ge H'(w_{*}) (w - w_{*}) \qquad \text{for all } w \in (-\infty ,w_{\mathrm{max}}], $$

and $H'(w_{*}) < 0$ by Assumption A.1. Thus we obtain

$$ \lim _{w \downarrow w_{*}}Q_{1}(w,a) = -\int _{w_{*}}^{a} \frac{d\zeta }{H(\zeta )} \ge -\frac{1}{H'(w_{*})} \int _{w_{*}}^{a} \frac{d\zeta }{ \zeta - w_{*}} = +\infty . $$

The proof of (b) is analogous; only the different sign of $H$ on $(-\infty ,w_{*})$ has to be taken into account. □

Theorem 5

Let$\kappa$be an$L_{1}$-kernel in the sense of Definition 2.2and$H$a convex function that satisfies Assumption A.1; in particular, $w_{*}$with$H(w_{*}) = 0$is its unique root in the interval$(-\infty ,w_{\mathrm{max}}]$. For any continuous function$a: \mathbb{R}_{+}\to (-\infty ,w_{ \mathrm{max}}]$, consider the nonlinear Volterra equation

$$ f(t) = a(t) + \int _{0}^{t} \kappa (t-s) H\big(f(s)\big) \mathit{ds}, \qquad t \in \mathbb{R}_{+}. $$

(A.4)

A function$f \in C(\mathbb{R}_{+};\mathbb{R})$that satisfies this equation is called a solution of (A.4).

• (a) If$a$is increasing with values in$(w_{*},w _{0}]$, then (A.4) has a unique global solution$f$which satisfies

  $$ w_{*} < r_{1}(t) \le f(t) < a(t), \qquad \forall \,t > 0, $$

  (A.5)

  where$r_{1}(t) = Q_{1}^{-1}(\int _{0}^{t} \kappa (s) \mathit{ds}, a(0))$and$Q_{1}$is given by (A.1).

• (b) If$a \equiv w_{*}$, then$f \equiv w_{*}$is the unique global solution of (A.4).

• (c) If$a$is decreasing with values in$(-\infty ,w _{*})$, then (A.4) has a unique global solution$f$which satisfies

  $$ a(t) < f(t) \le r_{2}(t) < w_{*}, \qquad \forall \,t > 0, $$

  (A.6)

  where$r_{2}(t) = Q_{2}^{-1}(\int _{0}^{t} \kappa (s) \mathit{ds},a(0))$and$Q_{2}$is given by (A.2).

In addition, case (a) can be extended to the following more general statement:

$\mathrm{(a^{\prime })}$ :

If$a$is increasing with values in$(w_{*},w_{\mathrm{max}}]$, then (A.4) has a unique global solution$f$which satisfies

$$ w_{*} < \overline{r}_{1}(t) \le f(t) < a(t), \qquad \forall \,t > 0, $$

where$\overline{r}_{1}(t) = \overline{Q}_{1}^{-1} (\int _{0}^{t} \kappa (s) \mathit{ds}, a(0))$and$\overline{Q}_{1}$is given by (A.3).

Remark 6

Clearly, if $H$ is decreasing (and hence $w_{0} = w_{\mathrm{max}}$), cases (a) and $\mathrm{(a^{\prime })}$ coincide. In the general case, (a) gives better bounds on $f$ than $\mathrm{(a^{\prime })}$, but is more restrictive in its assumptions on the function $a$.

Before proving the theorem, we add two corollaries that are used in the proofs of Theorems 2.6, 3.1 and 4.10.

Corollary 7

Under the assumptions of Theorem A.5, consider the nonlinear integral equation

$$ g(t) = H\left (a(t) + \int _{0}^{t} \kappa (t-s) g(s) \mathit{ds}\right ), \qquad t \in \mathbb{R}_{+}. $$

(A.7)

• (a) If$a$is increasing with values in$(w_{*},w _{0}]$, then (A.7) has a unique global solution$g$which satisfies

  $$ H\big(a(t)\big) < g(t) \le H\big(r_{1}(t)\big) < 0, \qquad \forall \,t > 0. $$

  (A.8)

• (b) If$a \equiv w_{*}$, then$g \equiv 0$is the unique global solution of (A.7).

• (c) If$a$is decreasing with values in$(-\infty ,w _{*})$, then (A.7) has a unique global solution$g$which satisfies

  $$ 0 < g(t) \le H\big(r_{2}(t)\big) < H\big(a(t)\big), \qquad \forall \,t > 0. $$

  (A.9)

In addition, case (a) can be extended to

$\mathrm{(a^{\prime })}$ :

If$a$is increasing with values in$(w_{*},w_{\mathrm{max}}]$, then (A.7) has a unique global solution$g$which satisfies

$$ g(t) < 0, \qquad \forall \,t > 0. $$

(A.10)

In any of the above cases, $g(t) = H(f(t))$, where$f$is the solution of (A.4).

Corollary 8

Let the assumptions of Theorem A.5hold with$w_{\mathrm{max}}= 0$. Let$\varDelta > 0$and let$h$be a piecewise continuous function from$[0,\varDelta )$to$\mathbb{R}_{-}$. Consider the nonlinear integral equation

$$\begin{aligned} g(t) &= H\left (\int _{0}^{t} \kappa (t-s) g(s) ds\right ), \qquad t \in [\varDelta , \infty ), \end{aligned}$$

(A.11)

with initial condition

$$\begin{aligned} g(t) &= h(t), \qquad t \in [0,\varDelta ). \end{aligned}$$

If$w_{*} < \int _{0}^{\varDelta }\kappa (\varDelta - s) h(s)\mathit{ds}$, then (A.11) has a unique global solution$g$taking values in$\mathbb{R}_{-}$, which satisfies

$$ w_{*} < \int _{0}^{t} \kappa (t - s) g(s)\mathit{ds} \qquad \textrm{for all}\ t \ge 0. $$

(A.12)

We start with the proof of Theorem A.5, which closely follows the account of Lakshmikantham’s comparison method in [4, Sect. II.7].

Proof of Theorem A.5

Clearly, $H$ can be extended to a continuous function on all of ℝ, and thus it follows from [14, Theorem 12.1.1] that (A.4) has a local continuous solution $f$ on an interval $[0,T_{\mathrm{max}})$ with $T_{\mathrm{max}}> 0$. In addition, $T_{\mathrm{max}}$ can be chosen maximal in the sense that the solution cannot be continued beyond $[0,T_{\mathrm{max}})$.

(a) By assumption, $a$ is increasing and takes values in $(w_{*},w _{0}]$. Set

$$ T_{*} := \inf \left \{t \in (0,T_{\mathrm{max}}): f(t) = w_{*} \text{ or } f(t) = a(T_{\mathrm{max}})\right \} $$

(A.13)

and note that $T_{*} > 0$. From (A.4), it is clear that

$$ f(t) = a(t) + \int _{0}^{t} \kappa (t-s) H\big(f(s)\big) \mathit{ds} < a(t) \le a(T_{\mathrm{max}}), \qquad \forall t \in [0,T_{*}), $$

(A.14)

i.e., the lower bound $w_{*}$ in (A.13) is always hit before the upper bound $a(T_{\mathrm{max}})$. In addition, using that the kernel $\kappa$ is decreasing, we obtain that

$$\begin{aligned} f(t) & = a(t) + \int _{0}^{t} \kappa (t-s) H\big(f(s)\big) \mathit{ds} \\ &\le a(T)+ \int _{0}^{t} \kappa (T-s) H\big(f(s)\big) \mathit{ds} =: v(t,T) \end{aligned}$$

(A.15)

for all $0 \le t \le T \le T_{*}$. The function $v(t,T)$ which we have just defined satisfies

$$\begin{aligned} v(t,t) &= f(t) , \end{aligned}$$

(A.16)

$$\begin{aligned} v(0,T) &= a(T) \ge a(0) \end{aligned}$$

(A.17)

and the differential inequality

$$ \frac{\partial }{\partial t} v(t,T) = \kappa (T-t) H\big(f(t)\big) \ge \kappa (T-t) H\big(v(t,T)\big). $$

(A.18)

Here, we have used (A.15) and the fact that $H$ is decreasing on $(w_{*},w_{0}]$. Together with the initial estimate (A.17), a standard comparison principle for differential inequalities (cf. [23, II.§9]) yields

$$\begin{aligned} v(t,T) &\ge r(t,T), \end{aligned}$$

(A.19)

where

$$\begin{aligned} \frac{\partial }{\partial t}r(t,T) &= \kappa (T-t) H\big(r(t,T)\big), \qquad r(0,T) = a(0). \end{aligned}$$

(A.20)

We claim that the differential equation (A.20) is solved by

$$ r(t,T) = Q_{1}^{-1}\left (\int _{0}^{t} \kappa (T-s) \mathit{ds},a(0)\right ). $$

(A.21)

Indeed, applying $Q_{1}(\cdot ,a(0))$ to both sides of (A.21) yields

$$ \int _{0}^{t} \kappa (T-s)\mathit{ds} = Q_{1}\big(r(t,T),a(0)\big) = - \int _{r(t,T)} ^{a(0)} \frac{d\zeta }{H(\zeta )}. $$

Taking partial derivatives $\frac{\partial }{\partial t}$, we obtain

$$ \kappa (T-t) = \frac{1}{H(r(t,T))} \frac{\partial }{\partial t} r(t,T) $$

which is equivalent to (A.20). From (A.15), (A.16) and (A.19), we obtain the bound

$$ r_{1}(t) := \lim _{T \downarrow t} r(t,T) \le \lim _{T \downarrow t} v(t,T) = f(t) $$

(A.22)

for all $t \in [0,T_{*})$. This implies that

$$ \lim _{t \to T_{*}} f(t) \ge r_{1}(T_{*}) > w_{*}, $$

(A.23)

which in light of (A.13) means that $T_{*} = T_{ \mathrm{max}}$, i.e., we have shown the bounds (A.5) to hold for all $t \in [0,T_{\mathrm{max}})$. However, by [14, Theorem 12.1.1], $\lim _{t \to T_{\mathrm{max}}} |f(t)| = \infty$ whenever $T_{ \mathrm{max}}< \infty$. We conclude that $T_{\mathrm{max}}= \infty$, and hence that $f$ is a global solution of (A.4). Uniqueness follows from [14, Theorem 13.1.2].

(b) By assumption, $a \equiv w_{*}$. Since $H(w_{*}) = 0$, it is clear that $f(t) \equiv w_{*}$ is a global solution of (A.4). Uniqueness follows from [14, Theorem 13.1.2].

(c) By assumption, $a$ is decreasing and takes values in $(-\infty , w _{*}]$. This case can be handled analogously to (a) with the following adaptations. The inequality signs in (A.14)–(A.19) have to be reversed; in (A.21), $Q_{1}$ has to be substituted by $Q_{2}$; and also in (A.22) and (A.23), the inequalities have to be reversed.

$\mathrm{(a^{\prime })}$ The proof of (a) applies, except for the following modification. (A.18) holds only when $v(t,T) \le w_{0}$, since $H$ is decreasing only on $(-\infty ,w_{0}]$. However, when $v(t,T) > w_{0}$, we can use the trivial estimate

$$ \frac{\partial }{\partial t} v(t,T) = \kappa (T-t) H\big(f(t)\big) \ge \kappa (T-t) H(w_{0}), $$

which can be combined with (A.18) into

$$ \frac{\partial }{\partial t} v(t,T) = \kappa (T-t) H\big(f(t)\big) \ge \kappa (T-t) \overline{H}\big(v(t,T)\big), $$

where $\overline{H}$ is the decreasing envelope of $H$ from Definition A.2. The remaining proof of (a) applies after substituting $H$ by $\overline{H}$ and $Q_{1}$ by $\overline{Q}_{1}$. □

Proof of Corollary A.7

Let $f$ be the global solution of (A.4). Applying $H$ to both sides of (A.4), we see that $g(t) := H(f(t))$ is a global solution of (A.7). To show uniqueness, assume that ${\widetilde{g}}$ is a local solution of (A.7) on $[0,T)$ and define

$$ {\widetilde{f}}(t) := a(t) + \int _{0}^{t} \kappa (t-s) {\widetilde{g}}(s)\mathit{ds}. $$

Clearly, ${\widetilde{g}}(t) = H({\widetilde{f}}(t))$ on $[0,T)$, and hence ${\widetilde{f}}$ is a local solution of (A.4). By [14, Theorem 13.1.2], this solution is unique, and we conclude that ${\widetilde{f}} = f$, and hence also ${\widetilde{g}} = g$. Finally, applying $H$ – which is decreasing on $(-\infty ,w_{0}]$ – to the inequalities (A.5) and (A.6) yields (A.8) and (A.9). In case $\mathrm{(a^{\prime })}$, monotonicity of $H$ is lost, but $H(w) < 0$ for all $w \in (w_{*},w_{\mathrm{max}}]$ yields (A.10). □

Proof of Corollary A.8

Set

$$ a(t) := \int _{0}^{\varDelta }\kappa (t + \varDelta - s)h(s)\mathit{ds} $$

and note that $a$ is increasing with values in $(w_{*},0]$. Consider the nonlinear Volterra equation

$$ f(t) = a(t) + \int _{0}^{t} \kappa (t-s) H\big(f(s)\big)\mathit{ds}, $$

(A.24)

which has a unique global solution $f$ by Theorem A.5 (a) or $\mathrm{(a^{\prime })}$. For $t' \in \mathbb{R}_{+}$, set

$$ g(t') = \textstyle\begin{cases} H(f(t'-\varDelta )), &\quad t' \in [\varDelta , \infty ), \\ h(t'), &\quad t' \in [0,\varDelta ). \end{cases} $$

For $t' \ge \varDelta$, we have

$$\begin{aligned} g(t') &= H\big(f(t'-\varDelta )\big) = H\bigg(\int _{0}^{\varDelta }\kappa (t' - s) h(s) \mathit{ds} + \int _{\varDelta }^{t'} \kappa (t'-s)g(s)\mathit{ds}\bigg) \\ &= H\bigg(\int _{0}^{t'} \kappa (t'-s)g(s)\mathit{ds}\bigg), \end{aligned}$$

showing that $g$ is a global solution of (A.11). From cases (a) or $\mathrm{(a^{\prime })}$ of Theorem A.5, we obtain the bound

$$ w_{*} < f(t' - \varDelta ) = \int _{0}^{t'} \kappa (t'-s)g(s)\mathit{ds} $$

as claimed. To show uniqueness, assume that ${\widetilde{g}}$ is a solution of (A.11). Setting

$$ {\widetilde{f}}(t) := a(t) + \int _{0}^{t} \kappa (t-s) {\widetilde{g}}(s+ \varDelta ), $$

we see that ${\widetilde{f}}$ is a solution of (A.24) and conclude from Theorem A.5 that ${\widetilde{f}} = f$ and hence also ${\widetilde{g}} = g$. □

Appendix B: Theorem 2.6 in the case $\rho > 0$

We provide the remaining part of the proof of Theorem 2.6 in the case $\rho > 0$. Our starting point is the quadratic equation (2.21), which has been obtained without any assumption on the sign of $\rho$. In the case $\rho > 0$, additional arguments are needed since this equation may have two negative solutions.

Proof of Theorem 2.6, ‘only if’ part in the case $\rho > 0$

On the set which is defined by $S = \left \{(t, \omega ): V_{t}(\omega ) \neq 0\right \}$, we set

$$ k_{t}(\tau ,\omega ) = \frac{1}{V_{t}(\omega )}\eta _{t}(t+\tau , \omega ) $$

for $\tau \ge 0$. Note that by Assumption 2.1, $\tau \mapsto k_{t}(\tau ,\omega )$ must be a decreasing $L_{2}$-kernel for a.e. $(t,\omega )$. Since (2.5) holds trivially if $S$ is a $\mathit{dt} \otimes d\mathbb{P}$-nullset, we may without loss of generality assume that $S$ is not a nullset and consider only $(t, \omega ) \in S$ in the remainder of the proof. Inserting into (2.16) yields

$$ h_{t}(t + \tau ,u) = \sqrt{V_{t}} \int _{0}^{\tau }g(\tau -s,u) k _{t}(s) \mathit{ds} = \sqrt{V_{t}} \, (g \star k)_{t}(\tau ,u). $$

Plugging into (2.21) and eliminating $V_{t}$ gives

$$ \frac{1}{2}(u^{2} - u) - g(\tau ,u) + u \rho (g \star k)_{t}(\tau ,u) + \frac{1}{2} (g \star k)_{t}(\tau ,u)^{2} = 0, $$

(B.1)

which is a quadratic equation in the variable $(g \star k)_{t}(\tau ,u)$ with two solutions

$$ q_{\pm }(\tau ,u) = -\rho u \pm \sqrt{u^{2}(\rho ^{2} - 1) + u + 2g( \tau ,u)}, $$

(B.2)

both of which may be negative. However, using continuity of $g(\cdot ,\tau )$ and evaluating (B.1) for $\tau \downarrow 0$ yields that $g(0,u) = \frac{1}{2}(u^{2} - u)$. Inserting into (B.2) and using $(g \star k)_{t}(0,u) = 0$, this selects the solution $q_{+}$ at $\tau = 0$ and shows that

$$ (g \star k)_{t}(\tau ,u) = q_{+}(\tau ,u) \qquad \text{for all } \tau \in \big[0,T_{*}(u)\big), $$

where $T_{*}(u)$ is the first collision time of $q_{+}$ and $q_{-}$, i.e.,

$$ T_{*}(u) := \inf \left \{\tau > 0: g(\tau ,u) = \frac{1}{2}(u^{2} - u) - \frac{u^{2}\rho ^{2}}{2}\right \}\!. $$

On the interval $[0,T_{*}(u))$, we can proceed as in the case of $\rho \le 0$ and obtain that

$$\begin{aligned} \eta _{t}(t+\tau ) & = \sqrt{V_{t}} \, k_{t}(\tau ) \\ &= \sqrt{V_{t}} \frac{\partial }{\partial \tau } \bigg( \int _{0} ^{\tau }q_{+}(\tau -s,u) \pi (\mathit{ds},u) \bigg), \qquad \tau \in \big[0,T_{*}(u)\big), \end{aligned}$$

(B.3)

where $\pi$ is the resolvent of the first kind of $g(\tau ,u)$. Therefore, to complete the proof, it suffices to show that $T_{*}(u)$ can be made arbitrarily large by choosing a suitable $u \in (0,1)$. To this end, note that (B.1) is a convolution Riccati equation for $g(\cdot ,u)$ with the kernel $k_{t}(\cdot )$, i.e.,

$$ g(\tau ,u) = R_{V}\big(u,(g \star k)_{t}(\tau ,u)\big). $$

Applying Lemma 2.13, we obtain that $g(\cdot ,u)$ is its unique continuous solution, which by Corollary A.7 can be written as $g(\tau ,u) = R_{V}(u,f(\tau ,u))$, where $f(\tau ,u)$ solves

$$ f(\tau ,u) = \int _{0}^{\tau }k_{t}(\tau -s) R_{V}\big(u,f(s,u)\big) \mathit{ds}. $$

(B.4)

Moreover, the collision time can be represented in terms of $f$ as

$$ T_{*}(u) := \inf \left \{t > 0: f(t,u) = -\rho u\right \}\!. $$

(B.5)

Using its convexity in $w$, we can estimate the function $R_{V}(u,w)$ from below as $R_{V}(u,w) \ge w\rho + \frac{1}{2}(u^{2} - u)$. Hence (B.4) yields the estimate

$$ f(\tau ,u) \ge \rho (k_{t} \star f)(\tau ,u) + \frac{1}{2}(u^{2} - u). $$

Let $r_{t}$ be the $\rho$-resolvent of $k_{t}$ and note that $r_{t}$ is again an $L_{2}$-kernel (in particular nonnegative) by Lemma 2.8. By the generalised Gronwall lemma of [14, Lemma 9.8.2], it follows that $f(\tau ,u) \ge \ell (\tau ,u)$, where $\ell$ solves the linear Volterra equation

$$ \ell (\tau ,u) = \rho (k_{t} \star \ell )(\tau ,u) + \frac{1}{2}(u ^{2} - u). $$

Moreover, using [14, Theorem 2.3.5], we can express $\ell (t,u)$ in terms of the $\rho$-resolvent $r_{t}$ and obtain $f(\tau ,u) \ge \ell (\tau ,u) = \frac{1}{2}(u^{2} - u)\int _{0}^{\tau }r_{t}(s) \mathit{ds}$ for all $u \in (0,1)$. Combining with (B.5), we finally obtain

$$ \int _{0}^{T_{*}(u)} r_{t}(s) \mathit{ds} \ge -\frac{\rho u}{\frac{1}{2}(u^{2} - u)} = \frac{2\rho }{1-u}. $$

Sending $u \uparrow 1$, the right-hand side can be made arbitrarily large, and we conclude that $\lim _{u \uparrow 1} T_{*}(u) = +\infty$, which together with (B.3) completes the proof. □