A Level-1 Limit Order Book with Time Dependent Arrival Rates

Abstract

We propose a simple stochastic model for the dynamics of a limit order book, extending the recent work of Cont and de Larrard (SIAM J Financial Math 4(1), 1–25 2013), where the price dynamics are endogenous, resulting from market transactions. We also show that the conditional diffusion limit of the price process is the so-called Brownian meander.

Appendices

Appendix A: Auxiliary Results

Proposition A.1

Suppose thatV_n = X₁ + ⋯ + X_n, where the variablesX_iare i.i.d. with$xP(X_{i} > x) \stackrel {x\to \infty }{\to } c \in (0,\infty )$.Then$\frac {V_{n}}{n\log {n}} \stackrel {Pr}{\to } c$, asn →∞.

Proof

First, for any s > 0 and T > 0,

$$ s {\int}_{T}^{\infty} \frac{e^{-sx}}{x} dx = s {\int}_{sT}^{\infty} \frac{e^{-y}}{y} dy = - s\log(Ts) e^{-Ts} + s {\int}_{Ts}^{\infty} \log(y) e^{-y}dy, $$

so as s → 0, $s {\int }_{T}^{\infty } \frac {e^{-sx}}{x} dx \sim -s\log {s}$. Next, for any non negative random variable X and any s ≥ 0,

$$ \mathbb{E}\left[e^{-s X} \right] = 1 - s{\int}_{0}^{\infty} P(X>x)e^{-sx} dx. $$

As a result, if P(X > x) ∼ c/x, as x →∞, then, as s → 0,

$$ \mathbb{E}\left[e^{-s X} \right] = 1 +c s \log{s} + o(s\log{s}). $$

Therefore, setting a_n = n log n, one obtains, for a fixed s > 0,

$$ \begin{array}{@{}rcl@{}} \mathbb{E}\left[e^{-s V_{n}/a_{n}} \right] &=& \left[ \mathbb{E}\left[e^{-s X_{1}/a_{n}} \right] \right]^{n} \\ &=& \left\{1 - \frac{sc}{a_{n}} \log(sa_{n}) + o\left( log(a_{n})/a_{n}\right) \right\}^{n} \\ &\stackrel{n\to\infty}{\to}& e^{-cs}, \end{array} $$

since $\frac {ns}{a_{n}} \log (sa_{n}) \to s$ as n →∞. Hence, $V_{n}/a_{n} \stackrel {Pr}{\to } c$, as n →∞. □

Proposition A.2

Suppose that$V_{n}/f(n) \stackrel {Pr}{\to } c$, asn →∞, wheref(n) →∞is regularly varying of orderα. DefineN_t = max{n ≥ 0; V_n ≤ t} and suppose that for some function g on (0,∞),f ∘ g(t) ∼ g ∘ f(t) ∼ t, ast →∞. Then$N_{t}/g(t) \stackrel {Pr}{\to } c^{-1/\alpha }$.

Proof

The proof is similar to the proof of the renewal theorem in Durrett (1996)[Theorem 7.3]. By definition, $V_{N_{t}} \le t < V_{N_{t}+1}$. As a result,

$$ \frac{V_{N_{t}}}{f(N_{t})} \le \frac{t}{f(N_{t})} < \frac{V_{N_{t}}}{f(N_{t}+1)} \frac{f(N_{t}+1)}{f(N_{t})}. $$

By hypothesis, V_n/f(n) converges in probability to c ∈ (0,∞), as n →∞. Also, since V_n is finite for any $n\in {\mathbb {N}}$, it follows that N_t converges in probability to + ∞ as t →∞. Next, since f(n + 1)/f(n) → 1 as n →∞, it follows that as t →∞, f(N_t)/t converges in probability to $\frac {1}{c}$. Also, g is regularly varying of order 1/α, so one may conclude that $N_{t}/g(t) \stackrel {Pr}{\to } c^{-1/\alpha }$. □

Remark A.3

If f(t) = t log t, then α = 1 and one can take g(t) = t/ log t.

Proposition A.4

Set$\psi _{\lambda }(t,x) ={\int }_{t}^{\infty } \frac {1}{u}I_{x}(2u\lambda )e^{-2u\lambda }du$, for anyt, x, λ > 0. Then there exists a constant C so that for anyx, λ > 0, and any$t\ge \frac {1}{2\lambda }$,$\psi _{\lambda }(t,x) \le \frac {C}{\sqrt {2\lambda t}}.$

Proof

First, note that ψ_λ(t, x) = ψ_1/2(2λt, x). It is well-known that

$$ \begin{array}{@{}rcl@{}} I_{x}(z) &=&\frac{1}{\pi}{\int}_{0}^{\pi} e^{z\cos\theta}\cos(x\theta)d\theta \le \frac{1}{\pi}{\int}_{0}^{\pi} e^{z\cos\theta}d\theta \\ & \le & \frac{1}{2}+ \frac{1}{\pi}{{\int}_{0}^{1}} \frac{e^{zs}}{\sqrt{1-s^{2}}}ds. \end{array} $$

Next, set $E_{1}(u):={\int }_{u}^{\infty }\frac {e^{-w}}{w} dw$, u > 0. Then

$$ \begin{array}{@{}rcl@{}} \psi_{1/2}(t,x) &\le & {\int}_{t}^{\infty} \frac{e^{-u}}{u} \left\{\frac{1}{2}+ \frac{1}{\pi}{{\int}_{0}^{1}} \frac{e^{us}}{\sqrt{1-s^{2}}}ds\right\}du\\ &=& \frac{1}{2}E_{1}(t) +\frac{1}{\pi}{\int}_{t}^{\infty}{{\int}_{0}^{1}} \frac{e^{-su}}{u\sqrt{s(2-s)}}ds du\\ &=& \frac{1}{2}E_{1}(t) +\frac{1}{\pi}{{\int}_{0}^{1}} \frac{E_{1}(st)}{\sqrt{s(2-s)}}ds.\\ &=& \frac{1}{2}E_{1}(t) +\frac{1}{\pi}{{\int}_{0}^{t}} \frac{E_{1}(s)}{\sqrt{s(2t-s)}}ds. \end{array} $$

According to Olver et al (2010, Section 6.8.1), E₁(u) ≤ e^−u ln (1 + 1/u) for any u > 0. Furthermore, ln(1 + x) ≤ x and ln(1 + x) ≤ x^2/5 for any x ≥ 0. As a result,

$$ \begin{array}{@{}rcl@{}} \psi_{1/2}(t,x) &\le & \frac{e^{-t}}{2t} +\frac{t^{-1/2}}{\pi}{{\int}_{0}^{t}} s^{-9/10} e^{-s}ds \le \frac{e^{-t}}{2t} +\frac{{\Gamma}\left( \frac{1}{10}\right)}{\pi t^{1/2}} \le Ct^{-1/2} \end{array} $$

for any t ≥ 1, where $C = \frac {e^{-1}}{2}+ \frac {{\Gamma }\left (\frac {1}{10}\right )}{\pi }$. □

Appendix B: Proofs

Proof of Lemma 3.4

From Olver et al (2010)[Formula 10.30.4], for fixed ν, $I_{\nu }(z)\sim \frac {e^{z}}{\sqrt {2\pi z}}$ as z →∞. Also, from Abramowitz and Stegun (1972, p. 376), $I_{n}(z)= \frac {1}{\pi }{\int }_{0}^{\pi } e^{z \cos {\theta }}\cos (n\theta )d\theta$ , so for any $x\in {\mathbb {N}}$, I_n(z) ≤ e^z. Thus, as T →∞,

$$ \begin{array}{@{}rcl@{}} \mathbb{P}_{x}[\sigma_{Y}> T]&= &\left( \frac{\mu}{\lambda}\right)^{x/2}{\int}_{T}^{\infty}\frac{x}{s}I_{x}\left( 2s\sqrt{\lambda\mu}\right)e^{-s(\lambda+\mu)}ds\\ &\sim &\left( \frac{\mu}{\lambda}\right)^{x/2}{\int}_{T}^{\infty}\frac{x}{s} \frac{e^{2s\sqrt{\lambda\mu}}}{\sqrt{4s\pi\sqrt{\lambda\mu}}}e^{-s(\lambda+\mu)}ds\\ &\sim &\left( \frac{\mu}{\lambda}\right)^{x/2} \frac{x}{2\sqrt{\pi\sqrt{\lambda\mu}}} {\int}_{T}^{\infty}s^{-3/2}e^{-s{\mathcal{C}}}ds. \end{array} $$

Also, for any $x\in {\mathbb {N}}$,

$$ \mathbb{P}_{x}[\sigma_{Y}> T] \le x \left( \frac{\mu}{\lambda}\right)^{x/2} {\int}_{T}^{\infty}s^{-1}e^{-s{\mathcal{C}}}ds. $$

(17)

Consequently, if λ = μ, ${\mathcal {C}}=0$ and

$$ \mathbb{P}_{x}[\sigma_{Y}> T] \sim \frac{x}{2\lambda\sqrt{\pi}} {\int}_{T}^{\infty}s^{-3/2}ds \sim \frac{x}{2\lambda\sqrt{\pi}} \frac{2}{\sqrt{T}} \sim \frac{x}{\lambda\sqrt{\pi T}}. $$

This agrees with the result proved in Cont and de Larrard (2013). However, if λ < μ, using the change of variable $u = s{\mathcal {C}}$, one gets

$$ \begin{array}{@{}rcl@{}} \mathbb{P}_{x}[\sigma_{Y}> T] &\sim& {\mathcal{C}}^{1/2}\left( \frac{\mu}{\lambda}\right)^{x/2} \frac{x}{2\sqrt{\pi\sqrt{\lambda\mu} } } {\int}_{T{\mathcal{C}}}^{\infty}{u^{-3/2}}e^{-u} du \\ &\sim& \left( \frac{\mu}{\lambda}\right)^{x/2} \frac{x}{\sqrt{\pi\sqrt{\lambda\mu}}} \left[\frac{e^{-T{\mathcal{C}}}}{\sqrt{T}} - \sqrt{{\mathcal{C}}} {\Gamma}\left( \frac{1}{2},T{\mathcal{C}}\right)\right]. \end{array} $$

To compute the expectation in the case where λ = μ, note that for large enough T, $\mathbb {E}_{x}\left [\sigma _{Y}\right ]= {\int }_{0}^{\infty } \mathbb {P}_{x}[\sigma _{Y}> t] dt \geq \frac {x}{2\lambda \sqrt {\pi }} {\int }_{T}^{\infty } \frac {1}{\sqrt {t}}dt = \infty$, whereas if λ < μ, for a sufficiently large T, there are finite constants C₁ and C₂ such that for any $0 \le \theta <{\mathcal {C}}$,

$$ \begin{array}{@{}rcl@{}} \mathbb{E}_{x}\left[e^{\theta\sigma_{Y}} \right]&=& 1+ \theta {\int}_{0}^{\infty} e^{\theta t} \mathbb{P}_{x}[\sigma_{Y}> t] dt \leq C_{1} + \theta C_{2}{\int}_{T}^{\infty} e^{-t({\mathcal{C}}-\theta)} dt\\ &=&C_{1} + C_{2} \frac{e^{-T({\mathcal{C}}-\theta)}}{({\mathcal{C}}-\theta)} <\infty. \end{array} $$

□

Proof of Proposition 4.1

Let F_{n, Q}(t; x, y) and F_{n, L}(t; x, y) denote the cdf of ${S^{n}_{Q}}$ and ${S^{n}_{L}}$, respectively, starting from z₀ = (x, y), with densities $f_{n,\mathcal {Q}}(t;z_{0})$ and $f_{n,\mathcal {L}}(t;z_{0})$, where F_{n, Q}(⋅; z₀) is the convolution of F_1,Q (n − 1) times with F_1,Q(⋅; z − 0). The result will be proven by induction. The base case n = 1 is given in Corollary 3.7. Assume the result is true for any $m\leq n\in \mathbb {N}$. Then by Corollary 3.7 and the induction hypothesis,

$$ F_{{\mathcal{L}}}(t;x,y)=F_{{\mathcal{Q}}}(A_{t};x,y) \text{ and } f_{n,{\mathcal{L}}}(t;x,y)=f_{n,{\mathcal{Q}}}(A_{t};x,y)\alpha_{t}. $$

(2)

Also, by the definition of τ_n and V_n, under Assumption 2, if z₀ = (x, y), then

$$ \begin{array}{@{}rcl@{}} F_{n,{\mathcal{L}}}(t;z_{0}) &=& \mathbb{P}_{{\mathcal{L}}}[V_{n+1} \leq t | q_{0} = z_{0} ]=\mathbb{P}_{{\mathcal{L}}}[V_{n}\leq t, \tau_{n+1}\leq t-V_{n} | q_{0}=z_{0}]\\ &=& \sum\limits_{z} f(z) {{\int}_{0}^{t}} \mathbb{P}_{{\mathcal{L}}}[\tau_{n+1}\leq t-u| q_{u}=z]f_{n,{\mathcal{L}}}(u;z_{0})du\\ &=& \sum\limits_{z} f(z){{\int}_{0}^{t}} \mathbb{P}_{{\mathcal{Q}}}\left[\tau_{n+1}\leq A_{t-u}^{(n+1)}|q_{u}=z\right]f_{n,{\mathcal{Q}}}(A_{u};z_{0})\alpha_{u} du\\ &=&{{\int}_{0}^{t}} F_{1,{\mathcal{Q}}}(A_{t}-A_{u})f_{n,{\mathcal{Q}}}(A_{u};z_{0})\alpha_{u} du = {\int}_{0}^{A_{t}} F_{1,{\mathcal{Q}}}(A_{t}-u)f_{n,{\mathcal{Q}}}(u;z_{0}) du\\ &=&{\int}_{0}^{A_{t}} F_{1,{\mathcal{Q}}}(A_{t}-u)dF_{n,{\mathcal{Q}}}(u;z_{0}) ={\int}_{0}^{A_{t}} F_{n,{\mathcal{Q}}}(A_{t}-u)dF_{1,{\mathcal{Q}}}(u;z_{0})\\ &=&\mathbb{P}_{{\mathcal{Q}}}\left[V_{n+1} \leq A_{t}| q_{0}=z_{0}\right], \end{array} $$

where we used the fact that for any s ≥ 0, α^(n+ 1)(s) = α(s + u) given V_n = u, so $A^{(n+1)}(t) = {{\int }_{0}^{t}} \alpha (s+u)ds = A_{t+u}-A_{u}$. Furthermore, in the last equality we used the fact that for X and Y, non-negative independent random variables,

$$ F_{X+Y}(t)=\mathbb{P}[X+Y\leq t]=F_{X}*F_{Y}(t)={{\int}_{0}^{t}}F_{X}(t-x)dF_{Y}(x), $$

with F_X and F_Y denoting the cdfs of X and Y. Furthermore, starting q₀ from distribution f, one obtains that $\mathbb {P}_{{\mathcal {L}}}[V_{n} \le t] = \mathbb {P}_{{\mathcal {Q}}}[V_{n} \le A_{t}]$. □