Strong Consistency of the Bayesian Estimator for the Ornstein–Uhlenbeck Process

Abstract

In the accompanying paper Kohatsu-Higa et al. (submitted, 2013), we have done a theoretical study of the consistency of a computational intensive parameter estimation method for Markovian models. This method could be considered as an approximate Bayesian estimator method or a filtering problem approximated using particle methods. We showed in Kohatsu-Higa (submitted, 2013) that under certain conditions, which explicitly relate the number of data, the amount of simulations and the size of the kernel window, one obtains the rate of convergence of the method. In this first study, the conditions do not seem easy to verify and for this reason, we show in this paper how to verify these conditions in the toy example of the Ornstein–Uhlenbeck processes. We hope that this article will help the reader understand the theoretical background of our previous studies and how to interpret the required hypotheses.

Appendix

Here we give some lemmas, which are used in the parameter tuning sections.

Lemma 6

For c>1, we have

• (i). $(x+y)^{2} \le\frac{c}{c-1}x^{2}+cy^{2}$,

• (ii). $\frac{c-1}{c}x^{2}+(c-1)y^{2} \le(x-y)^{2}$.

The proofs are based on Young’s lemma, which follows from simple calculations.

Lemma 7

For m≥2βΔ, we have $| (1-\frac{\theta\varDelta}{m})^{m} - e^{-\theta\varDelta} | \le e^{-\alpha \varDelta} (\beta\varDelta)^{2} \frac{1}{m}$.

From this lemma, we obtain

Lemma 8

For k=0,1 and m≥2βΔ, we have the following estimations;

• (i). $|\frac{\partial^{k}}{\partial\theta^{k}}( \sigma ^{2}_{\varDelta}(\theta)-\sigma^{2}(m,\theta,h) )|\le C(\alpha,\beta,\varDelta)\{ \frac{1}{m}+h^{2}{\bf1}(k=0)\}$,

• (ii). $|\frac{\partial^{k}}{\partial\theta^{k}}( e^{-\theta \varDelta}-\mu(m,\theta) )|\le C(\alpha,\beta,\varDelta) \frac{1}{m}$,

where C(α,β,Δ) is some positive constant.

Lemma 9

For m>βΔ, we have $\left( 1-\frac{\theta\varDelta}{m}\right )^{m} \le e^{-\theta\varDelta}$.

Proof

Set $f(x)=(1+\frac{1}{x})^{x}$. Then f(x) is an increasing function for −∞<x<−1 and lim _x→−∞ f(x)=e. The conclusion now follows. □

Lemma 10

For k∈N∪{0},

• (i).
  $$\begin{aligned} \sup_{m\ge\max(\frac{k}{2},\beta\varDelta )}\sup_{\theta\in\varTheta} \left| \frac{\partial^k}{\partial\theta^k}\mu (m,\theta) \right| &= \sup_{m\ge\max(\frac{k}{2},\beta\varDelta)}\sup _{\theta\in\varTheta}\left| \frac{\partial^k}{\partial\theta^k}(1-\frac {\theta\varDelta}{m})^m \right|\\ &\le (2\varDelta)^k3^{2\beta\varDelta} < +\infty, \end{aligned}$$
• (ii).
  $$\begin{aligned} & \sup_{0\le h\le1}\sup_{m\ge\max (\frac{k}{2},\beta\varDelta)}\sup_{\theta\in\varTheta}\left| \frac{\partial ^k}{\partial\theta^k}\sigma^2(m,\theta,h)\right| \\ &\quad = \sup_{0\le h\le1}\sup_{m\ge\max(\frac{k}{2},\beta \varDelta)}\sup_{\theta\in\varTheta} \left| \frac{\partial^k}{\partial\theta ^k}\frac{(1-\frac{\theta\varDelta}{m})^{2m}-1}{\theta(\frac{\theta }{m}-2)} + h^2{\bf1}_{\{k=0\}} \right| \\ &\quad \le C(k,\varDelta,\alpha)+1<+\infty, \end{aligned}$$
• (iii).
  $$\begin{aligned} & \inf_{0\le h\le1}\inf_{m\ge\max (\frac{k}{2},\beta\varDelta)}\inf_{\theta\in\varTheta}\left| \sigma^2(m,\theta ,h) \right| \\ &\quad = \inf_{0\le h\le1}\inf_{m\ge\max(\frac{k}{2},\beta \varDelta)}\inf_{\theta\in\varTheta} \left| \frac{(1-\frac{\theta\varDelta }{m})^{2m}-1}{\theta(\frac{\theta}{m}-2)}+h^2 \right| \ge\frac {2(1-e^{-2\alpha\varDelta})}{3\beta} > 0, \end{aligned}$$

where the positive constant C(k,Δ,α) is defined in the proof.

Proof

Now $\mu(m,\theta)=(1-\frac{\theta\varDelta}{m})^{m}$ and set $D^{k}_{\theta}=\frac{\partial^{k}}{\partial\theta^{k}}$. Note that from Lemma 9, we have 0≤μ(m,θ)≤e ^−θΔ≤sup _θ e ^−θΔ=e ^−αΔ. Note that

$$\begin{aligned} D^k_{\theta}\mu(m,\theta)=(2m)(2m-1)\cdots(2m-(k-1))\left(1-\frac {\theta\varDelta}{m}\right)^{2m-k}\left(-\frac{\varDelta}{m}\right)^k. \end{aligned}$$

Then

$$\begin{aligned} & D^{k+1}_{\theta} \mu(m,\theta) \\ & \quad = (2m)(2m-1)\cdots(2m-(k-1))(2m-k)\left(1-\frac{\theta\varDelta }{m}\right)^{2m-(k+1)}\left(-\frac{\varDelta}{m}\right)^{k+1}. \end{aligned}$$

Moreover, for 2m≥k, we have

$$\begin{aligned} \sup_{m} \sup_{\theta} | D^k_{\theta}\mu(m,\theta)| \le\sup_{m} \left \{ \frac{(2m\varDelta)^k}{m^k} \biggl( 1+\frac{\beta\varDelta}{m}\biggr)^{2m-k} \right\} \le(2\varDelta)^k3^{2\beta\varDelta}. \end{aligned}$$

Hence we obtain (i).

Recall that $\sigma^{2}(m,\theta)=\frac{\mu(m,\theta)^{2}-1}{\theta(\frac {\theta}{m}-2)}$. From the Leibnitz formula, we have

$$\begin{aligned} | D^k_{\theta}\sigma^2(m,\theta)| \le&\sum^k_{i=0} C_{k,i} \sup_{m} \sup_{\theta}\bigl| D^i_{\theta}(\mu (m,\theta)^2) \bigr| \sup_{m} \sup_{\theta} \biggl| D^{k-i}_{\theta}\frac {1}{\theta(\frac{\theta}{m}-2)}\biggr|\\ &{} + \sup_{m} \sup_{\theta} \left| D^k_{\theta}\frac{1}{\theta(\frac{\theta}{m}-2)}\right|. \end{aligned}$$

From the above, the Leibnitz formula and the binomial theorem, we obtain, for i=0,1,…,k,

$$\begin{aligned} \sup_{m} \sup_{\theta}\bigl| D^i_{\theta}(\mu(m,\theta)^2)\bigr| \le& \sup_{m} \sup_{\theta} \left| \varDelta^i \biggl(1-\frac{\theta\varDelta}{m}\biggr)^{2m-i} \sum ^i_{j=0}\left( \begin{array}{c} i \\ j \end{array} \right) \right| \\ \le& \varDelta^i e^{-\alpha\varDelta}\sum^i_{j=0}\left( \begin{array}{c} i \\ j \end{array} \right) < \infty. \end{aligned}$$

Moreover, for all i=0,1,…,k, we have, from the binomial theorem,

$$\begin{aligned} \sup_{m} \sup_{\theta} \left| D^i_{\theta}\frac{1}{\theta(\frac{\theta }{m}-2)} \right| \le\sum^i_{j=0} C_{i,j} \frac{j!}{\alpha^{j+1}} \frac {(i-j)!}{2^{i-j}}. \end{aligned}$$

Then we have

$$\begin{aligned} & \sup_{m} \sup_{\theta\in[\alpha,\beta]}\bigl| D^k_{\theta}\sigma ^2(m,\theta)\bigr| \\ &\quad \le\sum^k_{i=0}C_{k,i} \left\{ \varDelta^i e^{-\alpha\varDelta}\sum ^i_{j=0}\left( \begin{array}{c} i \\ j \end{array} \right)\right\} \left\{ \sum^{k-i}_{j=0} C_{k-i,j} \frac{j!}{\alpha ^{j+1}} \frac{(k-i-j)!}{2^{k-i-j}} \right\} \\ &\qquad {} + \sum^k_{j=0} C_{k,j} \frac{j!}{\alpha^{j+1}} \frac{(k-j)!}{2^{k-j}} =: C(k,\varDelta,\alpha) < \infty, \end{aligned}$$

(18)

so that (ii) holds.

Finally, for m≥βΔ, we have

$$\begin{aligned} \sigma^2(m,\theta) \ge\frac{1-e^{-2\theta\varDelta}}{\theta\left( 2-\frac {\theta}{2\theta}\right)}=\frac{2}{3\theta}\left( 1-e^{-2\theta\varDelta }\right) \ge\frac{2}{3\beta}\left( 1-e^{-2\alpha\varDelta}\right)>0 \end{aligned}$$

and thus (iii) is valid. Here for m≥βΔ, $0 \le\left( 1-\frac{\theta\varDelta}{m} \right)^{m} \le e^{-\theta\varDelta} \le e^{-\alpha\varDelta}$, and for m≥βΔ,

$$\begin{aligned} \frac{2(1-e^{-2\alpha\varDelta})}{3\beta} \le\frac{( 1-\frac{\theta\varDelta }{m} )^{2m}-1}{\theta(\frac{\theta}{m}-2)} \le\frac{1}{\alpha(2-\beta )}. \end{aligned}$$

Thus the proof is complete. □