Inference for continuous-time long memory randomly sampled processes

Abstract

From a continuous-time long memory stochastic process, a discrete-time randomly sampled one is drawn using a renewal sampling process. We establish the existence of the spectral density of the sampled process, and we give its expression in terms of that of the initial process. We also investigate different aspects of the statistical inference on the sampled process. In particular, we obtain asymptotic results for the periodogram, the local Whittle estimator of the memory parameter and the long run variance of partial sums. We mainly focus on Gaussian continuous-time process. The challenge being that the randomly sampled process will no longer be jointly Gaussian.

Appendix: Proof of Lemma 2

Proof

The proof is essentially based on Corollary 2 and a well known cumulant formula.

Without loss of generality, we can assume that the Poisson rate is 1. The process Y is 4th order stationary as the conditional joint distribution of $(Y_k,Y_{k+h},Y_{k+r},Y_{k+s})$ given $(T_1,\dots ,T_{k+\max (h,r,s)})$ is a multivariate normal with variance-covariance matrix $M(T_k,T_{k+h},T_{k+r},T_{k+s})$ given by

$$\begin{aligned}{} & {} M(T_k,T_{k+h},T_{k+r},T_{k+s}) \nonumber \\{} & {} \quad :=\left( {\begin{matrix} \sigma _X(0)&{}\sigma _X(T_{k+h}-T_k)&{}\sigma _X(T_{k+r}-T_k)&{}\sigma _X(T_{k+s}-T_k)\\ \sigma _X(T_{k+h}-T_k)&{}\sigma _X(0)&{}\sigma _X(T_{k+r}-T_{k+h})&{}\sigma _X(T_{k+s}-T_{k+h})\\ \sigma _X(T_{k+r}-T_k)&{}\sigma _X(T_{k+r}-T_{k+h})&{}\sigma _X(0)&{}\sigma _X(T_{k+s}-T_{k+r})\\ \sigma _X(T_{k+s}-T_k)&{}\sigma _X(T_{k+s}-T_{k+h})&{}\sigma _X(T_{k+s}-T_{k+r})&{}\sigma _X(0) \end{matrix}}\right) \nonumber \\ \end{aligned}$$

(41)

which is k free. Hence it is enough to establish the lemma when $k=0$. We apply the total law of cumulance formula (Brillinger 1969), which for the sake of clarity, we remind here: for all random vectors $Z=(Z_1,\ldots ,Z_n)'$ and W, we have

$$\begin{aligned} \textrm{cum}(Z)=\sum _{\pi }\textrm{cum}\left[ \textrm{cum}(X_{\pi _1}\vert W),\ldots ,\textrm{cum}(X_{\pi _b}\vert W)\right] \end{aligned}$$

(42)

where $X_{\pi _j}=(X_i,i\in \pi _j)$, and $\pi _1,\ldots ,\pi _b$, ($b=1,\ldots ,n$) are the blocks of the permutation $\pi$, and the sum is over all permutations $\pi$ of the set $\{1,2,\ldots ,n\}$.

But condition on T, the process $Y_t$ is jointly zero-mean Gaussian and therefore $\mathbb {E}(Y_t\vert T)=0$ as well as $\textrm{cum}(Y_i,Y_j,Y_k,Y_\ell \vert T)=\textrm{cum}(Y_i,Y_j,Y_k\vert T)=0$ for all $i,j,k,\ell$. Hence applying (42) to $Y_t$ with $W=T$, only the two-by-two partitions of $\{0,h,r,s\}$ will survive. and since $\textrm{cum}(U,V)=\text {Cov}(U,V)$, we get from (41)

$$\begin{aligned} \textrm{cum}(Y_0,Y_{h},Y_{r},Y_{s})=\,&\textrm{Cov}(\sigma _X(T_h),\sigma _X(T_r-T_s))+\textrm{Cov}(\sigma _X(T_r),\sigma _X(T_h-T_s))\nonumber \\&+\textrm{Cov}(\sigma _X(T_s),\sigma _X(T_r-T_h)). \end{aligned}$$

(43)

Note that for $h< \min (r,s)$, $\textrm{Cov}(\sigma _X(T_h),\sigma _X(T_r-T_s))=0$. Moreover

$$\begin{aligned}&\sum _{1\le r \le h \le s \le n} |\textrm{Cov}(\sigma _X(T_h),\sigma _X(T_r-T_s)) |\\&\quad \le \sum _{1\le r \le h \le s \le n} \textrm{Var}(\sigma _X(T_h))^{1/2} \textrm{Var}( \sigma _X(T_s-T_r))^{1/2}\\&\quad \le \sum _{1\le r \le h \le s \le n} h^{-\alpha /2} (1 +s-r)^{-\alpha /2} \\&\quad \le h^{-\alpha /2} \sum _{1\le r \le h } \sum _{t=1}^n t ^{-\alpha /2} \\&\quad \le h^{1-\alpha /2} {\left\{ \begin{array}{ll} n^{1-\alpha /2} = n^{2d -1/2 }&{} \text {if d<1/4} \\ \log (n) &{} \text {if}\,\, d\ge 1/4 \\ \end{array}\right. }\\&\le {\left\{ \begin{array}{ll} n^{4d -1 }&{} \text {if} \,\, d<1/4 \\ \log (n) &{} \text {if}\,\, d\ge 1/4 \\ \end{array}\right. }\\&\le Cn^{2d} \; \text { for all } \; 0< d <1/2. \end{aligned}$$

The last configuration is

$$\begin{aligned} \sum _{r,s=1}^h | \textrm{Cov}(\sigma _X(T_h),\sigma _X(T_r-T_s))|&= \sum _{r,s=1}^h | \textrm{Var}(\sigma _X(T_h))^{1/2} \textrm{Var}( \sigma _X(T_s-T_r))^{1/2}\\&\le h^{-\alpha /2} \sum _{t=1}^h (h-t) t^{-\alpha /2} \\&\le h^{-\alpha /2 } {\left\{ \begin{array}{ll} C h^{1-\alpha /2} = Ch^{2d -1/2 }&{} \text {if} \,\,d<1/4 \\ \log (h) &{} \text {if}\,\, d\ge 1/4 \\ \end{array}\right. } \\&\le Cn^{2d} \; \text { for all } \; 0< d <1/2. \end{aligned}$$

Therefore uniformly in h we have

$$\begin{aligned} \sum _{r,s=1}^n | \textrm{Cov}(\sigma _X(T_h),\sigma _X(T_r-T_s))| \le Cn^{2d}. \end{aligned}$$

For the remaining two terms in the right hand side of (43) we have, for fixed h,

$$\begin{aligned} \sum _{r,s=1}^n | \textrm{Cov}(\sigma _X(T_r),\sigma _X(T_h-T_s))|&= \sum _{r,s=1}^n | \textrm{Cov}(\sigma _X(T_s),\sigma _X(T_h-T_r))| \\&\le \sum _{r,s=1}^n \textrm{Var}(\sigma _X(T_s))^{1/2} \textrm{Var}( \sigma _X(T_h-T_r))^{1/2}\\&\le \sum _{r,s=1}^n s^{-\alpha /2} (1+|h-r|)^{-\alpha /2} \\&\le C {\left\{ \begin{array}{ll} n^{2-\alpha } = n^{4d -1 }&{} \text {if}\,\, d<1/4 \\ \log (n)^2 &{} \text {if}\,\, d\ge 1/4 \\ \end{array}\right. } \\ {}&\le Cn^{2d} \; \text { for all } \; 0< d <1/2. \end{aligned}$$

This concludes the proof of (37).

Let us now prove (38). Note that

$$\begin{aligned} \sum _{h,r,s=0}^n \textrm{cum}(Y_0,Y_{h},Y_{r},Y_{s})&= 3 \sum _{h,r,s=1}^n \textrm{Cov}(\sigma _X(T_h),\sigma _X(T_r-T_s)) \nonumber \\&= 6 \sum _{h=1}^n\sum _{r<s=1}^n \textrm{Cov}(\sigma _X(T_h),\sigma _X(T_r-T_s)). \end{aligned}$$

(44)

Moreover, we have

$$\begin{aligned} \sum _{h,r,s=1}^n |\textrm{Cov}(\sigma _X(T_h),\sigma _X(T_r-T_s)) |&\le C \sum _{h,r,s=1}^n h^{-\alpha /2} (1+|r-s|)^{-\alpha /2} \\&\le C \sum _{h=1}^n h^{-\alpha /2} \sum _{t=1}^n (n-t) t^{-\alpha /2} \\&\le C {\left\{ \begin{array}{ll} n n^{2-\alpha } = n^{4d }&{} \text {if}\,\, d<1/4 \\ n \log (n)^2 &{} \text {if}\,\, d> 1/4 \\ \end{array}\right. } \\&\le Cn^{4d} \end{aligned}$$

In the particular case $d=1/4$ (where we still have $\alpha =2$), a supplementary term $\log (n)$ is needed in the bound. Indeed we split the sum in the right hand side of (44) into 3 configurations. when $1 \le h \le r< s \le n$ the covariance $\textrm{Cov}(\sigma _X(T_h),\sigma _X(T_r-T_s))$ is zero. When the sum is over $1 \le r< h \le s \le n$, we get

$$\begin{aligned} \sum _{1 \le r< h \le s \le n} |\textrm{Cov}(\sigma _X(T_h),\sigma _X(T_r-T_s)) |&\le C \sum _{s=1}^n \sum _{h=1}^s h^{-1 } \sum _{r=1}^{h-1} (s-h + h-r)^{-1} \\&\sim C \sum _{s=1}^n \sum _{h=1}^s h^{-1 } \left( \log (s) - \log (s-h) \right) \\&= -C \sum _{s=1}^n \sum _{h=1}^s (h/s)^{-1 } \log (1-h/s) (1/s) \\&\sim -C \sum _{s=1}^n\left( \int _{0}^1 \frac{ \log (1-x)}{x} \, dx\right) = C\frac{\pi ^2}{6} n. \end{aligned}$$

For the last sum over $1 \le r < s \le h \le n$ (where we will need the $\log$ term) we have

$$\begin{aligned} \sum _{1 \le r <s \le h \le n} |\textrm{Cov}(\sigma _X(T_h),\sigma _X(T_r-T_s)) |&\le C \sum _{h=1}^n h^{-1 } \sum _{s=1}^h \sum _{r=1}^{s-1} (s-r)^{-1} \\&= \sum _{h=1}^n h^{-1 } \sum _{t=1}^h (h-t) t^{-1} \\&= \sum _{h=1}^n \sum _{t=1}^h (1-t/h) t^{-1} \\&\sim C \sum _{h=1}^n ( \log (h) -1 ) \sim C n \log (n). \end{aligned}$$

This completes the proof of (38) in Lemma 2. $\square$