Order Determination of Large Dimensional Dynamic Factor Model

Consider the following dynamic factor model: $\mathbf{R}_t=\sum_{i=0}^q \mathbf{\Lambda}_i \mathbf{f}_{t-i}+\mathbf{e}_t,t=1,...,T$, where $\mathbf{\Lambda}_i$ is an $n\times k$ loading matrix of full rank, $\{\mathbf{f}_t\}$ are i.i.d. $k\times1$-factors, and $\mathbf{e}_t$ are independent $n\times1$ white noises. Now, assuming that $n/T\to c>0$, we want to estimate the orders $k$ and $q$ respectively. Define a random matrix $$\mathbf{\Phi}_n(\tau)=\frac{1}{2T}\sum_{j=1}^T (\mathbf{R}_j \mathbf{R}_{j+\tau}^* + \mathbf{R}_{j+\tau} \mathbf{R}_j^*),$$ where $\tau\ge 0$ is an integer. When there are no factors, the matrix $\Phi_{n}(\tau)$ reduces to $$\mathbf{M}_n(\tau) = \frac{1}{2T} \sum_{j=1}^T (\mathbf{e}_j \mathbf{e}_{j+\tau}^* + \mathbf{e}_{j+\tau} \mathbf{e}_j^*).$$ When $\tau=0$, $\mathbf{M}_n(\tau)$ reduces to the usual sample covariance matrix whose ESD tends to the well known MP law and $\mathbf{\Phi}_n(0)$ reduces to the standard spike model. Hence the number $k(q+1)$ can be estimated by the number of spiked eigenvalues of $\mathbf{\Phi}_n(0)$. To obtain separate estimates of $k$ and $q$ , we have employed the spectral analysis of $\mathbf{M}_n(\tau)$ and established the spiked model analysis for $\mathbf{\Phi}_n(\tau)$.


Introduction
For a p × p random Hermitian matrix A with eigenvalues λ j , j = 1, 2, · · · , p, the empirical spectral distribution (ESD) of A is defined as The limiting distribution F of {F An } for a given sequence of random matrices {A n } is called the limiting spectral distribution (LSD). Let {ε it } be independent identically distributed (i.i.d) random variables with common mean 0, variance 1. Consider a high dimensional dynamic kfactor model with lag q, that is, R t = q i=0 Λ i f t−i + e t , t = 1, ..., T , where Λ i is an n×k loading matrix of full rank, {f t } are i.i.d. k × 1-factors with common mean 0, variance 1, whereas e t corresponds to the noise component with e t = (ε 1t , · · · , ε nt ) ′ . In addition, both components of e t and f t are assumed to have finite 4th moment.
The matrix M n (0) has been intensively studied in the literature and it is well known that the LSD has an MP law (Marčenko and Pastur, 1967). Readers may refer to Jin et al. (2014) and Wang et al. (2015) for more details about the model.
To estimate k and q, the following method can be employed. First, note that when τ = 0 and Cov(f t ) = Σ f , the population covariance matrix of R t is a spiked population model (Johnstone (2001), Baik and Silverstein (2006), Bai and Yao (2008)) with k(q+1) spikes. Therefore, k(q+1) can be estimated by counting the number of eigenvalues of Φ n (0) that are larger than some phase transition point. Next, the separated estimation of k and q can be achieved by investigating the spectral property of M n (τ ) for general τ ≥ 1, using the fact that the number of eigenvalues of Φ n (τ ) that lie outside the support of the LSD of M n (τ ) at lags 1 ≤ τ ≤ q is different from that at lags τ > q. Thus, the estimates of k and q can be separated by counting the number of eigenvalues of Φ n (τ ) that lie outside the support of the LSD of M n (τ ) from τ = 0, 1, 2, · · · , q, q + 1, · · · .
Note that for the above method to work, the LSD of M n (τ ) for general τ ≥ 1 must be known. This is derived in Jin et al. (2014). Moreover, it is required that no eigenvalues outside the the support of the LSD of M n (τ ) so that if an eigenvalue of Φ n (τ ) goes out of the support of the LSD of M n (τ ), it must come from the signal part. Wang et al. (2015) proved such phenomenon theoretically. Both results are included in Section 2 for readers' reference.
The rest of the paper is structured as follows: Some known results are given in Section 2.
Section 3 presents truncation of variables and Section 4 estimates k(q + 1). The estimation of q is provided in Section 5, from the which the estimation of k can also be obtained. Section 6 discusses the case when the variance of the noise part is unknown. A simulation study is shown in Section 7 and some proofs are presented in Appendix.
Regrading the norm used in this paper, the norm applied to a vector is the usual Euclidean norm, with notation * . For a matrix, two kinds of norm have been used. The operator norm, denoted by * o , is the largest singular value. For matrices of fixed dimension, the Kolmogorov norm, defined as the largest absolute value of all the entries, has been used, with notation * K .

Some known results
In this section, we present some known results.
(c) There exist K > 0 and a random variable X with finite fourth order moment such that, for

Truncation, centralization and standardization of variables
As proved in Wang et al.(2015), we may assume that the ε ij 's satisfy the conditions that for some C, M > 0.
For the truncation of variables in F τ , first note that for a random variable X with E|X| 4 < ∞, Then we have This completes the proof of truncation. Centralization and standardization can be justified in the same way as in Appendix A of Wang et al. (2015). In what follows, we may assume that 4 Estimation of k(q + 1) In this section, we will estimate k(q+1) by an investigation of the limiting properties of eigenvalues of Φ n (0). For simplicity, rewrite Φ n (0) = Φ(0), F 0 = F and e 0 = e. With these notations, we have Φ(0) = 1 T (ΛF + e)(ΛF + e) * and M(0) = 1 T ee * . When Λ = 0, Φ(0) reduces to M(0), which is a standard sample covariance matrix and thus its ESD tends to the famous MP law (Marčenko and Pastur, 1967).
Suppose ℓ is an eigenvalue of Φ(0), then we have Let B = (B 1 : B 2 ) be an n × n orthogonal matrix such that B 1 = Λ(Λ * Λ) −1/2 and thus Λ * B 2 = 0 k(q+1)×(n−k(q+1)) . Then (4.1) is equivalent to If we further assume that ℓ is not an eigenvalue of 1 T B * 2 ee * B 2 , then we have Next, we have Substitute (4.5) back to (4.4), and we have Therefore, we obtain Next, we want to show, with probability 1 that Note that Suppose ℓ is an eigenvalue of Φ(0) larger than (1 + √ c) 2 + 2δ. By the fact that M(0) and 1 T e * e have the same set of nonzero eigenvalues, we have, under A, Write F = (F 1 , · · · ,F k(q+1) ) ′ . For the first term, by Lemma 2.2, we have which is summable for r ≥ 1.
Hence, we have shown with probability 1 that Similarly, we have with probability 1 that Therefore, substituting into (4.3), we have Thus, if Λ * Λ → Q, then (4.7) can be further simplified as If α is an eigenvalue of Q, and there is an ℓ belonging to the complement of the support of the LSD of M(0) such that α = 1 ℓm(ℓ)m(ℓ) , then ℓ is a solution of (4.8). From (4.6), we have It is easy to verify that g ′ (ℓ) < 0, implying that ℓm(ℓ)m(ℓ) is decreasing. Also note that This recovers the result of Baik and Silverstein (2006).
is the support of the MP law. Hence, if all the eigenvalues of Q are greater than √ c, we have k(q + 1) sample eigenvalues of Φ n (0) goes outside the right boundary of the support of the MP law. Note that although the distribution of rest n − k(q + 1) sample eigenvalues follows the MP law with the largest sample eigenvalue converging to the right boundary, there is still a positive probability that the largest sample eigenvalue goes beyond the right boundary. Therefore, to completely separate the k(q + 1) spiked sample eigenvalues from the rest, the threshold is set In other words, k(q + 1) can be estimated by the number of sample Remark 4.1 For factor models, the loading matrix is unknown. This, however, is not a concern in our estimation because compared with the noise matrix, the loading matrix is denominating, making the condition easily satisfied that all the eigenvalues of Q are greater than √ c. has an order of n −2/3 . Therefore, it is good enough for us to choose (1 + 2n −2/3 ) to completely separate the spikes and the bulk eigenvalues.

Estimation of q
Next, we want to split k and q. Let τ ≥ 1 be given and assume that ℓ is an eigenvalue of Φ n (τ ).
For simplicity, write M n (τ ) = M and for t = 1, 2, · · · , T , define Define B, B 1 and B 2 the same as in the last section. Multiplying B * from left and B from right to the above matrix and by Λ * B 2 = 0 k(q+1)×(n−k(q+1)) , we have (5.1) equivalent to Therefore, if ℓ is not an eigenvalue of S 22 , by the factorization above, ℓ must be an eigenvalue of . By the assumptions of e t 's and F t 's, the random Chung 2001). It then follows with probability 1 that is of dimension k(q + 1) × k(q + 1) with two bands of 1's of kτ -distance from the main diagonal.
Note that Next, we give a lemma on the quadratic form of γ j .
Lemma 5.1 Let i, j ∈ N be given, we have almost surely and uniformly in i and j that The proof of the lemma is postponed in the Appendix.
First, we have Similarly, we have Here H L (τ ) and H U (τ ) denote the lower and upper part of H(τ ) with the rest entries being 0.
When p = q τ + 1, we have part of E(P5) is Next, we want to show that P i → E(P i ) a.s. Since all the P i 's are of finite dimension, it suffices to show the a.s convergence entry-wise. Denote the (u, v)-entry of a matrix A by A (u,v) . For Then for any positive integer s, applying Here a (u,v) denotes the matrix with the (u, v)-entry being a and 0 elsewhere. By the truncation of ε ij and the fact that Substituting the above back to (5.2) and choosing s ≥ 2, we have Again, by the almost sure and uniform convergence of α j and α j+τ to C 0 , we have Therefore, we have shown that , then we have C p = αβ p . Note that H(pτ ) = 0 for p > [q/τ ], and we have, with probability 1 that Note that Similarly, Therefore, we have Hence, we have a.s.
Last, we want to show that with probability 1, and that Hence, by WB 1 = o a.s. (1) and B * 1 W * = o a.s. (1), it suffices to show with probability 1 that, By law of large numbers, we have with probability 1 that, Hence, we have with probability 1 Similarly, Therefore, ℓ should satisfy Recall B 1 = Λ(Λ * Λ) −1/2 . Our next goal is to find the limit of

Finally, we have
When τ > q, one has H(τ ) = 0, G(τ ) = 2I and (5.3) reduces to Let λ be an eigenvalue of Q, then we have Writing one can easily verify that the eigenvalues of H(q) are 1, −1 and 0, with multiplicity k, k and k(q − 1), respectively.
Note that when x is outside the support [−d(c), d(c)] of the LSD of M n (τ ), m 2 (x) = 0. Hence, we have We will show that g j (ℓ) is increasing over (d(c), ∞) by showing so are g j1 and g j2 . By definition, over (d(c), ∞), m is an increasing function taking negative values and m 2 is a decreasing function taking positive values. Hence, it is easy to see that g j1 (ℓ) is increasing over (d(c), ∞).
iii. If a j ≤ g j (d(c)), then ( 6 Estimate of σ 2 The above estimation is based on the assumption that σ 2 , the variance of the noise part is given. In practice, it is often the case that σ 2 is unknown. To this end, we can estimate σ 2 by employing the properties of the MP law. More precisely, we first estimate the left boundary of the support of the MP law by the smallest sample eigenvalue of Φ n (0), sayλ 1 (the eigenvalues are arranged in ascending order), and estimate the right boundary by (1+ √ c) 2 (1− √ c) 2λ1 . An iteration is then applied.
The initial estimator of σ 2 , sayσ 2 (0) , is obtained as the sample mean of the sample eigenvalues of Φ n (0) that lie within the interval [λ m+1 ,λ n−m ], where m ≥ 0 is such thatλ n−m is the largest For i ≥ 1, we obtain the updated estimator σ 2 (i) by taking the sample mean of the sample eigenvalues of Φ n (0) that lie within the interval The iteration stops once we haveσ 2 (ℓ−1) =σ 2 (ℓ) for some ℓ and our estimatorσ 2 :=σ 2 (ℓ) . As shown in the simulation, our estimation of k and q still works well with such estimator. Table 1 presents a simulation about the result discussed above, displaying the largest 13 absolute values of the eigenvalues for lags τ from 0 to 5. Here

Simulation
where f t 's are factors of length k; Λ i , i = 0, ..., q is a constant time-invariant matrix of size n × k, e t is the error term and q is the lag of the model. In addition, assume that: e t are i.i.d.
random variables with e t ∼ N (0, σ 2 I n ) and f t are i.i.d. random variables with f t ∼ N 0, Eigenvalues of Q are from 95 to 285, making Q −1 ∼ 0.
When τ = 0, and σ 2 = 1 is known, using the phase transition point b c = (1 + √ c) 2 σ 2 = 3.7974, we see that the number of spotted spikes is 6, which estimates k(q + 1). When for In other words, the number of spikes first jumps to 2k(q + 1) at τ = 3 which estimates q + 1.
The estimation of k is obvious.
A.2 Proof of Lemma 5.2 By n j=1 |b j | 2 = A −1 k x 2 and Cauchy-Schwartz inequality, we have Here η := ℓ − d c > 0. Therefore, we have The proof of the lemma is complete.

A.3 Proof of Lemma 5.3
PROOF. Let x, y ∈ C n 1 be given. Define Using Lemma 2.1, we have, for i = 1, 2 Note that Hence, we have Next, we have Substitute back, we obtain Note that It is easy to see that work on α k11 and α k14 is the same as that on α k 2 .
Substituting all the above results into (A.1) and choosing l large enough, we have x * A −1 y − Ex * A −1 y → 0 a.s.
Next, we want to show the convergence of Ex * A −1 y.