Estimation and testing of the factor-augmented panel regression models with missing data

A.1.1 Step1: deriving the asymptotic expansion of N ( β ̂ CCEM − β )

This expansion depends on whether m < k + 1 or m = k + 1. Suppose first that m < k + 1. We introduce a full rank matrix H _j = [H _m,j , H _−m,j ] and a normalization matrix D N j = diag ( I m , N j I k + 1 − m ) , where H _j and D N j are both (k + 1) × (k + 1). Also, H _m,j and H _−m,j are (k + 1) × m and (k + 1) × (k + 1 − m) matrices, respectively. H m , j = Q ′ C ̄ j ′ C ̄ j Q Q ′ C ̄ j ′ − 1 and H _−m,j = G _j B _−m,j where G _j and B _−m,j are the same as G and B _−m in Westerlund et al. (2019). The useful properties of H _j and D N j are respectively C ̄ j Q H j = I m , 0 m × ( k + 1 − m ) and U ̄ j Q H j D N j = [ U ̄ j Q H m , j , N j U ̄ j Q H − m , j ] = U ̄ m , j 0 , U ̄ − m , j 0 = U ̄ j 0 , where U ̄ m , j 0 = O p N j − 1 / 2 and U ̄ − m , j 0 = O p ( 1 ) . We have

Moreover, since M Z ̄ j = M Z ̄ j 0 , we have

Consider the numerator,

In analogy to (A.46) in Westerlund et al. (2019), we have

where P − m , j = P M F j U ̄ − m , j 0 as m < k + 1, and P − m , j = 0 T j × T j as m = k + 1 (see also, Westerlund et al. 2019). Hence, the numerator of N ( β ̂ CCEM − β ) can be written as

Next, consider the denominator of N ( β ̂ CCEM − β ) . By analogy to Lemma A.2 in Westerlund et al. (2019), we can write

According to (A.1 & A.2), we can obtain the asymptotic expansion of N ( β ̂ CCEM − β ) as follows,

A.1.2 Step 2: proof of Theorem 1

Now, we try to derive the asymptotic distribution of N ( β ̂ CCEM − β ) , and first consider the case with m = k + 1. In analogy to Westerlund et al. (2019), we let ξ j i = V j i ′ ( M F j − P − m , j ) ε j i = V j i ′ M F j ε j i , ζ F j being the σ-field generated by F _j , and F j i being the corresponding σ-field generated by ζ F j and ( ξ j 1 , … , ξ j i ) . Then { ( ξ j i , F j i ) : i ≥ 1 } is a martingale difference sequence, and ξ j i is independent across j _i and E ( ξ j i | F j i − 1 ) = E ( ξ j i | ζ F j ) = 0 k × 1 . Hence, the outer sum has a mixed normal distribution (see also, Westerlund et al. 2019), which is given by

where

with Σ ε , j i = E ε j i ε j i ′ . By the law of large numbers of Hall and Heyde (1980), we can obtain

Hence, as N → ∞, we can also show that

Next, consider the case with m < k + 1. It follows that

where z ̄ j , t , s = N j − 1 ∑ i = 1 N j ε j i , s v j i , t , M F j , t , s and p_−m,j,t,s are the (t, s)-th element of M F j and p_−m,j respectively. In analogy to (A.56 & A.61) in Westerlund et al. (2019), we have

Hereafter, “ = d ” means equality in distribution, Σ u , j , t = lim N j → ∞ N j − 1 ∑ i = 1 N j ( Σ u , j i , t ) and Σ z , j , t , s = lim N j → ∞ N j − 1 ∑ i = 1 N j σ ε , j i , s 2 Σ v , j i , t . By using u ̄ − m , j , t 0 to signify the tth row of U ̄ − m , j 0 , we have

Then, we obtain

as N _j → ∞, which implies

According to (A.64) in Westerlund et al. (2019), we have E u j i , t z l r , n , s ′ = 0 ( k + 1 ) × k for all i, j, l, r, t, n and s. It means that n z j , n , s and M F j , t , s − p − m , j , t , s are independent of each other. Hence,

where S = ∑ j = 1 n w j S j , and

with Σ z j , t , l , s , r = lim N j → ∞ N j − 1 ∑ i = 1 N j ( σ ε , j i , s , r Σ v , j i , t , l ) , σ ε , j i , s , r = E ( ε j i , s ε j i , r ) and Σ v , j i , t , l = E v j i , t v j i , l ′ . For the denominator, we have

Hence, as N → ∞, we can show that

The proof of Theorem 1 is then complete. □