Purpose

This notebook is to demonstrate the derivation to obtain the limiting value of n(E((Subscript[((Overscript[ϕ, ⋀])), 1]) - (Subscript[ϕ, 1]))) as a function of roots a, b and c of the polynomial \(TraditionalForm\`ζ\^3 - \(ϕ\_1\) ζ\^2 - \(ϕ\_2\) ζ - ϕ\_3 = 0\).

Reformat the Burg Estimate (Subscript[((Overscript[ϕ, ⋀])), 1])

By the Levinson-Durbin recursion, the Burg estimate (Subscript[((Overscript[ϕ, ⋀])), 1]) is

ϕ[
        13] = ϕ[12] - ϕ[33] ϕ[22]

From Mathematica notebook, BurgAR(3)May_derivation, we know that

ϕ[
        22] = ((2  S[2, 3] - 2 S[1, 3] ϕ[11] - 2 S[1, 3] ϕ[11] + 2 S[0, 3] ϕ[11]^2)/(S[0, 3] + S[0, 3] - 2 S[1, 3] ϕ[11] - 2 S[1, 3] ϕ[11] + 2 S[0, 3] ϕ[11]^2)) /. {ϕ[11] ->  
              S[1, 2]/S[0, 2]} // Simplify

ϕ[
        12] = ((S[0, 2] S[1, 2] (S[0, 3] - S[2, 3]))/(S[0, 2]^2 S[0, 3] + S[0, 3] S[1, 2]^2 - 2 S[0, 2] S[1, 2] S[1, 3])) // 
        Simplify

ϕ[
        33] = (2 S[0, 2]^3 S[1, 
                  2] (-S[0, 4] S[2, 3]^2 - S[0, 3]^2 S[2, 4] + 
                    S[0, 3] (S[2, 3] (S[0, 4] + S[2, 4]) + 
                          2 S[1, 3] (S[1, 4] - S[3, 4]))) + 
              2 S[0, 2] S[0, 
                  3] S[1, 2]^3 ((S[0, 3] - S[2, 3]) (S[0, 4] - 
                          S[2, 4]) + 
                    2 S[1, 3] (S[1, 4] - S[3, 4])) + 
              S[0, 2]^2 S[1, 2]^2 (S[1, 4] S[2, 3]^2 - 
                    2 S[0, 
                        3] (S[1, 4] S[2, 3] + 
                          2 S[1, 3] (S[0, 4] - S[2, 4])) + 
                    4 S[1, 3] S[2, 3] (S[0, 4] - S[2, 4]) - 
                    S[0, 3]^2 (S[1, 4] - 2 S[3, 4]) - 
                    4 S[1, 3]^2 (S[1, 4] - S[3, 4])) + 
              S[0, 3]^2 S[1, 2]^4 (-S[1, 4] + S[3, 4]) + 
              S[0, 2]^4 (-2 S[0, 3] S[1, 4] S[2, 3] + 
                    S[1, 4] S[2, 3]^2 + 
                    S[0, 3]^2 S[3, 
                        4]))/(2 S[0, 3]^2 S[1, 2]^4 (S[0, 4] - 
                    S[2, 4]) + 
              8 S[0, 2] S[0, 3] S[1, 2]^3 S[1, 
                  3] (-S[0, 4] + S[2, 4]) + 
              S[0, 2]^4 (S[0, 4] (S[0, 3]^2 + S[2, 3]^2) - 
                    2 S[0, 3] S[2, 3] S[2, 4]) + 
              S[0, 2]^2 S[1, 2]^2 (S[0, 
                        4] (3 S[0, 3]^2 + 8 S[1, 3]^2 + 
                          S[2, 3]^2) - 
                    2 (4 S[1, 3]^2 + 
                          S[0, 3] (S[0, 3] + S[2, 3])) S[2, 4]) + 
              2 S[0, 2]^3 S[1, 
                  2] (-S[0, 3]^2 S[1, 4] - 
                    S[2, 3] (S[1, 4] S[2, 3] + 
                          2 S[1, 3] (S[0, 4] - S[2, 4])) + 
                    S[0, 3] (2 S[1, 4] S[2, 3] + 
                          2 S[1, 3] (-S[0, 4] + S[2, 4])))) // 
        Simplify

The Burg estimate (Subscript[((Overscript[ϕ, ⋀])), 1]) in AR(3) models is

phi13 = ϕ[12] - ϕ[33] ϕ[22] // FullSimplify

phi1[t1_, t2_, t3_, t4_, t5_, t6_, t7_, t8_, t9_] := 
      phi13 /. {S[0, 4] ->  t1, S[1, 4] ->  t2, 
            S[2, 4] ->  t3, S[3, 4] ->  t4, S[0, 3] ->  t5, 
            S[1, 3] ->  t6, S[2, 3] ->  t7, S[0, 2] ->  t8, 
            S[1, 2] ->  t9} // FullSimplify

phi1[t1, t2, t3, t4, t5, t6, t7, t8, t9]

Define Two Operators

The autocovariance at lag k in AR(3) is

R[k_] := A1 (a^k) + A2 (b^k) + A3 (c^k)

The operator of \(TraditionalForm\`\(\(lim\+\(n -> \ ∞\)\)\(Cov\)\(\ \)\)\)(S[p,i], S[q,j]), the formula (6) and (7) in the paper, where p = m - k and q = f - g, is

LCoV[p_, q_] := 
      Module[{a, b, c, d, e}, 
          a = Sum[R[-h] R[-(h - p + q)] + 
                R[-(h + q)] 
                  R[-(h - 
                          p)], {h, -∞, -q}]; 
b = 
            Sum[R[-h] R[-(h - p + q)] + 
                R[(h + q)] R[-(h - p)], {h, -q + 1, 
                0}]; 
c = 
            Sum[R[h] R[-(h - p + q)] + 
                R[(h + q)] R[-(h - p)], {h, 1, 
                p - q}]; 
d = 
            Sum[R[h] R[(h - p + q)] + 
                R[(h + q)] R[-(h - p)], {h, p - q + 1, 
                p}]; 
e = 
            Sum[R[h] R[(h - p + q)] + R[(h + q)] R[(h - p)], {h, 
                p + 1, ∞}]; 
a + b + c + d + e] // 
        Simplify

The operator of (lim_(n ->  ∞))n E(
            S[p, i] - R[p]) , the formula (5) in the paper, where p = m - k, is

LB[p_, 
          i_] := -(i - 1) R[
                p] - (Sum[R[-t], {t, -∞, 0}] + 
                Sum[R[t], {t, 1, ∞}]) // Simplify;

Large Sample Bias for the Unknown Mean Case

Obtaining  the large  sample  variances  and covariance coefficients of S[0,  4], S[1, 4], S[2, 4], S[3, 4], S[0, 3], 
      S[1, 3], S[2, 3], S[0, 2] and S[1, 2]

ExpSquare = {LCoV[0, 0], LCoV[1, 1], LCoV[2, 2], LCoV[3, 3], 
            LCoV[0, 0], LCoV[1, 1], LCoV[2, 2], LCoV[0, 0], LCoV[1, 1]} // 
          Simplify;

ExpCrossS0 = {LCoV[1, 0], LCoV[2, 0], LCoV[3, 0], LCoV[0, 0], 
            LCoV[1, 0], LCoV[2, 0], LCoV[0, 0], LCoV[1, 0]} // 
          Simplify;

ExpCrossS1 = {LCoV[2, 1], LCoV[3, 1], LCoV[1, 0], LCoV[1, 1], 
            LCoV[2, 1], LCoV[1, 0], LCoV[1, 1]} // Simplify;

ExpCrossS2 = {LCoV[3, 2], LCoV[2, 0], LCoV[2, 1], LCoV[2, 2], 
            LCoV[2, 0], LCoV[2, 1]} // Simplify;

ExpCrossS3 = {LCoV[3, 0], LCoV[3, 1], LCoV[3, 2], LCoV[3, 0], 
            LCoV[3, 1]} // Simplify;

ExpCrosss0 = {LCoV[1, 0], LCoV[2, 0], LCoV[0, 0], LCoV[1, 0]} // 
          Simplify;

ExpCrosss1 = {LCoV[2, 1], LCoV[1, 0], LCoV[1, 1]};

ExpCrosss2 = {LCoV[2, 0], LCoV[2, 1]} // Simplify;

ExpCrosss00 = LCoV[1, 0] // Simplify;

Large sampe bias coefficients of S[0,  4], S[1, 4], S[2, 4], S[3, 4], S[0, 3], 
      S[1, 3], S[2, 3], S[0, 2] and S[1, 2]

FirstExp = { LB[0, 4], LB[1, 4], LB[2, 4], LB[3, 4], LB[0, 3], 
            LB[1, 3], LB[2, 3], LB[0, 2], LB[1, 2]} // Simplify;

The expectations  of S[0,  4], S[1, 4], S[2, 4], S[3, 4], S[0, 3], 
      S[1, 3], S[2, 3], S[0, 2] and S[1, 2]:

(Subscript[γ, 0]) = R[0]; (Subscript[γ, 1]) = R[1]; (Subscript[γ, 2]) = 
      R[2]; (Subscript[γ, 3]) = R[3]; (Subscript[γ, 4]) = R[0]; (Subscript[γ, 5]) = 
      R[1]; (Subscript[γ, 6]) = R[2]; (Subscript[γ, 7]) = R[0]; (Subscript[γ, 8]) = 
      R[1];

The derivatives  in the Taylor expansion

FirstTerm = (D[
                  phi1[t1, t2, t3, t4, t5, t6, t7, t8, t9], #] & /@ {t1, 
                t2, t3, t4, t5, t6, t7, t8, 
                t9}) /. {t1 ->  (Subscript[γ, 0]), 
            t2 ->  (Subscript[γ, 1]), t3 ->  (Subscript[γ, 2]), 
            t4 ->  (Subscript[γ, 3]), t5 ->  (Subscript[γ, 4]), 
            t6 ->  (Subscript[γ, 5]), t7 ->  (Subscript[γ, 6]), 
            t8 ->  (Subscript[γ, 7]), t9 ->  (Subscript[γ, 8])};

SquareTerm = (D[
                  phi1[t1, t2, t3, t4, t5, t6, t7, t8, t9], #, #] & /@ {t1, 
                t2, t3, t4, t5, t6, t7, t8, 
                t9}) /. {t1 ->  (Subscript[γ, 0]), 
            t2 ->  (Subscript[γ, 1]), t3 ->  (Subscript[γ, 2]), 
            t4 ->  (Subscript[γ, 3]), t5 ->  (Subscript[γ, 4]), 
            t6 ->  (Subscript[γ, 5]), t7 ->  (Subscript[γ, 6]), 
            t8 ->  (Subscript[γ, 7]), t9 ->  (Subscript[γ, 8])};

t = {t1, t2, t3, t4, t5, t6, t7, t8, t9}

Crosst1 = 
        Table[D[phi1[t1, t2, t3, t4, t5, t6, t7, t8, t9], t1, 
              t[[j]]],  {j, 2, 
              9}] /. {t1 ->  (Subscript[γ, 0]), t2 ->  (Subscript[γ, 1]), 
            t3 ->  (Subscript[γ, 2]), t4 ->  (Subscript[γ, 3]), 
            t5 ->  (Subscript[γ, 4]), t6 ->  (Subscript[γ, 5]), 
            t7 ->  (Subscript[γ, 6]), t8 ->  (Subscript[γ, 7]), 
            t9 ->  (Subscript[γ, 8])};

Crosst2 = 
        Table[D[phi1[t1, t2, t3, t4, t5, t6, t7, t8, t9], t2, 
              t[[j]]],  {j, 3, 
              9}] /. {t1 ->  (Subscript[γ, 0]), t2 ->  (Subscript[γ, 1]), 
            t3 ->  (Subscript[γ, 2]), t4 ->  (Subscript[γ, 3]), 
            t5 ->  (Subscript[γ, 4]), t6 ->  (Subscript[γ, 5]), 
            t7 ->  (Subscript[γ, 6]), t8 ->  (Subscript[γ, 7]), 
            t9 ->  (Subscript[γ, 8])};

Crosst3 = 
        Table[D[phi1[t1, t2, t3, t4, t5, t6, t7, t8, t9], t3, 
              t[[j]]],  {j, 4, 
              9}] /. {t1 ->  (Subscript[γ, 0]), t2 ->  (Subscript[γ, 1]), 
            t3 ->  (Subscript[γ, 2]), t4 ->  (Subscript[γ, 3]), 
            t5 ->  (Subscript[γ, 4]), t6 ->  (Subscript[γ, 5]), 
            t7 ->  (Subscript[γ, 6]), t8 ->  (Subscript[γ, 7]), 
            t9 ->  (Subscript[γ, 8])};

Crosst4 = 
        Table[D[phi1[t1, t2, t3, t4, t5, t6, t7, t8, t9], t4, 
              t[[j]]],  {j, 5, 
              9}] /. {t1 ->  (Subscript[γ, 0]), t2 ->  (Subscript[γ, 1]), 
            t3 ->  (Subscript[γ, 2]), t4 ->  (Subscript[γ, 3]), 
            t5 ->  (Subscript[γ, 4]), t6 ->  (Subscript[γ, 5]), 
            t7 ->  (Subscript[γ, 6]), t8 ->  (Subscript[γ, 7]), 
            t9 ->  (Subscript[γ, 8])};

Crosst5 = 
        Table[D[phi1[t1, t2, t3, t4, t5, t6, t7, t8, t9], t5, 
              t[[j]]],  {j, 6, 
              9}] /. {t1 ->  (Subscript[γ, 0]), t2 ->  (Subscript[γ, 1]), 
            t3 ->  (Subscript[γ, 2]), t4 ->  (Subscript[γ, 3]), 
            t5 ->  (Subscript[γ, 4]), t6 ->  (Subscript[γ, 5]), 
            t7 ->  (Subscript[γ, 6]), t8 ->  (Subscript[γ, 7]), 
            t9 ->  (Subscript[γ, 8])};

Crosst6 = 
        Table[D[phi1[t1, t2, t3, t4, t5, t6, t7, t8, t9], t6, 
              t[[j]]],  {j, 7, 
              9}] /. {t1 ->  (Subscript[γ, 0]), t2 ->  (Subscript[γ, 1]), 
            t3 ->  (Subscript[γ, 2]), t4 ->  (Subscript[γ, 3]), 
            t5 ->  (Subscript[γ, 4]), t6 ->  (Subscript[γ, 5]), 
            t7 ->  (Subscript[γ, 6]), t8 ->  (Subscript[γ, 7]), 
            t9 ->  (Subscript[γ, 8])};

Crosst7 = 
        Table[D[phi1[t1, t2, t3, t4, t5, t6, t7, t8, t9], t7, 
              t[[j]]],  {j, 8, 
              9}] /. {t1 ->  (Subscript[γ, 0]), t2 ->  (Subscript[γ, 1]), 
            t3 ->  (Subscript[γ, 2]), t4 ->  (Subscript[γ, 3]), 
            t5 ->  (Subscript[γ, 4]), t6 ->  (Subscript[γ, 5]), 
            t7 ->  (Subscript[γ, 6]), t8 ->  (Subscript[γ, 7]), 
            t9 ->  (Subscript[γ, 8])};

Crosst8 = 
        D[phi1[t1, t2, t3, t4, t5, t6, t7, t8, t9], t8, 
            t9] /. {t1 ->  (Subscript[γ, 0]), t2 ->  (Subscript[γ, 1]), 
            t3 ->  (Subscript[γ, 2]), t4 ->  (Subscript[γ, 3]), 
            t5 ->  (Subscript[γ, 4]), t6 ->  (Subscript[γ, 5]), 
            t7 ->  (Subscript[γ, 6]), t8 ->  (Subscript[γ, 7]), 
            t9 ->  (Subscript[γ, 8])};

Clear[A1]; Clear[A2]; Clear[A3]; Clear[a]; Clear[b]; 
    Clear[c];

LSBiasPhi1 = 
        Flatten[FirstExp . FirstTerm] + (1/2) 
            Flatten[SquareTerm] . Flatten[ExpSquare] + 
          Flatten[Crosst1] . Flatten[ExpCrossS0] + 
          Flatten[Crosst2] . Flatten[ExpCrossS1] + 
          Flatten[Crosst3] . Flatten[ExpCrossS2] + 
          Flatten[Crosst4] . Flatten[ExpCrossS3] + 
          Flatten[Crosst5] . Flatten[ExpCrosss0] + 
          Flatten[Crosst6] . Flatten[ExpCrosss1] + 
          Flatten[Crosst7] . Flatten[ExpCrosss2] + 
          Flatten[Crosst8 ExpCrosss00];

The limiting value of \(TraditionalForm\`n\ \((E\ \((\(ϕ\&⋀\)\_1 - ϕ\_1)\))\)\) for the unknown means case is

LSBiasPhi1