Determining whether the statistic is a pivot

For bootstrap analysis of any test statistic the only requirement is that the dis-tribution under the null should not be affected by any model specific features, see MacKinnon [2002] andMacKinnon [2006] for the philosophical discussion on inference on bootstrap. That is, we do not need to know the true model structure to identify the distribution of the test statistic under the null. For instance, we have an unbiased estimate of the mean of a random variable and its variance.

If the variable is IID normally distributed, then the distribution of the mean divided by the squareroot of the variance under the asumption that the mean is zero can be determined by drawing random samples of equivalent length data from a distribution with a zero mean and variance identical to the sample variance and constructing the same ratio. Inference is then based on the 100α percentiles as needed. Furthermore, as the series is presumed to be normal, be can sample with replacement the series, subtract the resampled mean and compute the re-sampled variance, to give a distribution of the test statistic. In both cases the test statistic is generated under the null of the mean being zero. In both cases

the resampling is normally distributed, in the first case by construction and in the second case using the presumed properties of the actual data.

Unfortunately, bootstrapping a z-score is one of the few tests where derivation of the pivot under the null is simple enough to illustrate without recourse to the underlying stochastic properties. Theorems 9 and 10 are re-derived from Muirhead[1982] in our notation, to show the pivot features indeed, the intention of Muirhead [1982] was to derive the Neyman-Pearson form of the statistic, but this is useful as it serves the same purpose for the bootstrap.

Theorem 8. Muirhead [1982] ,when the null hypothesis H_k is true, the limiting distribution, as n → ∞, of the statistic

P_k = −

Proof. Proof of Theorem 8 [Muirhead,1982]

Ec(T_k)

= − ∂

∂h[Ec(e^−hTk)]_h=0

We can exchange the order of differentiation, seeApostol[1969, Volume 2, Chap-ter 14] and integration because in a neighborhood of h = 0

h⁻¹ 1 −

This can obviously be done by finding

EN

Since Pm and finally, to get the expectation of quadratic variation we need to decompose the following expectation:

This problem is addressed in the following lemma using the seminal result of Balakrishnan [2006, Volume 2, Chapter 4].



where

q, from which it follows easily that

E(¯l_q^r) = 1 where we have used the fact that

m

Now

the accumulation of second-order (2×2) principal minors of S. Since the principal minors all give the same expectation, I need only to find the expectation involving the first one to find the bound

∆ = det

We can then multiply by

 upper-triangular matrix and by construction t²₁₁ are independent χ²_n0

−i+1 random

and

=

which completes the proof of the Theorem 9.

Let us now explain the proof of Theorem 8 :

Ec(e^−hTk) = θ(h)

Substitution gives

E0(T_k) = d

n − k − (2q²+ q + 2)/6q − αd

n² + O(n⁻³) (4.137)

from which it follows that if P_k is the statistic defined then

Ec(P_k) = d + O(n⁻²) (4.138)

and the proof is complete.

 It follows from Theorem 3 that if n is large an approximate test of size α of the null hypothesis

H_k : λ_k+1 = · · · = λ_m (4.139)

is to reject H_k if P_k > c(α; (q + 2)(q − 1)/2), where P_k is given, q = m − k and c(α; r) is the upper 100α% point of the χ²_r distribution. An estimate of λ is provided by

¯l_q = q⁻¹

m

X

i=k+1

l_i (4.140)

and it is easy to show, for example, that as n → ∞ the asymptotic distribution of ¹₂nq
1/2

(¯l_q− λ)/λ is standard normal N(0, 1). Let zα be the upper 100α%

point of the N(0, 1) distribution, that is, such that Φ(zα) = 1 − α, where Φ(.) denotes the standard normal distribution function. Then asymptotically,

P

nq 2

1/2¯l_q− λ λ



≥ −z_α



= 1 − α (4.141)

which appoints to a one-sided confidence interval for λ, namely,

λ ≤

¯l_q

1 − z_α(2/nq)^1/2 (4.142)

with asymptotic confidence coefficient 1 − α. If the upper limit of this confidence interval is sufficiently small we might decide that λ is negligible and study only the first k principal components.

Even if we cannot conclude that some of the smallest latent roots of Σ are equal, it still may be possible that the variation explained by the last q = m − k principal components, namely Pm

i=k+1λ_i, is small compared with the total variationPm

i=1λ_i, in which case we might decide to study only the first k principal components. Thus it is of interest to consider the null hypothesis

H_k^∗ : Pm

i=k+1λ_i Pm

i=1λ_i = h (4.143)

where h(0 < h < 1) is a number to be specified by the experimenter. This can be tested using the statistic

M_k ≡

Assuming the latent roots of Σ are distinct, Corollary 2 shows that the limiting distribution as n → ∞ of

is normal with mean 0 and variance approximate test of H_k^∗ and to give confidence intervals for

m

Finally, let me derive an asymptotic test for a given principal component. Let H^∗∗ be the null hypothesis that the vector of coefficients h₁ of the first principal components is equal to an specified m × 1 vector h⁰₁,i.e.,

H^∗∗ : h₁ = h⁰₁, h⁰₁⁰h⁰₁ = 1 (4.148)

Recall that h₁ is the eigenvector of Σ corresponding to the largest latent root λ₁; we will assume that λ₁ is a distinct root. A test of H^∗∗ can be constructed using the result of Theorem 7, namely, that if q₁ is the normalized eigenvector of the sample covariance matrix S corresponding to the largest latent root l₁ of S then the asymptotic distribution of y = n^1/2(q₁− h₁) is N_m(0, Γ), where

with H₂ = [h₂. . . h_m] and

Note that the covariance matrix Γ in this asymptotic distribution is singular, as is to expected. Put z = B⁻¹H₂⁰y; then the limiting distribution of z is Nm−1(0, I_m−1), and hence the limiting distribution of z⁰z is χ²_m−1. Now note that

z⁰z = y⁰H₂B⁻²H₂⁰y (4.151)

and the matrix of this quadratic form in y is

H₂B⁻²H₂⁰

Putting ∧ = diag(λ₁, . . . , λ_m) and using

Hence the limiting distribution of

n(q1− h1)⁰ can be substituted for Σ, Σ⁻¹, and λ₁ without affecting the limiting distribution.

Hence, when H^∗∗ : h₁ = h⁰₁ is true, the limiting distribution of

W

= n(q₁− h⁰₁)⁰



l₁S⁻¹− 2I + 1 l₁S



(q₁− h⁰₁)

= n



l₁h⁰₁⁰S⁻¹h⁰₁+ 1 l1

h⁰₁⁰Sh⁰₁− 2



is χ²_m−1. It follows that a test of H^∗∗ of asymptotic size α is to reject H^∗∗ if W > c(α; m − 1), where c(α; m − 1) is the upper 100α% point of the χ²_m−1 distribution.