Robust Bootstrap for Location and Scale

Assume the location model (2) with unknown scale, and let T_n and S_n be defined by (4) and (5). Under mild assumptions (see Maronna and Yohai, 1981) we have

$$\sqrt{n} \left( \left( \begin{array}{c} T_n \\ S_n \end{... ...) \right) \longrightarrow N_2 \left( {\bf0}, \Sigma \right),$$ (32)

where N₂ denotes a normal distribution in ${\mbox{I}\!\mbox{R}}^2$, and the asymptotic variance-covariance matrix is

$$\Sigma \ \ = \ \ D^{-1} \ C \ \left( D^{-1} \right)',$$ (33)

where

$$C = \left( \begin{array}{cc} E_F \psi^2 \left( Y \right) ... ...t] & E_F \chi^2 \left( Y \right) - b^2 \end{array} \right),$$ (34)

$$D = -\frac{1}{S_\infty} \ \left( \begin{array}{cc} E_F \p... ...\left[ \chi' \left( Y \right) Y \right] \end{array} \right),$$ (35)

and $Y = ( X - T_\infty) / S_\infty$. Let the weights and be defined as

$$\begin{eqnarray*}\omega_i \left( T_n, S_n \right) & = & \left. \frac{ \psi \lef... ...( \frac{ x_i - T_n }{S_n} \right) \hspace{.5in} 1 \le i \le n. \end{eqnarray*}$$

Similarly define and for . We have

$$\left( \begin{array}{c} T_n \\ S_n \end{array} \right) = \... ...n, S_n \right) \left( x_i - T_n \right) \end{array} \right).$$

To simplify the notation call

$$\theta_n \equiv \left( \begin{array}{c} \theta_{n1} \\ \thet... ...ft( \begin{array}{c} T_\infty\\ S_\infty\end{array} \right),$$

and

$$\hat{\theta}_n \equiv \left( \begin{array}{c} \hat{\thet... ...\right) \left( x_i - T_\infty\right) \end{array} \right).$$ (36)

The motivation for theorem 5 below is as follows. We have the identity

$$\theta_n = \left( \begin{array}{c} \theta_{n1} \\ \theta_... ...n, S_n \right) \left( x_i - T_n \right) \end{array} \right).$$ (37)

Rearranging the terms after a Taylor expansion around we get

$$\sqrt{n} \left( \theta_n - \theta_\infty \right) = \left[... ...1} \ \sqrt{n} \left( \hat{\theta}_n - \theta_\infty \right)$$ (38)

where

$$A_n \left( \tilde{\theta}_n \right)= \left( \begin{array}{cc}... ...ial S_n \left( \tilde{\theta}_n \right) \end{array} \right),$$

and $\tilde{\theta}_n$ is a point between $\theta_n$ and $\theta_\infty$. Equation (37) motivates our bootstrap, and (38) suggests the correction needed to account for the use of fixed weights. As before, we can complete our argument by proving that the sequence of matrices $\left[ I - A_n \left( \tilde{\theta}_n \right) \right]$ converges almost surely to a non-singular matrix. The necessary regularity conditions for this proof are very restrictive. The next theorem is the location-scale extension of theorem 4.

Theorem 5   Assume that (32) and (33) to (35) hold, and that $0 < \left\vert E_F \left( \psi \left( Y \right) / Y \right) \right\vert < +\infty$, with $Y = \left( X - T_\infty\right) / S_\infty$.

1.

Let $\hat{\theta}_n$ be defined by (36). Then

$$\sqrt{n} \left( \hat{\theta}_n - \theta_\infty \right) \longrightarrow N_2 \left( {\bf0}, \Sigma_2 \right),$$

where

$$\Sigma_2 = \left( \begin{array}{cc} \frac{ E_F \psi^2 \... ...F \chi^2 \left( Y \right) - b ^2 }{b^2} \end{array} \right),$$ (39)

and $Y = ( X - T_\infty) / S_\infty$.

2.

There exists a matrix $A \in {\mbox{I}\!\mbox{R}}^{2 \times 2}$ such that

$$\Sigma = A \ \Sigma_2 \ A^t$$ (40)

where $\Sigma$ is given by (33) to (35) and $\Sigma_2$ is given by (39).

3.

If the functions $\psi'\left( t \right)$, $\psi' \left( t \right)t$, $\chi' \left( t \right)$, $\chi' \left( t \right)t$ and $\psi \left( t \right) / t$ are uniformly continuous, then there exists a sequence of matrices $\left\{ A_n \right\}_{n \ge 1} = \left\{ A_n \left( \theta_n \right) \right\}_{n \ge 1}$ such that

$$A_n \longrightarrow A \mbox{ almost surely},$$

where A satisfies (40).

Proof of theorem 5:

1.

The following argument will show that there exists a function $g:{\mbox{I}\!\mbox{R}}^3 \rightarrow {\mbox{I}\!\mbox{R}}^2$, and n independent and identically distributed random vectors ${\bf y}_i \in {\mbox{I}\!\mbox{R}}^3, \ 1 \le i \le n$, with mean $\mu = E_F \left( {\bf y}_i \right)$, such that

$$\sqrt{n} \left( \hat{\theta}_n - \theta_\infty \right) = \sq... ...eft( \bar{{\bf y}} \right) - g \left( \mu \right) \right).$$

This representation will allow us to calculate the asymptotic variance-covariance matrix of $\sqrt{n} \left( \hat{\theta}_n - \theta_\infty \right)$. By Slutzky's Theorem it will be given by

$$\nabla g \left( \mu \right) \ \Sigma_{\bf y} \ \nabla g \left( \mu \right)'$$

where denotes the matrix of first derivatives of g at $\mu$, and $\Sigma_{\bf y}$ is the matrix of variance-covariance of the random vectors ${\bf y}_i$. Note that

$$\begin{eqnarray*}\hat{\theta}_{n1} & = & \sum_{i=1}^nx_i \ \frac{ \Psi \left( \... ...nfty} \right) / \left( X- T_\infty\right) \right) } + T_\infty \end{eqnarray*}$$

and

$$\begin{eqnarray*}\hat{\theta}_{n2} & = & \frac{1}{nb} \sum \chi \left( \frac{x... ...\frac{x-T_\infty}{ S_\infty} \right) \right) \ \right/ \ b \ . \end{eqnarray*}$$

For each $i=1, \ldots, n$ define the random vectors as

$${\bf y}_i = \left( \Psi \left( \frac{x_i-T_\infty}{ S_\infty... ... \chi \left( \frac{x_i -T_\infty}{S_\infty} \right) \right).$$

Their sample mean is

$$\bar{{\bf y}} = \left( E_{F_n} \Psi \left( \frac{X-T_\infty... ...F_n} \chi \left( \frac{X-T_\infty}{S_\infty} \right) \right).$$

Define by

$$g \left( x,\ y,\ z \right) \ = \ \left( x / y + T_\infty, \ \ S_\infty\ z / b \right),$$

and note that

$$\hat{\theta}_n = g \left( \bar{{\bf y}} \right).$$

Now let . We have $\mu_1 = 0$, $0 < \left\vert \mu_2 \right\vert < \infty$, and $\mu_3 = b$. This implies

$$g \left( \mu \right) = \left( \begin{array}{c} T_\infty\\ S_\infty\end{array} \right) = \theta_\infty.$$

We have then expressed $\hat{\theta}_n - \theta_\infty$ as a smooth function of means:

$$\sqrt{n} \left( \hat{\theta}_n - \theta_\infty \right) = \sq... ...eft( \bar{{\bf y}} \right) - g \left( \mu \right) \right).$$

It is easy to check that for any ${\bf x}= \left( x_1, x_2, x_3 \right) \in {\mbox{I}\!\mbox{R}}^3$ with $x_2 \ne 0$ we have

$$\nabla g \left( {\bf x}\right) = \left( \begin{array}{ccc} 1... ...x_1 / x_2^2 & 0 \\ 0 & 0 & S_\infty/ b \end{array} \right),$$

so that

$$\begin{eqnarray*}\nabla g \left( \mu \right) & = & \left( \begin{array}{ccc} 1... ...ight)^{-1}& 0 & 0 \\ 0 & 0 & S_\infty/ b \end{array} \right), \end{eqnarray*}$$

where $y = \left( X - T_\infty\right) / S_\infty$. From here, part 1 of the theorem follows.

2.

Consider the matrix A such that

$$A^{-1} = \left( \begin{array}{cc} \left. E_F \psi' \left( Y ... ... \left( Y \right) \ Y \right] \right/ b \end{array} \right),$$

where $Y = \left( X - T_\infty\right) / S_\infty$. Note that A is non-singular if and only if D in equation (35) is non-singular. It is easy to see that A satisfies (40).

3.

From Lemma A in page 253 of Serfling (1980) and the assumed regularity conditions we get the result with

$$A_n^{-1} = \left( \begin{array}{cc} \left. E_{F_n} \psi' \... ...} \right) \ \tilde{Y} \right] \right/ b \end{array} \right),$$

where $\tilde{Y} = \left( X - T_n \right) / S_n$. This finishes the proof of the theorem.

The questions that remain to be answered are:

• Can we bootstrap ?
• When we replace $\left( T_\infty, S_\infty\right)'$ by $\left( T_n, S_n \right)'$, is the bootstrap distribution close to the one we want to estimate? This question can be seen as a matter of ``continuity'' or qualitative robustness of the bootstrap (see Cuevas, et. al., 1993).

The representation

$$\sqrt{n} \left( \hat{\theta}_n - \theta_\infty \right) = \s... ...ft( \bar{{\bf y}} \right) - g \left( \mu \right) \right),$$

we get from the proof of Part 1 in Theorem 5 together with Theorem 3 positively answer the first question. We are still working on the second question. Empirical results obtained are very encouraging.