## Example Problems on Asymptotics

ST793 Homework 5 on Asymptotics

5.32, 5.33, 5.37, 5.51, 5.52

# 5.32

Suppose that $\widehat \theta_1 , \dots \widehat \theta_k$ each satisfy the assumptions of Theorem 5.23 (p. 242):

$\widehat \theta_i - \theta_i = \frac{ 1 }{ n } \sum_{j=1}^n h_i(X_j) + R_{in},$

where $\sqrt{ n } R_{in} \stackrel{ \text{p}}{\rightarrow} 0$ as $n \rightarrow \infty$ and $E(h_i(X_1))=0$ and $\text{Var}( h_i(X_1) ) = \sigma^2_{hi}<\infty$. Let $T = \sum_{i=1}^k c_i \widehat \theta_i$ for any set of constants $c_1 , \dots , c_k$. Find the correct approximating function $h_T$ for $T$, show that Theorem 5.23 may be used (verify directly without using later theorems), and find the limiting distribution of $T$.

Let’s rewrite $T$.

\begin{align} T & = \sum_{i=1}^k c_i \widehat \theta_i \\ & = \sum_{i=1}^k c_i(\theta_i + \frac{ 1 }{ n } \sum_{i=1}^n h_i(X_j) + R_{in}) \\ & = \sum_{i=1}^k c_i \theta_i + \sum_{i=1}^k c_i \frac{ 1 }{ n } \sum_{i=1}^n h_i(X_j) + \sum_{i=1}^k c_i R_{in} \\ \end{align}

Now we will show that this is in Bahadur Representation. First take $T_\infty = \sum_{i=1}^k c_i \theta_i$. Then,

\begin{align} \sum_{i=1}^k c_i R_{in} & = c_1 R_{1n} + \dots c_k R_{kn} \\ & = op(n^{-1/2}) + \dots + op(n^{-1/2}) & \text{CMT} \\ & = op(n^{-1/2}) & \text{page 234 property 5} \end{align}

Thus, the last term is asymptotically negligible. Further take $h_T =\sum_{i=1}^n c_i h_i(X_j)$.

\begin{align} E(\sum_{i=1}^k c_i h_i(X_j)) & = \sum_{i=1}^k c_i E(h_i(X_j)) \\ & = 0 \\ \text{Var}( \sum_{i=1}^k c_i h_i(X_j) )& =\sum_{i=1}^k c_i^2 \text{Var}( h_i(X_j)) \\ \sum_{i=1}^k c_i^2 \sigma^2_{ih} < \infty \end{align}

Thus,

$T \sim N\Big( \sum_{i=1}^k c_i \theta_i , \frac{ 1 }{ n }\sum_{i=1}^k c_i^2 \sigma_{hi}^2\Big)$

# 5.33

Let $(X_1, Y_1) , \dots , (X_n, Y_n)$ be iid pairs with $E(X_1) = \mu_1$, $E(Y_1) = \mu_2$, $\text{Var}( X_1 ) = \sigma^2_1$, $\text{Var}( Y_1 ) = \sigma^2_2$ and $\text{Cov}( X_1, Y_1 ) = \sigma_{12}$.

## a

What can we say about the asymptotic distribution of $(\overline{ X }, \overline{ Y })^T$?

Take

$\mu = \begin{bmatrix} \mu_1 \\ \mu_2 \end{bmatrix}. \Sigma = \begin{bmatrix} \sigma_1^2 & \sigma_{12} \\ \sigma_{12} & \sigma_2^2. \end{bmatrix}.$

We have appropriate conditions to apply the multivariate central limit theorem.

\begin{align} \sqrt{ n } \Big( \begin{bmatrix} \overline{ X } \\ \overline{ Y } \end{bmatrix} - \begin{bmatrix} \mu_1 \\ \mu_2 \end{bmatrix}\Big) & \stackrel{ \text{d}}{\rightarrow} N_2(0 \Sigma) \\ \begin{bmatrix} \overline{ X } \\ \overline{ Y } \end{bmatrix} & \sim AN_2(\mu, \Sigma) \end{align}

## b

Suppose that $\mu_1 = \mu_2 = 0$ and let $T=(\overline{ X })(\overline{ Y })$. Show that

\begin{align} nT & \stackrel{ \text{d}}{\rightarrow} Q & \text{as } n \rightarrow \infty, \end{align}

describe the random variable $Q$.

By the WLLN we know

$\overline{ X } \rightarrow \mu_1 = 0 , \ \overline{ Y } \rightarrow \mu_2 = 0.$

Take $g(\overline{ X }, \overline{ Y }) = n \overline{ X } \overline{ Y }$ Then by the CMT,

$g(\overline{ X }, \overline{ Y }) \stackrel{ \text{p}}{\rightarrow} n \mu_1 \mu_2 = 0.$

Since convergence in probability implies convergence in distribution, $n T \stackrel{ \text{d}}{\rightarrow} Q = 0$

## c

Suppose that $\mu_1 =0, \mu_2 \neq 0$ and let $T=(\overline{ X })(\overline{ Y })$. Show that

\begin{align} \sqrt{ n }T & \stackrel{ \text{d}}{\rightarrow} R & \text{as } n \rightarrow \infty, \end{align}

describe the random variable $R$.

Choose $b_n = \frac{ 1 }{ \sqrt{ n } }$. Then we can apply the Delta Method (Theorem 5.19). Take

\begin{align} g\Big( \begin{bmatrix} \overline{ X } \\ \overline{ Y } \end{bmatrix} \Big) & = \sqrt{ n } \overline{ X } \overline{ Y }\\ g'\Big( \begin{bmatrix} \overline{ X } \\ \overline{ Y } \end{bmatrix} \Big) & = \sqrt{ n } \begin{bmatrix} \overline{ Y } \\ \overline{ X } \end{bmatrix}. \end{align}

Thus,

$g\Big( \begin{bmatrix} \overline{ X } \\ \overline{ Y } \end{bmatrix} \Big) \sim AN\Bigg(0 \mu_2,\frac{ 1 }{ n } \begin{bmatrix} \sqrt{ n } \mu_2 & 0 \end{bmatrix}) \Sigma \begin{bmatrix} \sqrt{ n } \mu_2 \\ 0 \end{bmatrix}\Bigg) = AN(0, \mu_2^2 \sigma_1^2).$

# 5.37

Find the approximating $h_T$ function for the sample skewness $T = m_3 / (m_2)^{3/2}$

Take $m_2’$ and $m_3’$ be the sample noncentral moments. Then,

\begin{align} T & = \frac{ \frac{ 1 }{ 2 } \sum_{i=1}^n (X_i - \overline{ X })^3}{ \Big( \frac{ 1 }{ n } \sum_{i=1}^n (X_i - \overline{ X })^2\Big)^{3/2} } \\ & = \frac{ m_3' - 3 m_2' \overline{ X } + 2 \overline{ X }^3 }{ \Big( \frac{ 1 }{ n } \sum_{i=1}^n X_i^2 - 2X_i \overline{ X } + \overline{ X }^2 \Big)^{3/2} } \\ & = \frac{ m_3' - 3 m_2' \overline{ X } + 2 \overline{ X }^3 }{ \Big( m_2' - \overline{ X }^2 \Big)^{3/2} } \\ & = g(\overline{ X }, m_2', m_3'). \end{align}

Now we will take the derivatives of $g$

\begin{align} \frac{ \partial g(y_1, y_2, y_3) }{\partial y_1} & = \frac{ 3 y_1 y_3 - 3 y_2^2}{ (y_2 - y_1^2)^{5/2} } \\ \frac{ \partial g(y_1, y_2, y_3) }{\partial y_2} & = \frac{ 3(y_1 y_2 - y_3) }{ 2 (y_2 - y_1^2)^{5/2} } \\ \frac{ \partial g(y_1, y_2, y_3) }{\partial y_3} & = \frac{ 1 }{ (y_2 - y_1^2)^{3/2} }. \end{align}

Now take $\mu_1$, $\mu_2’$, and $\mu_3’$ to be the population noncentral moments.

$h_T = \frac{ 3 \mu_1 \mu'_3 - 3 \mu_2 ' ^2}{ (\mu'_2 - \mu_1^2)^{5/2} } (X_i - \mu_1) + \frac{ 3(\mu_1 \mu'_2 - \mu'_3) }{ 2 (\mu'_2 - \mu_1^2)^{5/2} } (X_i^2 - \mu'_2) + \frac{ 1 }{ (\mu'_2 - \mu_1^2)^{3/2} } (X_i^3 - \mu_3')$

# 5.51

Suppose that $B, B_1, B_2, \dots$ are iid random variables with $P(B = -1) = P(B = 1) = 1/2$. Define $X_i = 2^{-i/2}B_i$ and $S_n = X_1 + \dots + X_n$. Prove that the standardized sum $S_n / \sqrt{ \text{Var}( S_n ) }$ does not converge in distribution to a standard normal. In doing so show that $\text{Var}( X_1 )$ contributes to one-half the of the variability of $S_n$ asymptotically. Hint: Show that $S_{n-1}$ and $\sqrt{ 2 }(X_2 + \dots + X_n)$ are identically distributed.

We will start by showing the hint.

\begin{align} P(\sqrt{ 2 } (X_2 + \dots + X_n) \leq y) & =P ( \sqrt{ 2 }(\frac{ 1 }{ 2 }B_2 + \dots + (\frac{ 1 }{ 2 })^{n/2} B_n )\leq y ) \\ & = P ( (\frac{ 1 }{ 2 })^{1/2} B_2 + \dots + (\frac{ 1 }{ 2 })^{(n-1)/2} B_n \leq y ) \\ & = P ( (\frac{ 1 }{ 2 })^{1/2} B_1 + \dots + (\frac{ 1 }{ 2 })^{(n-1)/2} B_{n-1} \leq y ) & B_i \text{ iid} \\ & = P(X_1 + \dots + X_{n-1} \leq y) \\ & = P(S_{n-1} \leq y) \end{align}

Notice that

$E(X_i) = 0 \ \text{Var}( X_i ) = 2^{-i}.$

Now $E(S_n) = 0$ and we can find the variance of $S_n$ using a finite geometric sum.

$\text{Var}( S_n ) = \sum_{i=1}^n 2^{-i} = \frac{ 1-(1/2)^{n+1} }{ 1-1/2 } - 1 = 1-2^{-n} = \nu_n^2$

Notice that

$\lim_{n \rightarrow \infty} \frac{ \text{Var}( X_1 ) }{ \text{Var}( S_n ) } = \frac{ 1/2 }{ 1- 2^{-n} } = \frac{ 1/2 }{ 1 } = \frac{ 1 }{ 2 }.$

This shows that asymptotically $X_1$ contributes half of the variance of $S_n$.

Take $S_{2:n} = S_n - X_1$ and $Y_n = S_n / \nu_n$ is the standardized sum.

We will proceed by contradiction. Assume that $Y_n$ does converge in distribution to $N(0,1)$. Then $Y_{n-1}$ must also converge in distribution of $N(0,1)$.

\begin{align} Y_n & = \frac{ S_n }{ \nu_n } \\ & = \frac{ X_1 }{ \nu_n } + \frac{ S_{2:n} }{ \nu_n } \\ & = \frac{ X_1 }{ \nu_n } + \frac{ \frac{ 1 }{ \sqrt{ 2 } } S_{n-1} }{ \nu_n } & \text{by the hint} \\ & = \frac{ \frac{ 1 }{ \sqrt{ 2 } } B_1}{ \nu_n } + \frac{ 1 }{ \sqrt{ 2 } } \frac{ \nu_{n-1} }{ \nu_n } Y_{n-1} \\ & = \frac{ 1 }{ \sqrt{ 2 } } B + \frac{ 1 }{ \sqrt{ 2 } } \frac{ \nu_{n-1} }{ \nu_n } Y_{n-1} \end{align}

Notice that $\lim_{n\rightarrow \infty} \frac{ \nu_{n-1} }{ \nu_n } = 1$. Thus,

$Y_n \stackrel{ \text{d}}{\rightarrow} \frac{ 1 }{ \sqrt{ 2 } } B + \frac{ 1 }{ \sqrt{ 2 } } Z$

where $Z$ and $B$ are independent and $Z \sim N(0,1)$. However, the sum of a normal random variable $Z / \sqrt{ 2 }$ and a discrete random variable $B / \sqrt{ 2 }$ does not have a normal distribution. This contradicts the assumption that $Y_n \stackrel{ \text{d}}{\rightarrow}N(0,1)$.

# 5.52

Consider a Gauss-Markov linear model

$Y = X \overline{ \beta } + \epsilon$

where $Y$ and $n \times 1$, the components of $e = (e_1, \dots e_n)^T$ are iid$(0, \sigma^2)$, and $X$ is $n \times p_{n}$. Note that the number of predictors $p_n$ depends on $n$. Let $H = X(X^TX)^{-1} X^T$ denote the projection (or “hat”) matrix with entries $h_{i,j}$. Note that the $h_{i,j}$ also depend on $n$. We are interested in the asymptotic properties $n \rightarrow \infty$ or the $i$th residual $Y_i - \widehat Y_i$, from this regression model, for a fixed $i$. Prove that if $n$ and $p_n \rightarrow \infty$ such that

\begin{align} h_{i,i} \rightarrow c_i & \text{ for some } 0 \leq c_i < 1 & \text{ and } & \max_{1 \leq j \leq n ,\ j\neq 1} | h_{i,j} \rightarrow 0, \end{align}

then

$Y_i - \widehat Y_i \stackrel{ \text{d}}{\rightarrow} (1-c_i)e_i + [ (1-c_i)c_i ]^{1/2} \sigma Z,$

where $Z$ is a standard normal random variable independent of $e_i$. Hint: First prove that $\text{Var}( \sum_{j = 1, j \neq i}^n h_{i,j} e_j ) = (h_{i,i} - h_{i,i}^2) \sigma^2$. Then formulate and verify the appropriate Lindeberg condition. (This result illustrates that least square residuals tend to have a distribution closer to a normal distribution than that of the original error distribution is not normal. Thus, tests for nonnormality may have little power when many predictors are used.)

First, let’s expand $Y_i - \widehat Y_i$.

\begin{align} Y_i - \widehat Y_i & = Y_i - (HY)_i \\ & = Y_i - \begin{bmatrix} h_{11} & \dots & h_{1n} \\ \vdots \\ h_{n1} & \dots & h_{nn} \end{bmatrix} \begin{bmatrix} Y_{1} \\ \vdots \\ Y_n \end{bmatrix}_i \\ & = Y_i - \sum_{j=1}^n h_{ij} Y_i \\ & = Y_i - h_{ii} Y_i - \sum_{j\neq i} h_{ij} Y_j \\ & = (1- h_{ii}) Y_i - \sum_{j\neq i} h_{ij} Y_j \\ & = (1- h_{ii}) (X_i\beta + e_i) - \sum_{j\neq i} h_{ij} (X_j\beta + e_j) \\ & = (1-h_{ii}) X_i \beta + (1-h_{ii}) e_i -\sum_{j \neq i} h_{ij} X_j \beta - \sum_{j \neq i} h_{ij} e_j \end{align}

Notice

\begin{align} \lim_{n\rightarrow \infty} (1-h_{ii}) e_i & = (1-c_i) e_i. \end{align}

Now let’s group two terms together,

\begin{align} (1- h_{ii}) X_i \beta - \sum_{j \neq i} h_{ij} X_j \beta & = X_i \beta - h_{ii} X_i \beta - \sum_{j \neq i} h_{ij} X_j \beta \\ & = X_i \beta - \sum_{j=1}^n h_{ij} X_j \beta \\ & = X_i \beta - \Big( \sum_{j=1}^n h_{ij} X_j \Big) \beta \\ & = X_i \beta - \Big( H_i X\Big) \beta \\ & = X_i \beta - \Big( H X\Big)_i \beta \\ & = X_i \beta - X_i \beta \\ & = 0 \end{align}

Recall that $H$ is idempotent, or $H = H^2$.

\begin{align} H & = \begin{bmatrix} h_{11} & \dots & h_{1n} \\ \vdots \\ h_{n1} & \dots & h_{nn} \end{bmatrix} \\ & = \begin{bmatrix} h_{11} & \dots & h_{1n} \\ \vdots \\ h_{n1} & \dots & h_{nn} \end{bmatrix} \begin{bmatrix} h_{11} & \dots & h_{1n} \\ \vdots \\ h_{n1} & \dots & h_{nn} \end{bmatrix} \\ & = \begin{bmatrix} \sum_{j=1}^n h_{1j}^2 & \sum_{j=1}^n h_{1j} h_{2j} & \dots & \sum_{j=1}^n h_{1j} h_{nj} \\ \vdots \\ \sum_{j=1}^n h_{nj} h_{1j} & \dots & & \sum_{j=1}^n h_{nj}^2 \end{bmatrix} \\ & = H^2 \end{align}

Thus,

$\sum_{j=1}^n h_{ij}^2 = h_{ii} \Rightarrow \sum_{j\neq i}^n h_{ij}^2 + h_{ii}^2 = h_{ii} \Rightarrow \sum_{j\neq i}^n h_{ij}^2 = h_{ii} - h_{ii}^2.$

This gives us that

$Var(\sum_{j \neq i} h_{ij} e_j) = (h_{ii} - h_{ii}^2) \sigma^2.$

Since $e_i$ has mean 0 and they are iid (and we no longer have any cross terms of $H$), by the CLT

$\sum_{j \neq i} h_{ij} e_{j} \stackrel{ \text{d}}{\rightarrow} \sqrt{ (1-c_i)c_i } \sigma Z$

where $Z\sim N(0,1)$.

Thus,

$Y_i - \widehat Y_i \stackrel{ \text{d}}{\rightarrow} (1-c_i)e_i + [ (1-c_i)c_i ]^{1/2} \sigma Z.$