3.2 Some Convergence Results
We develop many results needed to assess the asymptotic distributions of \(u\)-statistics.
- (a)
- The sequence of random variables \(\{W_n\}^{\infty }_1\) is said to converge in probability to a constant \(c\), if for every \(\varepsilon > 0\) \[\lim _{n \rightarrow \infty } P\left \{\left |W_n - c\right | < \varepsilon \right \} = 1.\]
- (b)
- The sequence \(\{W_n\}^{\infty }_1\) of estimators of a parameter \(\gamma \) is said to be consistent if the sequence \(\{W_n\}^{\infty }_1\) converges in probability to \(\gamma \) for every value of \(\gamma \) in the parameter space.
Definition 3.2.2. The sequence \(\{W_n\}^{\infty }_1\) of random variables is said to converge in quadratic mean to a constant \(c\) if \[\lim _{n\rightarrow \infty } E\left [(W_n - c)^2\right ] = 0.\]
Proof. We use Chebyshev’s inequality \[P\left (\left |W_n - c\right | < \varepsilon \right ) \geq 1 - \frac {E(W_n - c)^2}{\varepsilon ^2}\] \begin {align*} \lim _{n\rightarrow \infty }P\left (\left |W_n - c\right | < \varepsilon \right ) & \geq 1 - \frac {1}{\varepsilon ^2}\, \lim _{n\rightarrow \infty } E(W_n - c)^2\\\\ & = 1 - \frac {0}{\varepsilon ^2} = 1.\\\\ \lim _{n\rightarrow \infty }P\left (\left |W_n - c\right | < \varepsilon \right ) & = 1. \end {align*}
\(\implies \hspace {0.3cm} W_n\) converges in probability to \(c\). □
Example 3.2.4. \(E(\overline {X}_n - \mu )^2 = \dfrac {\sigma ^2}{n}\)
\[\lim _{n\rightarrow \infty }E(\overline {X}_n - \mu )^2 = 0\]
\(\implies \hspace {0.3cm} \overline {X}_n\,\) converges in quadratic mean to \(\mu \).
\(\implies \hspace {0.3cm} \overline {X}_n\,\) converges in probability to \(\mu \).
Corollary 3.2.5. Let \(X_1,\, X_2, \, \cdots \, , \, X_n\) be a random sample from a population with cdf \(F(x)\) and let \(\gamma \) be an estimable parameter of degree \(K\). For \(n\geq k\), let \(U(X_1, \, X_2, \, \cdots \, , \, X_n)\) denote the \(U-\)statistics estimator for \(\gamma \). Then \(U(X_1, \, X_2, \, \cdots \, , \, X_n)\) converges in quadratic mean to \(\gamma \) provided \(E\left (h^2(X_1, \, X_2, \, \cdots \, , \, X_K)\right ) < \infty \).
Proof. \[E\left (U(X_1, \, X_2, \, \cdots \, , \, X_n) - \gamma \right )^2 = var\left (U(X_1, \, X_2, \, \cdots \, , \, X_n)\right )\] \begin {align*} \lim _{n\rightarrow \infty }E\left [U(X_1, \, X_2, \, \cdots \, , \, X_n) - \gamma \right ]^2 & = \lim _{n\rightarrow \infty }\frac {n\, var\left (U(X_1, \, X_2, \, \cdots \, , \, X_n)\right )}{n}\\\\ & = \frac {\lim _{n\rightarrow \infty } n var\left (U(X_1, \, X_2, \, \cdots \, , \, X_n)\right )}{\lim _{n\rightarrow \infty } n}\\\\ & = \frac {K^2\, \zeta _1}{\infty } = 0. \end {align*}
\(\therefore \, U\) converges in quadratic mean to \(\gamma \), which implies that \(U\,\) converges in probability to \(\gamma \). □
Theorem 3.2.6. If \(\{W_n\}^{\infty }_1\) is a sequence of random variables that converges in probability to a constant \(c\), and \(L(\cdot )\) is a function that is continuous at \(c\), then \(\{L(W_n)\}^{\infty }_1\) converges in probability to \(L(c)\).
Definition 3.2.7. A sequence of random variables \(\{W_n\}^{\infty }_1\) is said to have a limiting distribution \(F(w)\) (or to be asymptotically distributed) if \[\lim _{n\rightarrow \infty }P\left (W_n \leq w\right ) = F(w)\] for all \(w\) values a which the cdf \(F(w)\) is continuous.
Remark 3.2.8. \(\{W_n\}^{\infty }_{1}\) converges in probability to \(c\) if and only if \(\{W_n\}^{\infty }_1\) has a limiting distribution that is degenerate at \(c\), i.e it puts probability 1 at \(c\). \[\text {i.e}\hspace {0.3cm} F(w) = \begin {cases} 0 & \text {if}\hspace {0.2cm} w\neq c\\ 1 & \text {if}\hspace {0.2cm} w =c\\ \end {cases}\]
Theorem 3.2.9 (Slutsky’s Thoerem). Let \(\{W_n\}^{\infty }_1\) be a sequence of random variables with limiting distribution \(F(w)\). Let \(\{X_n\}^{\infty }_1\) denote a sequence of random variables that converges in probability to a constant \(C\). Then
- (i)
- \(\{W_n + X_n\}\) and \(\{W_n + C\}\) have the same limiting distribution.
- (ii)
- \(\{W_n\, X_n\}\) and \(\{C\, W_n\}\) have the same limiting distribution.
- (iii)
- \(\{W_n/X_n\}\) and \(\{W_n/C\}\) have the same limiting distribution provided \(C\neq 0\).
Example 3.2.11. Let \(X_1, \, X_2, \, \cdots \,, \, X_n\) be a random sample of size \(n\) from a distribution with mean \(\mu \) and \(\sigma ^2 > 0\) and finite fourth moment. Let \(S\) be the sample standard deviation then \(\dfrac {S}{\sigma }\overset {P}{\longrightarrow }1.\) \[\frac {\sqrt {n}\, (\overline {X}_n -\mu _0)}{S} = \frac {\sqrt {n}(\overline {X}_n - \mu _0)}{\sigma \times S/\sigma } = \frac {W_n}{X_n}\]
\[W_n = \frac {\sqrt {n}\, (\overline {X}_n - \mu _0)}{\sigma }\hspace {0.4cm}, \hspace {0.4cm} X_n = \frac {S}{\sigma }\]
\[W_n \overset {L}{\longrightarrow } N(0,1)\hspace {0.4cm}\text {and}\hspace {0.4cm} X_n\overset {P}{\longrightarrow }1.\]
Theorem 3.2.12. If the sequence of random variables \(\{V_n\}^{\infty }_1\) has asymptotic distribution with cdf \(F(v)\) and if \(\{W_n\}^{\infty }_1\) is a sequence of random variables such that \(\{W_n - V_n\}^{\infty }_1\) converges in probability to zero, then the limiting distribution of \(\{W_n\}^{\infty }_1\) has cdf \(F(w)\).
Proof. We note that \(W_n = V_n + W_n - V_n\) by Slutsky’s theorem part(i) with \(c = 0\) for \(W_n - V_n\), then \(V_n + W_n - V_n\) have the same limiting distribution as \(V_n\). □
Remark 3.2.13. The condition of theorem 3.2.9 are often verified by showing that \(W_n - V_n\) converges in quadratic mean to zero. Since convergence in quadratic mean implies convergence in probability.
Example 3.2.14. Let \(X_1, \, X_2, \, \cdots \, , \, X_n\) denote a random sample from a population with mean \(\mu \), variance \(\sigma ^2>0\), and \(\tau ^2 = E\left [(X_i - \mu )^4\right ] - \sigma ^4\) satisfying \(0 < \tau ^2 < \infty \). We want the limiting distribution of \[W_n = \sqrt {n}\, \left \{\frac {1}{n}\sum \left (X_i - \overline {X}\right )^2 - \sigma ^2\right \}\] \[V_n = \sqrt {n}\, \left \{\frac {1}{n}\sum ^n_{i = 1} (X_i - \mu )^2 - \sigma ^2\right \}\] \begin {align*} W_n - V_n & = \sqrt {n} \left (\frac {1}{n}\sum (X_i - \mu )^2 - (\overline {X} - \mu )^2 - \frac {1}{n}\sum (X_i - \mu )^2\right )\\ & = -\sqrt {n}\, (\overline {X} - \mu )^2\\ & = - \left (n^{1/4}(\overline {X}- \mu )\right )^2 \end {align*}
Let \begin {align*} E\left (n^{1/4}(\overline {X} - \mu )\right )^2 & = n^{1/2}\, var(\overline {X}) = n^{1/2}\cdot \frac {\sigma ^2}{n} = \frac {\sigma ^2}{\sqrt {n}} \end {align*}
\[\lim _{n\rightarrow \infty }E\left (n^{1/4}(\overline {X} - \mu )\right )^2 = \lim _{n\rightarrow \infty }\frac {\sigma ^2}{\sqrt {n}} = 0.\]
\(n^{1/4}(\overline {X} - \mu )\, \) converges in quadratic mean to zero which implies \(n^{1/4}(\overline {X} - \mu )\,\) converges in probability to zero.
\(\implies \, W_n - V_n\) converges in probability to zero.
\(\implies \, W_n\) and \(V_n\) have the same limiting distribution.
\[V_n = \sqrt {n}\, \left (\frac {1}{n}\sum ^n_{i = 1} (X_i - \mu )^2 - \sigma ^2\right )\]
Let \(Y_i = (X_i - \mu )^2\) then \(E(Y_i) = \sigma ^2\) and \(var(Y_i) = E(X_i - \mu )^4 - \sigma ^4 = \tau ^2\)
\[\implies \, V_n = \sqrt {n}\, \left (\frac {1}{n}\sum Y_i - \sigma ^2\right ) = \sqrt {n}\, \left (\overline {Y}_n - \sigma ^2\right )\]
\(E(\overline {Y}_n - \sigma ^2) = 0\hspace {0.3cm}, \hspace {0.3cm} var(\overline {Y}_n) = \dfrac {\tau ^2}{n}\,\) by central limit theorem
\[\therefore \hspace {0.3cm} V_n \thicksim N\left (0,\frac {\tau ^2}{n}\right ).\]
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