Let X1, X2, ..., Xn be a random sample from an unknown distribution with absolutely continuous cumulative distribution function (cdf) F. Let F0 be a specified absolutely continuous cdf. For testing H0 : F(x) = F0(x) for all x against H1 : F(x) ≠ F0(x) for some x, consider the following two test statistics: \(\displaystyle T_{1, n}=\sup _{x \in \mathbb{R}}\left|\frac{1}{n} \sum_{i=1}^n I_{\left\{X_i \leq x\right\}}-F_0(x)\right| \), and \(\displaystyle T_{2, n}=\sup _{x \in \mathbb{R}} n\left|\frac{1}{n} \sum_{i=1}^n I_{\left\{X_i \leq x\right\}}-F_0(x)\right|\), where \(I_{\left\{X_i \leq x\right\}}=\left\{\begin{array}{ll}1, & \text { if } X_i \leq x \\ 0, & \text { if } X_i>x\end{array}\right.\) for i = 1, 2, ..., n. Then which of the following statements are true?
The question asks about the convergence properties of two test statistics, \(T_{1,n}\) and \(T_{2,n}\), used for testing the hypothesis \(H_0: F(x) = F_0(x)\) against \(H_1: F(x) \ne F_0(x)\), where \(F\) is the unknown distribution function and \(F_0\) is a specified distribution function.
Test Statistics Defined
The test statistics are given by:
\(\displaystyle T_{1, n}=\sup _{x \in \mathbb{R}}\left|\frac{1}{n} \sum_{i=1}^n I_{\left\{X_i \leq x\right\}}-F_0(x)\right|\)
\(\displaystyle T_{2, n}=\sup _{x \in \mathbb{R}} n\left|\frac{1}{n} \sum_{i=1}^n I_{\left\{X_i \leq x\right\}}-F_0(x)\right|\)
where \(I_{\left\{X_i \leq x\right\}}\) is an indicator function:
\(I_{\left\{X_i \leq x\right\}}=\left\{\begin{array}{ll}1, & \text { if } X_i \leq x \\ 0, & \text { if } X_i>x\end{array}\right\}\)
The term \(\frac{1}{n} \sum_{i=1}^n I_{\left\{X_i \leq x\right\}}\) is the empirical distribution function, denoted as \(F_n(x)\). So the statistics can be written as:
\(T_{1,n} = \sup_{x \in \mathbb{R}} |F_n(x) - F_0(x)|\)
\(T_{2,n} = \sup_{x \in \mathbb{R}} n|F_n(x) - F_0(x)| = n T_{1,n}\)
Convergence of T1n under H0
Consider the first test statistic, \(T_{1,n}\), under the null hypothesis \(H_0: F(x) = F_0(x)\). Under \(H_0\), the true distribution of the random sample \(X_1, \ldots, X_n\) is \(F_0\). The statistic \(T_{1,n}\) measures the maximum vertical distance between the empirical distribution function \(F_n(x)\) and the hypothesized distribution function \(F_0(x)\).
The Glivenko-Cantelli theorem states that for any distribution function \(F\), the empirical distribution function \(F_n(x)\) converges uniformly to \(F(x)\) almost surely. That is,
\(\displaystyle \sup_{x \in \mathbb{R}} |F_n(x) - F(x)| \stackrel{a.s.}{\rightarrow} 0 \quad \text{as } n \rightarrow \infty\)
Under \(H_0\), \(F(x) = F_0(x)\). Applying the Glivenko-Cantelli theorem under \(H_0\), we have:
\(\displaystyle \sup_{x \in \mathbb{R}} |F_n(x) - F_0(x)| \stackrel{a.s.}{\rightarrow} 0 \quad \text{as } n \rightarrow \infty\)
This means \(T_{1,n} \stackrel{a.s.}{\rightarrow} 0\). Almost sure convergence implies convergence in probability. Therefore,
\(\displaystyle T_{1, n} \stackrel{P}{\rightarrow} 0 \quad \text{as } n \rightarrow \infty \text{ under } H_0\)
Thus, the statement that \(T_{1, n} \stackrel{P}{\rightarrow} 0\) as \(n \rightarrow \infty\) under \(H_0\) is true.
Convergence of T2n under H0
Now consider the second test statistic, \(T_{2,n} = n T_{1,n}\), under \(H_0\). We know that \(T_{1,n} \stackrel{P}{\rightarrow} 0\). However, multiplying by \(n\) can change the limit behavior.
The Kolmogorov-Smirnov theorem is relevant here. Under \(H_0: F(x) = F_0(x)\), the theorem states that \(\sqrt{n} \sup_{x \in \mathbb{R}} |F_n(x) - F_0(x)|\) converges in distribution to a non-degenerate distribution (the Kolmogorov-Smirnov distribution).
Let \(D_n = \sup_{x \in \mathbb{R}} |F_n(x) - F_0(x)| = T_{1,n}\). The theorem says \(\sqrt{n} D_n \stackrel{D}{\rightarrow} D\), where \(D\) is a random variable with the KS distribution, which is not concentrated at 0.
The statistic \(T_{2,n} = n D_n = \sqrt{n} (\sqrt{n} D_n)\). Since \(\sqrt{n} D_n\) converges in distribution to a non-degenerate variable, multiplying by \(\sqrt{n}\) (which goes to infinity) makes the product diverge in probability.
Specifically, for any constant \(M > 0\), \(P(T_{2,n} > M) = P(\sqrt{n} (\sqrt{n} D_n) > M)\). Let \(Y_n = \sqrt{n} D_n\). \(Y_n \stackrel{D}{\rightarrow} D\). So, \(P(\sqrt{n} Y_n > M) = P(Y_n > M/\sqrt{n})\). As \(n \rightarrow \infty\), \(M/\sqrt{n} \rightarrow 0\). Since \(Y_n \stackrel{D}{\rightarrow} D\) and \(P(D > 0) = 1\) (the KS distribution is positive), we have \(P(Y_n > M/\sqrt{n}) \rightarrow P(D > 0) = 1\).
Thus, \(P(T_{2,n} > M) \rightarrow 1\) for any \(M > 0\) under \(H_0\). This means \(T_{2,n} \stackrel{P}{\rightarrow} \infty\) under \(H_0\).
Therefore, the statement \(T_{2, n} \stackrel{P}{\rightarrow} 0\) as \(n \rightarrow \infty\) under \(H_0\) is false.
Probability P_F(T2n > 1) as n → ∞ for all F
Consider the limit of \(P_F(T_{2, n}>1)\) as \(n \rightarrow \infty\) for all possible distributions \(F\).
Case 1: Under \(H_0\) ($F = F_0$). As shown above, \(T_{2,n} \stackrel{P}{\rightarrow} \infty\) under \(H_0\). This implies that for any constant \(c\), \(P_{F_0}(T_{2,n} > c) \rightarrow 1\) as \(n \rightarrow \infty\). Taking \(c=1\), we get \(P_{F_0}(T_{2,n} > 1) \rightarrow 1\).
Case 2: Under \(H_1\) ($F \ne F_0$). Since \(F \ne F_0\) and both are absolutely continuous cdfs, there exists at least one point \(x\) where \(F(x) \ne F_0(x)\). This means the maximum difference is strictly positive: \(\delta = \sup_{x \in \mathbb{R}} |F(x) - F_0(x)| > 0\).
We know that \(F_n(x) \stackrel{a.s.}{\rightarrow} F(x)\) for each fixed \(x\) by the Strong Law of Large Numbers. By the Glivenko-Cantelli theorem, \(\sup_{x \in \mathbb{R}} |F_n(x) - F(x)| \stackrel{a.s.}{\rightarrow} 0\).
Consider \(T_{1,n} = \sup_{x \in \mathbb{R}} |F_n(x) - F_0(x)| = \sup_{x \in \mathbb{R}} |(F_n(x) - F(x)) + (F(x) - F_0(x))|\). Let \(A_n = \sup_{x \in \mathbb{R}} |F_n(x) - F(x)|\) and \(\delta = \sup_{x \in \mathbb{R}} |F(x) - F_0(x)|\). We have \(A_n \stackrel{a.s.}{\rightarrow} 0\).
Using the inequality \(|\sup |a_x+b_x| - \sup |b_x|| \le \sup |a_x|\), we get \(|T_{1,n} - \delta| \le \sup_x |(F_n(x) - F(x))| = A_n\). Since \(A_n \stackrel{a.s.}{\rightarrow} 0\) and \(|T_{1,n} - \delta| \le A_n\), it follows that \(T_{1,n} \stackrel{a.s.}{\rightarrow} \delta\).
Under \(H_1\), \(\delta > 0\). Since almost sure convergence implies convergence in probability, \(T_{1,n} \stackrel{P}{\rightarrow} \delta > 0\) under \(H_1\). Then \(T_{2,n} = n T_{1,n} \stackrel{P}{\rightarrow} n \delta\). As \(n \rightarrow \infty\) and \(\delta > 0\), \(n \delta \rightarrow \infty\).
So, under \(H_1\), \(T_{2,n} \stackrel{P}{\rightarrow} \infty\). This means that for any constant \(c\), \(P_F(T_{2,n} > c) \rightarrow 1\) as \(n \rightarrow \infty\). Taking \(c=1\), we get \(P_F(T_{2,n} > 1) \rightarrow 1\) for any \(F \ne F_0\).
Combining Case 1 and Case 2, we conclude that \(\lim_{n \rightarrow \infty} P_F(T_{2, n}>1)=1\) for all F. This statement is true.
Convergence in Distribution of T2n under H0
Consider the convergence in distribution of \(T_{2,n}\) under \(H_0\). A degenerate random variable is a variable that takes a single specific value with probability 1 (e.g., \(P(Y=c)=1\) for some constant \(c\)).
We found that under \(H_0\), \(T_{2,n} \stackrel{P}{\rightarrow} \infty\). A sequence of random variables that converges in probability to infinity cannot converge in distribution to a degenerate random variable with a finite value. If a variable converges in distribution to a constant \(c\), it must converge in probability to \(c\). Since \(T_{2,n}\) converges in probability to infinity, it cannot converge in distribution to a finite, degenerate value.
Therefore, the statement that \(T_{2, n}\) converges in distribution to a degenerate real valued random variable under \(H_0\) is false.
Summary
Based on the analysis:
- \(T_{1, n} \stackrel{P}{\rightarrow} 0\) as \(n \rightarrow \infty\) under \(H_0\) is true.
- \(T_{2, n} \stackrel{P}{\rightarrow} 0\) as \(n \rightarrow \infty\) under \(H_0\) is false.
- \(\displaystyle \lim _{n \rightarrow \infty} P_F\left(T_{2, n}>1\right)=1\) for all F is true.
- \(T_{2, n}\) converges in distribution to a degenerate real valued random variable under \(H_0\) is false.
Important Questions from Sampling Distributions
Let X1, X2, ..., X6 be a random sample from a gamma distribution with the probability density function
\(f(x \mid \lambda)=\left\{\begin{array}{cl} \frac{\lambda^4}{6} e^{-\lambda x} x^3, & \text { if } x>0 \\ 0, & \text { if } x \leq 0 \end{array},\right.\)
where λ > 0 is unknown. Let \(T=\sum_{i=1}^6 X_i\) and ψ be the uniformly most powerful test of size α = 0.05 for testing null hypothesis H0 : λ = 1 against alternative hypothesis H1 : λ > 1. For any positive integer v, let \(\chi_{v, α}^2\) denote the (1 - α)th quantile of \(\chi_v^2\) distribution. Then the test ψ rejects H0 if and only if
For n ≥ 2, let X1, X2, ..., Xn be a random sample from a distribution with the probability density function
\(f(x \mid θ)=\left\{\begin{array}{cc} θ x^{θ-1}, & 0<x<1 \\ 0, & \text { otherwise } \end{array},\right.\)
where θ > 0 is an unknown parameter. Then which of the following is the uniformly minimum variance unbiased estimator for \(\frac{1}{\theta}\) ?
Let X1, ..., Xn be a random sample from N(μ, 1) distribution, where μ ∈ ℝ is unknown. In order to test H0 : μ = μ0 against H1 : μ > μ0, where μ0 ∈ ℝ is some specified constant, consider the following two tests:
(A) Reject H0 if and only if X̅n > c1, where c1 is such that \(P_{μ_0}\) (X̅n > c1) = α ∈ (0, 1) and X̅n = \(\frac{1}{n} \sum_{i=1}^n X_i\).
(B) Reject H0 if and only if Median {X1, ..., Xn} > c2, where c2 is such that \(P_{μ_0}\)(Median{X1, ..., Xn} > c2) = α ∈ (0, 1).
Then which of the following statements are true?
- Which one of the following is the formula to calculate standard error of mean differences of two sample groups ?
- Standard deviation of a sampling distribution of a statistic is termed as,
