$X_t = \alpha + \alpha^2 t + \epsilon_t$, for $t = 1,..., n$,
where, $\epsilon_1, \epsilon_2,..., \epsilon_n$ are independent and identically distributed $N(0, \sigma^2)$ random variables. Here $\alpha \in R$ and $\sigma > 0$ are unknown parameters. Which of the following statements are true?
Solution Analysis
We analyze each statement regarding the random variables $X_t = \alpha + \alpha^2 t + \epsilon_t$, where $\epsilon_t \sim N(0, \sigma^2)$ are independent and $\alpha, \sigma$ are unknown parameters.
Statement A: Sufficiency of the Sample
- The joint probability density function (PDF) for $X_1, ..., X_n$ depends on $\alpha$ and $\sigma$.
- The entire sample $(X_1, ..., X_n)$ contains all the information about the parameters $(\alpha, \sigma)$.
- By definition, the sample itself is always a sufficient statistic for the parameters it depends on.
- Therefore, Statement A is True.
Statement B: Minimal Sufficiency
- The model $X_t = \alpha + \alpha^2 t + \epsilon_t$ involves a non-linear term ($\alpha^2$) in the parameter $\alpha$.
- Minimal sufficiency is guaranteed for statistics derived from linear exponential families. This model is not a linear exponential family.
- The proposed statistic $(\sum X_t, \sum tX_t, \sum t^2X_t)$ involves sums related to moments, but establishing minimal sufficiency for non-linear models requires more rigorous conditions, which are not met here.
- Therefore, Statement B is False.
Statement C: Ancillarity of $X_4 - X_3 - X_2 + X_1$
- An ancillary statistic is one whose probability distribution does not depend on the unknown parameters $(\alpha, \sigma)$.
- Let $A = X_4 - X_3 - X_2 + X_1$. Substituting the model equation: $ A = (\alpha + 4\alpha^2 + \epsilon_4) - (\alpha + 3\alpha^2 + \epsilon_3) - (\alpha + 2\alpha^2 + \epsilon_2) + (\alpha + \alpha^2 + \epsilon_1) $
- Simplifying the terms: $ A = (\alpha - \alpha - \alpha + \alpha) + \alpha^2(4 - 3 - 2 + 1) + (\epsilon_4 - \epsilon_3 - \epsilon_2 + \epsilon_1) $ $ A = 0 + \alpha^2(0) + (\epsilon_1 - \epsilon_2 - \epsilon_3 + \epsilon_4) $ $ A = \epsilon_1 - \epsilon_2 - \epsilon_3 + \epsilon_4 $
- Since $\epsilon_t \sim N(0, \sigma^2)$ are independent, $A$ follows a normal distribution: $A \sim N(0, Var(A))$. $ Var(A) = Var(\epsilon_1) + Var(-\epsilon_2) + Var(-\epsilon_3) + Var(\epsilon_4) = \sigma^2 + \sigma^2 + \sigma^2 + \sigma^2 = 4\sigma^2 $
- The distribution of $A$ is $N(0, 4\sigma^2)$, which depends on the parameter $\sigma$.
- Therefore, Statement C is False.
Statement D: Ancillarity of the Ratio
- Let $Y = \frac{X_4+X_1-X_2 - X_3}{X_5+ X_2 - X_3 - X_4}$.
- Consider the numerator $N = X_4+X_1-X_2 - X_3$. This is a linear combination of $X_t$. The coefficients $(1, -1, -1, 1)$ for $X_1, X_2, X_3, X_4$ sum to zero ($\sum c_t=0$), which eliminates the $\alpha$ term. They also satisfy $\sum c_t t = 0$, eliminating the $\alpha^2$ term. $ N = (\epsilon_1 - \epsilon_2 - \epsilon_3 + \epsilon_4) $ $N$ follows $N(0, 4\sigma^2)$.
- Consider the denominator $D = X_5+ X_2 - X_3 - X_4$. This is also a linear combination. The coefficients $(1, 1, -1, -1)$ for $X_2, X_3, X_4, X_5$ sum to zero ($\sum c'_t=0$), eliminating $\alpha$. They also satisfy $\sum c'_t t = 0$, eliminating $\alpha^2$. $ D = (\epsilon_2 - \epsilon_3 - \epsilon_4 + \epsilon_5) $ $D$ follows $N(0, 4\sigma^2)$.
- The statistic is $Y = N/D$. Let $N = \sigma N'$ and $D = \sigma D'$, where $N'$ and $D'$ are linear combinations of standard normal variables ($\epsilon_t/\sigma$). $ Y = \frac{\sigma N'}{\sigma D'} = \frac{N'}{D'} $
- The distribution of the ratio $N'/D'$ involves only the relationships between the standard normal variables and is independent of $\sigma$. Since the $\alpha$ and $\alpha^2$ terms were eliminated in both $N$ and $D$, the distribution is also independent of $\alpha$.
- Therefore, $Y$ is an ancillary statistic. Statement D is True.
Conclusion
Statements A and D are true.
Important Questions from Gauss-Markov Models
Consider the linear statistical model
$y_{ij} = \mu + \tau_i + \varepsilon_{ij}; \quad i = 1, 2, \cdots, a; \quad j = 1, 2, \cdots, n$
where $\mu$ is unknown, $\tau_i$ are independently and identically distributed as $N(0, \sigma_\tau^2)$, $\varepsilon_{ij}$ are independently and identically distributed as $N(0, \sigma^2)$; $\tau_i$ and $\varepsilon_{ij}$ are independent for all $i$ and $j$. Note that $\tau_i$ is the $i^{th}$ treatment effect. Suppose $SS_{total}, SS_{treatment}, SS_{error}$ are total sum of squares, total treatment sum of squares and error sum of squares, respectively. To test :$H_0 : \sigma_\tau^2 = 0$ vs $H_A : \sigma_\tau^2 > 0$ which of the following statements is not true?
Consider the model $Y = X\beta + \epsilon$, where $Y = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix}, X = \left( (x_{ij}) \right)_{n \times p}$ and $\beta = \begin{bmatrix} \beta_1 \\ \vdots \\ \beta_p \end{bmatrix}, \epsilon = \begin{bmatrix} \epsilon_1 \\ \epsilon_2 \\ \vdots \\ \epsilon_n \end{bmatrix}$. $E(\epsilon) = 0$ and $D(\epsilon) = \sigma^2 I_n, p < n$. Let $\hat{\beta}$ be the solution of $X^T X \beta = X^T Y$. Which of the following are true?
Let $X_1, X_2, \ldots$ be a random sample from uniform $(0, 3\theta), \theta > 0$. Define
$T_n = \frac{1}{3} \max\{X_1, X_2, \ldots, X_n\}$. Which of the following is NOT true?
- Consider the following regression problem $y_i = \beta_1 e^i + \beta_2 e^{-i} + \epsilon_i; \ i = 1, \ldots, n$. Here $\epsilon_1, \ldots, \epsilon_n$ are i.i.d. $N(0, \sigma^2)$ random variables. If $\hat{\beta}_1$ and $\hat{\beta}_2$ are the least square estimators of $\beta_1$ and $\beta_2$, respectively, then which of the following statements are correct?
Consider the multiple linear regression model
$Y_i = \beta_1 x_{i1} + \dots + \beta_p x_{ip} + \epsilon_i$, with $E(\epsilon_i) = 0$, $Cov(\epsilon_i, \epsilon_k) = 0$,
if $i \ne k$ and $Var(\epsilon_i) = \sigma^2$, for $i, k = 1, \dots, n$. If $\hat{y}_i$ is the least squares fit of $y_i$ and $\hat{\epsilon}_i$ is the corresponding estimated residual for $i = 1, \dots, n$, then which of the following statements are always correct?
