Theory of Statistics - RASIO 2025 - answers

Question 4

Step 1: Write the Likelihood Function

The support of each \(X_i\) is \((\theta - 1/2, \theta + 1/2)\). The joint density is: \[ L(\theta) = \prod_{i=1}^n f(x_i; \theta) = \begin{cases} 1, & \text{if } \theta - \frac{1}{2} < x_i < \theta + \frac{1}{2} \text{ for all } i \\ 0, & \text{otherwise} \end{cases} \] So, \(L(\theta) = 1\) if and only if all \(x_i\) are in \((\theta - 1/2, \theta + 1/2)\).

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Step 2: Rewrite the Condition in Terms of Order Statistics

The condition that all \(x_i \in (\theta - 1/2, \theta + 1/2)\) is equivalent to: \[ \theta - \frac{1}{2} < Y_1 \quad \text{and} \quad Y_n < \theta + \frac{1}{2} \] Rearranging: \[ Y_n - \frac{1}{2} < \theta < Y_1 + \frac{1}{2} \]

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Step 3: Interpret the Likelihood

The likelihood is 1 for all \(\theta\) in the interval \((Y_n - 1/2, Y_1 + 1/2)\), and 0 otherwise. Therefore, every\(\theta\) in this interval maximizes the likelihood (since the likelihood is constant and maximal on this interval). Hence, any value in this interval is an MLE.

A statistic \(u(X_1, \dots, X_n)\) is an MLE if it lies in this interval, i.e., \[ Y_n - \frac{1}{2} \leq u \leq Y_1 + \frac{1}{2} \]

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Step 4: Examples

Given examples: - \(\frac{4Y_1 + 2Y_n + 1}{6}\) - \(\frac{Y_1 + Y_n}{2}\) - \(\frac{2Y_1 + 4Y_n - 1}{6}\)

We can check that each lies between \(Y_n - 1/2\) and \(Y_1 + 1/2\). For instance, \(\frac{Y_1 + Y_n}{2}\) is the midpoint and clearly in the interval.

Therefore, these are all MLEs, and the MLE is not unique.

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Question 5

Step 1: Write the Likelihood Function

\[ L(\theta) = \prod_{i=1}^n f(x_i | \theta) = \prod_{i=1}^n \theta x_i^{-2} = \theta^n \prod_{i=1}^n x_i^{-2} \]

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Step 2: Incorporate the Support

The support is \(\theta \leq x_i < \infty\) for all \(i\). So, for the likelihood to be positive, we require: \[ \theta \leq \min(x_1, \dots, x_n) = Y_1 \] where \(Y_1\) is the smallest order statistic.

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Step 3: Maximize the Likelihood

The likelihood is: \[ L(\theta) = \theta^n \cdot \left(\prod_{i=1}^n x_i^{-2}\right) \] The term \(\prod x_i^{-2}\) does not depend on \(\theta\). So, to maximize \(L(\theta)\), we need to maximize \(\theta^n\) subject to \(\theta \leq Y_1\).

Since \(\theta^n\) is increasing in \(\theta\), the maximum occurs at the largest possible \(\theta\), which is: \[ \theta = Y_1 = \min(X_1, \dots, X_n) \]

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Step 4: Final Answer

\[ \hat{\theta}_{\text{MLE}} = \min(X_1, \dots, X_n) \]

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Question 6

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Step 1: Joint Density of \((X, Y)\)

Since \(X\) and \(Y\) are independent: \[ f_{X,Y}(x,y) = \frac{1}{\Gamma(r)} x^{r-1} e^{-x} \cdot \frac{1}{\Gamma(s)} y^{s-1} e^{-y}, \quad x>0, y>0 \]

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Step 2: Change of Variables

Let: \[ Z_1 = X + Y, \quad Z_2 = \frac{X}{X+Y} \] Then: \[ X = Z_1 Z_2, \quad Y = Z_1 (1 - Z_2) \] The Jacobian is: \[ J = \left| \begin{matrix} \frac{\partial x}{\partial z_1} & \frac{\partial x}{\partial z_2} \\ \frac{\partial y}{\partial z_1} & \frac{\partial y}{\partial z_2} \end{matrix} \right| = \left| \begin{matrix} z_2 & z_1 \\ 1-z_2 & -z_1 \end{matrix} \right| = -z_1 z_2 - z_1(1-z_2) = -z_1 \] So \(|J| = z_1\).

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Step 3: Joint Density of \((Z_1, Z_2)\)

\[ f_{Z_1, Z_2}(z_1, z_2) = f_{X,Y}(x,y) \cdot |J| = \left[ \frac{1}{\Gamma(r)} (z_1 z_2)^{r-1} e^{-z_1 z_2} \right] \left[ \frac{1}{\Gamma(s)} (z_1(1-z_2))^{s-1} e^{-z_1(1-z_2)} \right] \cdot z_1 \] Simplify: \[ = \frac{1}{\Gamma(r)\Gamma(s)} z_1^{r-1} z_2^{r-1} z_1^{s-1} (1-z_2)^{s-1} e^{-z_1} \cdot z_1 \] \[ = \frac{1}{\Gamma(r)\Gamma(s)} z_1^{r+s-1} e^{-z_1} \cdot z_2^{r-1} (1-z_2)^{s-1} \]

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Step 4: Factor the Joint Density

Notice: \[ f_{Z_1, Z_2}(z_1, z_2) = \underbrace{\frac{1}{\Gamma(r+s)} z_1^{r+s-1} e^{-z_1}}_{\text{Gamma}(r+s,1) \text{ density}} \cdot \underbrace{\frac{\Gamma(r+s)}{\Gamma(r)\Gamma(s)} z_2^{r-1} (1-z_2)^{s-1}}_{\text{Beta}(r,s) \text{ density}} \]

So: \[ f_{Z_1, Z_2}(z_1, z_2) = f_{Z_1}(z_1) \cdot f_{Z_2}(z_2) \] where: - \(Z_1 \sim \text{Gamma}(r+s, 1)\) - \(Z_2 \sim \text{Beta}(r, s)\)

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Step 5: Independence and Distributions

Since the joint density factors, \(Z_1\) and \(Z_2\) are independent. Their distributions are: - \(Z_1 = X + Y \sim \text{Gamma}(r+s, 1)\) - \(Z_2 = \frac{X}{X+Y} \sim \text{Beta}(r, s)\)

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