Let's tackle your question step by step. --- ## **Given:** The pdf is: \[ f(y|\theta) = \begin{cases} \frac{2}{\theta^2} (\theta - y) & 0 \leq y \leq \theta \\ 0 & \text{elsewhere} \end{cases} \] Suppose \( Y_1, Y_2, ..., Y_n \) is a random sample from this distribution. ### **a) Is the estimator for \(\theta\) unbiased?** **But which estimator?** You didn't specify which estimator (e.g., sample mean, sample maximum, etc). Common estimators for the upper bound of a distribution are: - The **maximum** of the sample, \( Y_{(n)} = \max(Y_1, ..., Y_n) \) - Some function of the sample mean, \( \bar{Y} \) - The method of moments estimator. Let's consider two common estimators: 1. **Sample maximum \( Y_{(n)} \)** 2. **Method of moments estimator (\( \hat{\theta}_{\text{MOM}} \))** --- ## **Step 1: Calculate the Mean** First, let's compute the expected value \( E[Y] \): \[ E[Y] = \int_0^\theta y \cdot \frac{2}{\theta^2} (\theta - y) \, dy \] Expand: \[ = \frac{2}{\theta^2} \int_0^\theta y(\theta - y) dy = \frac{2}{\theta^2} \int_0^\theta (\theta y - y^2) dy \] \[ = \frac{2}{\theta^2} \left[\theta \int_0^\theta y dy - \int_0^\theta y^2 dy\right] \] \[ = \frac{2}{\theta^2} \left[\theta \cdot \frac{y^2}{2} \Big|_0^\theta - \frac{y^3}{3}\Big|_0^\theta\right] \] \[ = \frac{2}{\theta^2} \left[\theta \cdot \frac{\theta^2}{2} - \frac{\theta^3}{3}\right] \] \[ = \frac{2}{\theta^2} \left[\frac{\theta^3}{2} - \frac{\theta^3}{3}\right] \] \[ = \frac{2}{\theta^2} \left[\frac{3\theta^3 - 2\theta^3}{6}\right] \] \[ = \frac{2}{\theta^2} \cdot \frac{\theta^3}{6} \] \[ = \frac{\theta^3}{3\theta^2} \] \[ = \frac{\theta}{3} \] --- ### **Method of Moments Estimator** Set sample mean equal to population mean: \[ \bar{Y} = E[Y] = \frac{\theta}{3} \implies \hat{\theta}_{\text{MOM}} = 3\bar{Y} \] Now, **is this unbiased?** \[ E[\hat{\theta}_{\text{MOM}}] = E[3\bar{Y}] = 3E[\bar{Y}] = 3E[Y] = 3 \times \frac{\theta}{3} = \theta \] So **YES, this is unbiased**. --- ### **Maximum Likelihood Estimator (MLE)** Let's find the likelihood: Joint pdf: \[ L(\theta) = \prod_{i=1}^n f(y_i|\theta) = \prod_{i=1}^n \frac{2}{\theta^2} (\theta - y_i) \] with \( 0 \leq y_i \leq \theta \), so \( \theta \geq \max_i y_i \). \[ L(\theta) = \left(\frac{2}{\theta^2}\right)^n \prod_{i=1}^n (\theta - y_i) \] The likelihood is maximized for the **smallest \(\theta\) such that \(\theta \geq \max_i y_i\)**, i.e., at \( \theta = \max_i y_i \). But let's check if this is unbiased. #### **Compute \( E[Y_{(n)}] \) where \( Y_{(n)} = \max(Y_1, ..., Y_n) \)** First, get CDF: \[ F(y|\theta) = \int_0^y \frac{2}{\theta^2} (\theta - t) dt \] \[ = \frac{2}{\theta^2} \int_0^y (\theta - t) dt = \frac{2}{\theta^2} \left[\theta t - \frac{t^2}{2}\right]_0^y = \frac{2}{\theta^2} (\theta y - \frac{y^2}{2}) \] So, \[ F(y|\theta) = \frac{2\theta y - y^2}{\theta^2} \] for \( 0 \leq y \leq \theta \). The CDF of \( Y_{(n)} \) is \[ F_{Y_{(n)}}(y) = [F(y)]^n \] The pdf is: \[ f_{Y_{(n)}}(y) = n [F(y)]^{n-1} f(y) \] So, \[ f_{Y_{(n)}}(y) = n \left(\frac{2\theta y - y^2}{\theta^2}\right)^{n-1} \frac{2}{\theta^2} (\theta - y) \] Now, \[ E[Y_{(n)}] = \int_0^\theta y f_{Y_{(n)}}(y) dy \] This integral is a bit messy. But generally, for this kind of triangular distribution, the MLE \( \max(Y_i) \) is **biased**. #### **Conclusion for (a):** - **Method of Moments estimator \( 3\bar{Y} \) is unbiased.** - **MLE estimator \( Y_{(n)} \) is biased.** --- ## **b) Is the estimator for \(\theta\) consistent?** ### **For MOM Estimator:** \[ \hat{\theta}_{\text{MOM}} = 3\bar{Y} \] As \( n \to \infty \), \( \bar{Y} \to E[Y] = \theta/3 \), so \( \hat{\theta}_{\text{MOM}} \to \theta \). **So it is consistent.** ### **For MLE Estimator:** \( Y_{(n)} \) converges in probability to \(\theta\) as \( n \to \infty \). **So it is consistent.** --- ## **c) Is the estimator for \(\theta\) efficient?** - **Efficiency** means it achieves the Cramér–Rao lower bound (minimum possible variance). - For uniform-type distributions, the MLE (max) is often efficient (asymptotically), but for the triangular distribution, we need to compute variances and compare to the Cramér–Rao lower bound. ### **Let’s compute the Fisher Information for one observation:** \[ f(y|\theta) = \frac{2}{\theta^2} (\theta - y) \] \[ \log f(y|\theta) = \log 2 - 2\log \theta + \log (\theta - y) \] \[ \frac{\partial}{\partial \theta} \log f(y|\theta) = -\frac{2}{\theta} + \frac{1}{\theta - y} \] \[ \left( \frac{\partial}{\partial \theta} \log f(y|\theta) \right)^2 = \left(-\frac{2}{\theta} + \frac{1}{\theta - y}\right)^2 \] Fisher information: \[ I(\theta) = E\left[\left( \frac{\partial}{\partial \theta} \log f(y|\theta) \right)^2\right] \] This is a bit of a messy integral, but you can see that the MOM estimator (which is unbiased and has variance decreasing as \( 1/n \)), is not necessarily the **most efficient**. The MLE is typically more efficient for large \( n \), even if it is slightly biased for small \( n \). #### **Conclusion for (c):** - **MLE is asymptotically efficient.** - **MOM estimator is consistent and unbiased but not as efficient as MLE.** --- ## **Summary Table** | Estimator | Unbiased? | Consistent? | Efficient? | |-----------------------------------|-----------|-------------|----------------------| | MOM (\( 3\bar{Y} \)) | Yes |