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根据提示完成题目，要求：尽可能完整的给我所有的步骤 Q1. Bias and MSE of an MLE (a) (1 mark) Show that $g_{...
Oct 1, 2024
根据提示完成题目，要求：尽可能完整的给我所有的步骤
Solution by Steps
step 1
To show that gλ(y) g_{\lambda}(y) is a probability density function (pdf), we need to verify two conditions: non-negativity and normalization. The function is defined as gλ(y)=e−(y−λ) g_{\lambda}(y) = e^{-(y-\lambda)} for y≥λ y \geq \lambda and 0 0 otherwise. Since e−(y−λ) e^{-(y-\lambda)} is always positive for y≥λ y \geq \lambda , the non-negativity condition is satisfied
step 2
Next, we check the normalization condition by computing the integral of gλ(y) g_{\lambda}(y) over its entire range: ∫−∞∞gλ(y) dy=∫λ∞e−(y−λ) dy. \int_{-\infty}^{\infty} g_{\lambda}(y) \, dy = \int_{\lambda}^{\infty} e^{-(y-\lambda)} \, dy. This integral evaluates to 1 1 , confirming that the total probability is 1 1
step 3
For part (b), to find the maximum likelihood estimator T(Y) T(Y) for λ \lambda , we need to maximize the likelihood function L(λ)=∏i=1ngλ(Yi) L(\lambda) = \prod_{i=1}^{n} g_{\lambda}(Y_i) . Taking the logarithm gives us the log-likelihood: log⁡L(λ)=∑i=1nlog⁡gλ(Yi)=∑i=1n(−(Yi−λ)). \log L(\lambda) = \sum_{i=1}^{n} \log g_{\lambda}(Y_i) = \sum_{i=1}^{n} (-(Y_i - \lambda)). Maximizing this with respect to λ \lambda leads to T(Y)=min⁡(Y1,Y2,…,Yn) T(Y) = \min(Y_1, Y_2, \ldots, Y_n)
step 4
For part (c), we compute \mathbb{P}_{\lambda}(T(Y) > t) . Since T(Y)=min⁡(Y1,Y2,…,Yn) T(Y) = \min(Y_1, Y_2, \ldots, Y_n) , we have: \mathbb{P}_{\lambda}(T(Y) > t) = \mathbb{P}(Y_1 > t, Y_2 > t, \ldots, Y_n > t) = \left(1 - e^{-(t-\lambda)}\right)^n \text{ for } t \geq \lambda.
step 5
For part (d), the cumulative distribution function (cdf) Fλ(t) F_{\lambda}(t) is given by: F_{\lambda}(t) = 1 - \mathbb{P}_{\lambda}(T(Y) > t) = 1 - \left(1 - e^{-(t-\lambda)}\right)^n.
step 6
For part (e), the pdf fλ(t) f_{\lambda}(t) can be found by differentiating the cdf: fλ(t)=ddtFλ(t)=ne−(t−λ)(1−e−(t−λ))n−1 for t≥λ. f_{\lambda}(t) = \frac{d}{dt} F_{\lambda}(t) = n e^{-(t-\lambda)} \left(1 - e^{-(t-\lambda)}\right)^{n-1} \text{ for } t \geq \lambda.
step 7
For part (f), the bias of the estimator T(Y) T(Y) is computed as: Bias(T(Y))=E[T(Y)]−λ. \text{Bias}(T(Y)) = \mathbb{E}[T(Y)] - \lambda. Using the known mean of the minimum of n n iid exponential random variables, we find E[T(Y)]=λ−1n \mathbb{E}[T(Y)] = \lambda - \frac{1}{n} . Thus, the bias is −1n -\frac{1}{n}
step 8
For part (g), the Mean Square Error (MSE) of the estimator T(Y) T(Y) is given by: MSE(T(Y))=Var(T(Y))+Bias(T(Y))2. \text{MSE}(T(Y)) = \text{Var}(T(Y)) + \text{Bias}(T(Y))^2. Using the known variance of the minimum of n n iid exponential random variables, we find Var(T(Y))=1n2 \text{Var}(T(Y)) = \frac{1}{n^2} and thus the MSE is 1n2+(−1n)2=2n2 \frac{1}{n^2} + \left(-\frac{1}{n}\right)^2 = \frac{2}{n^2}
Answer
The answers to the questions are as follows: (a) gλ(y) g_{\lambda}(y) is a pdf. (b) T(Y)=min⁡(Y1,Y2,…,Yn) T(Y) = \min(Y_1, Y_2, \ldots, Y_n) . (c) \mathbb{P}_{\lambda}(T(Y) > t) = \left(1 - e^{-(t-\lambda)}\right)^n . (d) Fλ(t)=1−(1−e−(t−λ))n F_{\lambda}(t) = 1 - \left(1 - e^{-(t-\lambda)}\right)^n . (e) fλ(t)=ne−(t−λ)(1−e−(t−λ))n−1 f_{\lambda}(t) = n e^{-(t-\lambda)} \left(1 - e^{-(t-\lambda)}\right)^{n-1} . (f) Bias is −1n -\frac{1}{n} . (g) MSE is 2n2 \frac{2}{n^2} .
Key Concept
The key concept involves understanding the properties of probability density functions, maximum likelihood estimation, and the relationships between bias, variance, and mean square error in statistical estimation.
Explanation
The solution outlines the necessary steps to verify the properties of the pdf, derive the maximum likelihood estimator, and compute the bias and MSE, utilizing known results from the exponential distribution.
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