| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the Cramer-Rao lower bound for the variance of an unbiased estimator $\hat{\theta}(X)$ of $\theta$ for the distribution family $f_\theta(x) = \theta e^{-\theta x}$, we need to compute the Fisher information $I(\theta)$ and then take its reciprocal. The distribution family given is an exponential distribution with parameter $\theta$. The probability density function (pdf) is $f_\theta(x) = \theta e^{-\theta x}$ for $x \geq 0$. The Fisher information $I(\theta)$ is defined as the expected value of the negative second derivative of the log-likelihood function with respect to $\theta$. The log-likelihood function is given by: \[ \ell(\theta) = \log f_\theta(x) = \log(\theta e^{-\theta x}) = \log(\theta) - \theta x \] Taking the derivative with respect to $\theta$: \[ \frac{\partial \ell}{\partial \theta} = \frac{1}{\theta} - x \] Taking the second derivative: \[ \frac{\partial^2 \ell}{\partial \theta^2} = -\frac{1}{\theta^2} \] The Fisher information $I(\theta)$ is the expected value of the square of the score (the first derivative of the log-likelihood): \[ I(\theta) = E\left[\left(\frac{\partial \ell}{\partial \theta}\right)^2\right] = E\left[\left(\frac{1}{\theta} - x\right)^2\right] = E\left[\frac{1}{\theta^2} - \frac{2x}{\theta} + x^2\right] \] Since $E[x] = \frac{1}{\theta}$ for the exponential distribution, and $E[x^2] = \frac{2}{\theta^2}$ (which can be derived from the moment generating function or directly from the definition of the expected value), we have: \[ I(\theta) = \frac{1}{\theta^2} - \frac{2}{\theta^2} + \frac{2}{\theta^2} = \frac{1}{\theta^2} \] The Cramer-Rao lower bound for the variance of an unbiased estimator $\hat{\theta}(X)$ is the reciprocal of the Fisher information: \[ \text{CRLB} = \frac{1}{I(\theta)} = \theta^2 \] Therefore, the Cramer-Rao lower bound on $E_\theta(\hat{\theta}(X)-\theta)^2$ is $\boxed{\theta^2}$, which corresponds to option (b). |
(b) |
| deepseekr1 | To find the Cramer-Rao lower bound (CRLB) on the variance of an unbiased estimator \(\hat{\theta}(X)\) for the distribution family \(f_\theta(x) = \theta e^{-\theta x}\) (an exponential distribution with rate parameter \(\theta\)), we need to compute the Fisher information \(I(\theta)\). 1. **Log-likelihood function**: \[ \log f_\theta(x) = \log \theta - \theta x \] 2. **First derivative of the log-likelihood with respect to \(\theta\)**: \[ \frac{d}{d\theta} \log f_\theta(x) = \frac{1}{\theta} - x \] 3. **Second derivative of the log-likelihood with respect to \(\theta\)**: \[ \frac{d^2}{d\theta^2} \log f_\theta(x) = -\frac{1}{\theta^2} \] 4. **Fisher information \(I(\theta)\)** is the negative expectation of the second derivative: \[ I(\theta) = -E\left[ -\frac{1}{\theta^2} \right] = \frac{1}{\theta^2} \] 5. **Cramer-Rao lower bound (CRLB)** is the reciprocal of the Fisher information: \[ \text{CRLB} = \frac{1}{I(\theta)} = \frac{1}{\frac{1}{\theta^2}} = \theta^2 \] Thus, the Cramer-Rao lower bound on \(E_\theta(\hat{\theta}(X) - \theta)^2\) is \(\theta^2\). \[ \boxed{B} \] |
(b) |
| gpto1 | To solve this problem, we need to find the Cramér-Rao lower bound (CRLB) for the variance of any unbiased estimator \(\hat{\theta}(X)\) of \(\theta\) for the given exponential distribution \(f_\theta(x) = \theta e^{-\theta x}\), where \(x \geq 0\). **Step 1: Recall the Cramér-Rao Lower Bound** For an unbiased estimator \(\hat{\theta}\) of \(\theta\), the CRLB is given by: \[ \operatorname{Var}(\hat{\theta}) \geq \frac{1}{I(\theta)} \] where \(I(\theta)\) is the Fisher Information. **Step 2: Compute the Fisher Information \(I(\theta)\)** First, write the log-likelihood function for a single observation \(X\): \[ \ell(\theta) = \ln f_\theta(x) = \ln \theta - \theta x \] Compute the first derivative with respect to \(\theta\): \[ \frac{\partial}{\partial \theta} \ell(\theta) = \frac{1}{\theta} - x \] Compute the second derivative: \[ \frac{\partial^2}{\partial \theta^2} \ell(\theta) = -\frac{1}{\theta^2} \] The Fisher Information is: \[ I(\theta) = -E_\theta\left[\frac{\partial^2}{\partial \theta^2} \ell(\theta)\right] = -\left(-\frac{1}{\theta^2}\right) = \frac{1}{\theta^2} \] **Step 3: Apply the CRLB** Using the Fisher Information, we find: \[ \operatorname{Var}(\hat{\theta}) \geq \frac{1}{I(\theta)} = \frac{1}{1/\theta^2} = \theta^2 \] So, the CRLB on the variance of any unbiased estimator of \(\theta\) is \(\theta^2\). **Answer: (b) θ²** |
(b) |
| GPT-4o | To determine the Cramer-Rao lower bound (CRLB) for the given problem, we will follow these steps: 1. **Identify the likelihood function**: The probability density function (pdf) given is: \[ f_\theta(x) = \theta e^{-\theta x}, \quad x \geq 0. \] This is the pdf of an exponential distribution with parameter \(\theta\). 2. **Compute the Fisher information**: The Fisher information \(I(\theta)\) for a single observation from the given distribution can be computed as: \[ I(\theta) = E_\theta \left[ \left( \frac{\partial}{\partial \theta} \log f_\theta(X) \right)^2 \right]. \] First, we find the log-likelihood function: \[ \log f_\theta(x) = \log(\theta) - \theta x. \] Taking the derivative with respect to \(\theta\): \[ \frac{\partial}{\partial \theta} \log f_\theta(x) = \frac{1}{\theta} - x. \] Squaring and then taking the expectation with respect to \(X\): \[ \left( \frac{\partial}{\partial \theta} \log f_\theta(x) \right)^2 = \left( \frac{1}{\theta} - x \right)^2. \] Now, we need to find the expectation \(E_\theta \left[ \left( \frac{1}{\theta} - X \right)^2 \right]\): \[ E_\theta \left[ \left( \frac{1}{\theta} - X \right)^2 \right] = E_\theta \left[ \frac{1}{\theta^2} - \frac{2X}{\theta} + X^2 \right]. \] Using the properties of the exponential distribution: - \(E_\theta[X] = \frac{1}{\theta}\), - \(\text{Var}(X) = \frac{1}{\theta^2}\), - \(E_\theta[X^2] = \left( E_\theta[X] \right)^2 + \text{Var}(X) = \frac{1}{\theta^2} + \frac{1}{\theta^2} = \frac{2}{\theta^2}\). Substituting these expectations: \[ E_\theta \left[ \frac{1}{\theta^2} - \frac{2X}{\theta} + X^2 \right] = \frac{1}{\theta^2} - \frac{2}{\theta} \cdot \frac{1}{\theta} + \frac{2}{\theta^2} = \frac{1}{\theta^2} - \frac{2}{\theta^2} + \frac{2}{\theta^2} = \frac{1}{\theta^2}. \] Therefore, the Fisher information is: \[ I(\theta) = \frac{1}{\theta^2}. \] 3. **Apply the Cramer-Rao lower bound**: The CRLB for an unbiased estimator \(\hat{\theta}(X)\) of \(\theta\) is given by: \[ \text{Var}(\hat{\theta}(X)) \geq \frac{1}{I(\theta)}. \] Substituting the Fisher information: \[ \text{Var}(\hat{\theta}(X)) \geq \frac{1}{\frac{1}{\theta^2}} = \theta^2. \] Therefore, the Cramer-Rao lower bound on \(E_\theta[(\hat{\theta}(X)-\theta)^2]\) is \(\theta^2\). The correct choice is: (b) \(\theta^2\). |
(b) |
| llama3.1-70B-Instruct | ## Step 1: Identify the distribution and its properties The given distribution is $f_\theta(x) = \theta e^{-\theta x}$, which is a probability density function (pdf) for an exponential distribution with parameter $\theta$. This distribution is defined for $x \geq 0$. ## Step 2: Recall the Cramer-Rao lower bound formula The Cramer-Rao lower bound (CRLB) is a lower bound on the variance of any unbiased estimator $\hat{\theta}(X)$ of a parameter $\theta$. It is given by the formula: \[ \text{Var}(\hat{\theta}(X)) \geq \frac{1}{nE\left[\left(\frac{\partial \ln f_\theta(X)}{\partial \theta}\right)^2\right]} \] where $n$ is the sample size, and $E$ denotes the expected value. ## Step 3: Calculate the derivative of the log-likelihood function To apply the Cramer-Rao lower bound formula, we first need to calculate the derivative of the log-likelihood function with respect to $\theta$. The log-likelihood function is given by: \[ \ln f_\theta(x) = \ln \theta - \theta x \] Taking the derivative with respect to $\theta$, we get: \[ \frac{\partial \ln f_\theta(x)}{\partial \theta} = \frac{1}{\theta} - x \] ## Step 4: Calculate the expected value of the squared derivative Next, we need to calculate the expected value of the squared derivative: \[ E\left[\left(\frac{\partial \ln f_\theta(X)}{\partial \theta}\right)^2\right] = E\left[\left(\frac{1}{\theta} - X\right)^2\right] \] Using the properties of the exponential distribution, we know that $E[X] = \frac{1}{\theta}$ and $\text{Var}(X) = \frac{1}{\theta^2}$. Therefore: \[ E\left[\left(\frac{1}{\theta} - X\right)^2\right] = \text{Var}(X) = \frac{1}{\theta^2} \] ## Step 5: Apply the Cramer-Rao lower bound formula Now we can apply the Cramer-Rao lower bound formula: \[ \text{Var}(\hat{\theta}(X)) \geq \frac{1}{nE\left[\left(\frac{\partial \ln f_\theta(X)}{\partial \theta}\right)^2\right]} = \frac{1}{n\frac{1}{\theta^2}} = \frac{\theta^2}{n} \] Since we are interested in the expected squared error, we can write: \[ E_\theta(\hat{\theta}(X) - \theta)^2 \geq \frac{\theta^2}{n} \] However, since the question does not specify the sample size $n$, we can assume $n=1$ for a single observation, which is a common assumption in such problems. The final answer is: $\boxed{\theta^2}$ |
(b) |
| llama3.1-405B-Instruct | To find the Cramer-Rao lower bound (CRLB) on the variance of an unbiased estimator $\hat{\theta}(X)$ of $\theta$, we first need to compute the Fisher information $I(\theta)$ for the given distribution family $f_\theta(x) = \theta e^{-\theta x}$, $x \geq 0$. The Fisher information is defined as: \[I(\theta) = E_\theta \left[ \left( \frac{\partial \ln f_\theta(X)}{\partial \theta} \right)^2 \right]\] Given $f_\theta(x) = \theta e^{-\theta x}$, we find $\ln f_\theta(x) = \ln \theta - \theta x$. Therefore, \[\frac{\partial \ln f_\theta(x)}{\partial \theta} = \frac{1}{\theta} - x\] Now, we compute the Fisher information: \[I(\theta) = E_\theta \left[ \left( \frac{1}{\theta} - X \right)^2 \right]\] \[I(\theta) = E_\theta \left[ \frac{1}{\theta^2} - 2\frac{X}{\theta} + X^2 \right]\] \[I(\theta) = \frac{1}{\theta^2} - 2\frac{E_\theta[X]}{\theta} + E_\theta[X^2]\] For the exponential distribution, $E_\theta[X] = \frac{1}{\theta}$ and $E_\theta[X^2] = \frac{2}{\theta^2}$. Substituting these values: \[I(\theta) = \frac{1}{\theta^2} - 2\frac{1/\theta}{\theta} + \frac{2}{\theta^2}\] \[I(\theta) = \frac{1}{\theta^2} - \frac{2}{\theta^2} + \frac{2}{\theta^2} = \frac{1}{\theta^2}\] The Cramer-Rao lower bound (CRLB) is the reciprocal of the Fisher information: \[CRLB = \frac{1}{I(\theta)} = \frac{1}{1/\theta^2} = \theta^2\] Therefore, the Cramer-Rao lower bound on $E_\theta(\hat{\theta}(X)-\theta)^2$ for the given distribution family is $\theta^2$. The correct answer is (b) $\theta^2$. |
(b) |
| DeepSeek-Coder-V2 | To solve the problem, we need to find the Cramer-Rao lower bound (CRLB) for the variance of an unbiased estimator \(\hat{\theta}(X)\) of \(\theta\) for the given probability density function (PDF) \(f_\theta(x) = \theta e^{-\theta x}\), \(x \geq 0\). The CRLB is given by: \[ \text{CRLB} = \frac{1}{I(\theta)} \] where \(I(\theta)\) is the Fisher information. First, we compute the Fisher information \(I(\theta)\). The Fisher information for a single observation is: \[ I(\theta) = E_\theta\left[\left(\frac{\partial \ln f_\theta(x)}{\partial \theta}\right)^2\right] \] We start by finding the log-likelihood function: \[ \ln f_\theta(x) = \ln(\theta) - \theta x \] Next, we take the derivative of the log-likelihood with respect to \(\theta\): \[ \frac{\partial \ln f_\theta(x)}{\partial \theta} = \frac{1}{\theta} - x \] Now, we square this derivative: \[ \left(\frac{\partial \ln f_\theta(x)}{\partial \theta}\right)^2 = \left(\frac{1}{\theta} - x\right)^2 = \frac{1}{\theta^2} - \frac{2x}{\theta} + x^2 \] We then take the expected value of this squared derivative under the distribution \(f_\theta(x)\): \[ I(\theta) = E_\theta\left[\frac{1}{\theta^2} - \frac{2x}{\theta} + x^2\right] \] We need to compute each term separately: \[ E_\theta\left[\frac{1}{\theta^2}\right] = \frac{1}{\theta^2} \] \[ E_\theta\left[-\frac{2x}{\theta}\right] = -\frac{2}{\theta} E_\theta[x] \] \[ E_\theta[x^2] \] For the exponential distribution \(f_\theta(x) = \theta e^{-\theta x}\), we know: \[ E_\theta[x] = \frac{1}{\theta} \] \[ E_\theta[x^2] = \frac{2}{\theta^2} \] Substituting these into the Fisher information expression, we get: \[ I(\theta) = \frac{1}{\theta^2} - \frac{2}{\theta} \cdot \frac{1}{\theta} + \frac{2}{\theta^2} = \frac{1}{\theta^2} - \frac{2}{\theta^2} + \frac{2}{\theta^2} = \frac{1}{\theta^2} \] Thus, the Fisher information \(I(\theta)\) is: \[ I(\theta) = \frac{1}{\theta^2} \] The Cramer-Rao lower bound (CRLB) is therefore: \[ \text{CRLB} = \frac{1}{I(\theta)} = \frac{1}{\frac{1}{\theta^2}} = \theta^2 \] Thus, the Cramer-Rao lower bound on \(E_\theta(\hat{\theta}(X) - \theta)^2\) is \(\theta^2\). The correct answer is: \[ \boxed{B} \] |
(b) |
| Qwen2-72B-Instruct | To find the Cramer-Rao lower bound, we first need to compute the Fisher information $I(\theta)$. The Fisher information is given by the formula: \[I(\theta) = E_\theta\left[\left(\frac{\partial}{\partial\theta}\log f_\theta(X)\right)^2\right]\] Given the probability density function $f_\theta(x) = \theta e^{-\theta x}$, we can compute the derivative of the log-likelihood function with respect to $\theta$: \[\frac{\partial}{\partial\theta}\log f_\theta(x) = \frac{\partial}{\partial\theta}(\log\theta - \theta x) = \frac{1}{\theta} - x\] Squaring this and taking the expectation with respect to $\theta$, we get: \[I(\theta) = E_\theta\left[\left(\frac{1}{\theta} - x\right)^2\right] = E_\theta\left[\frac{1}{\theta^2} - 2\frac{x}{\theta} + x^2\right]\] Since $f_\theta(x) = \theta e^{-\theta x}$ is an exponential distribution, we know that $E_\theta[x] = \frac{1}{\theta}$ and $E_\theta[x^2] = \frac{2}{\theta^2}$, so: \[I(\theta) = \frac{1}{\theta^2} - 2\frac{1}{\theta^2} + 2\frac{1}{\theta^2} = \frac{1}{\theta^2}\] The Cramer-Rao lower bound for the variance of an unbiased estimator $\hat{\theta}(X)$ is given by: \[Var_\theta(\hat{\theta}(X)) \geq \frac{1}{I(\theta)}\] Substituting $I(\theta) = \frac{1}{\theta^2}$, we get: \[Var_\theta(\hat{\theta}(X)) \geq \theta^2\] However, the question asks for the Cramer-Rao lower bound on the mean squared error of the estimator, which is given by: \[E_\theta(\hat{\theta}(X)-\theta)^2 = Var_\theta(\hat{\theta}(X)) + Bias(\hat{\theta}(X))^2\] Since $\hat{\theta}(X)$ is unbiased, $Bias(\hat{\theta}(X))^2 = 0$, and thus: \[E_\theta(\hat{\theta}(X)-\theta)^2 = Var_\theta(\hat{\theta}(X)) \geq \theta^2\] Therefore, the answer is (b) $\theta^2$. |
(b) |
| baichuan4 | To find the Cramer-Rao lower bound for the given distribution, we first need to calculate the Fisher information $I(\theta)$. The Fisher information for a single observation $x$ from the distribution $f_\theta(x)=\theta e^{-\theta x}$ is defined as: $$ I(\theta) = E_\theta\left[\left(\frac{\partial}{\partial \theta} \log f_\theta(X)\right)^2\right] $$ First, we find the log-likelihood: $$ \log f_\theta(x) = \log(\theta) - \theta x $$ Then we take its derivative with respect to $\theta$: $$ \frac{\partial}{\partial \theta} \log f_\theta(x) = \frac{1}{\theta} - x $$ Next, we square this derivative and calculate its expectation: $$ I(\theta) = E_\theta\left[\left(\frac{1}{\theta} - X\right)^2\right] = E_\theta\left[\frac{1}{\theta^2} - \frac{2X}{\theta} + X^2\right] $$ Now we need to find the expectations $E_\theta[X]$ and $E_\theta[X^2]$: $$ E_\theta[X] = \int_0^\infty x \theta e^{-\theta x} dx = \frac{1}{\theta} $$ $$ E_\theta[X^2] = \int_0^\infty x^2 \theta e^{-\theta x} dx = \frac{2}{\theta^2} $$ Substituting these into the expression for $I(\theta)$: $$ I(\theta) = \frac{1}{\theta^2} - \frac{2}{\theta^2} + \frac{2}{\theta^2} = \frac{1}{\theta^2} $$ The Cramer-Rao lower bound is then: $$ E_\theta(\hat{\theta}(X)-\theta)^2 \geq \frac{1}{I(\theta)} = \theta^2 $$ So the correct answer is (b) $\theta^2$. |
(b) |
| doubao | The Fisher information is given by $I(\theta)=-E_\theta[\frac{\partial^2}{\partial\theta^2}\ln f_\theta(x)]$. For $f_\theta(x)=\theta e^{-\theta x}$, we have $\ln f_\theta(x)=\ln\theta-\theta x$. Then $\frac{\partial}{\partial\theta}\ln f_\theta(x)=\frac{1}{\theta}-x$ and $\frac{\partial^2}{\partial\theta^2}\ln f_\theta(x)=-\frac{1}{\theta^2}$. So $I(\theta)=\frac{1}{\theta^2}$. The Cramer-Rao lower bound is $\frac{1}{I(\theta)}=\theta^2$. So the answer is (b). |
(b) |
| minimax | The Cramer-Rao lower bound provides a lower bound on the variance of any unbiased estimator of a parameter $\theta$. The bound is given by the inverse of the Fisher information $I(\theta)$, where the Fisher information for a single observation is defined as: $$ I(\theta) = -E\left[\frac{\partial^2}{\partial \theta^2} \ln f_\theta(X)\right] $$ Given the probability density function $f_\theta(x) = \theta e^{-\theta x}$ for $x \geq 0$, we first need to find the Fisher information for this distribution. First, we calculate the log-likelihood function $\ell(\theta)$: $$ \ell(\theta) = \ln f_\theta(x) = \ln(\theta) - \theta x $$ Now, we take the first derivative of the log-likelihood with respect to $\theta$: $$ \frac{\partial \ell(\theta)}{\partial \theta} = \frac{1}{\theta} - x $$ Next, we take the second derivative: $$ \frac{\partial^2 \ell(\theta)}{\partial \theta^2} = -\frac{1}{\theta^2} $$ The Fisher information is the expected value of the negative second derivative of the log-likelihood: $$ I(\theta) = -E\left[\frac{\partial^2 \ell(\theta)}{\partial \theta^2}\right] = -E\left[-\frac{1}{\theta^2}\right] = \frac{1}{\theta^2} $$ The Cramer-Rao lower bound for the variance of any unbiased estimator $\hat{\theta}(X)$ of $\theta$ is then: $$ \text{Var}(\hat{\theta}(X)) \geq \frac{1}{I(\theta)} = \frac{1}{\frac{1}{\theta^2}} = \theta^2 $$ Therefore, the Cramer-Rao lower bound on $E_\theta(\hat{\theta}(X)-\theta)^2$ is $\theta^2$. The correct answer is (b) $\theta^2$. |
(b) |
| kimi | To find the Cramer-Rao lower bound (CRLB) for the variance of an unbiased estimator, we need to calculate the Fisher information number, which is defined as: \[ I(\theta) = -E\left[\frac{\partial^2}{\partial \theta^2} \ln f_\theta(X)\right] \] Given the distribution family \( f_\theta(x) = \theta e^{-\theta x} \) for \( x \geq 0 \), let's first find the log-likelihood function \( \ln f_\theta(x) \). \[ \ln f_\theta(x) = \ln(\theta e^{-\theta x}) = \ln \theta - \theta x \] Now, we find the first derivative of the log-likelihood with respect to \( \theta \): \[ \frac{\partial}{\partial \theta} \ln f_\theta(x) = \frac{1}{\theta} - x \] And the second derivative: \[ \frac{\partial^2}{\partial \theta^2} \ln f_\theta(x) = -\frac{1}{\theta^2} \] Now, we calculate the expectation of the second derivative, which is the Fisher information number: \[ I(\theta) = -E\left[-\frac{1}{\theta^2}\right] = \frac{1}{\theta^2} \] The Cramer-Rao lower bound states that the variance of an unbiased estimator \( \hat{\theta}(X) \) is at least as large as the reciprocal of the Fisher information number: \[ \text{Var}(\hat{\theta}(X)) \geq \frac{1}{I(\theta)} \] Substituting the Fisher information number we found: \[ \text{Var}(\hat{\theta}(X)) \geq \theta^2 \] So, the Cramer-Rao lower bound on \( E_\theta(\hat{\theta}(X) - \theta)^2 \) is \( \theta^2 \), which corresponds to option (b). |
(b) |
| wenxinyiyan | 首先，我们需要计算Fisher信息量$I(\theta)$。 对于给定的分布族$f_\theta(x)=\theta e^{-\theta x}$，其对数似然函数为$l(\theta)=\log(\theta)-\theta x$。 计算对数似然函数关于$\theta$的一阶导数，得到 $l'(\theta)=\frac{1}{\theta}-x$。 接着计算对数似然函数关于$\theta$的二阶导数，得到 $l''(\theta)=-\frac{1}{\theta^2}$。 然后，我们计算Fisher信息量$I(\theta)$，根据定义有 $I(\theta)=-E_\theta(l''(\theta))$。 将$l''(\theta)$代入上式，得到 $I(\theta)=-E_\theta(-\frac{1}{\theta^2})=\frac{1}{\theta^2}$。 最后，根据Cramer-Rao不等式，有 $E_\theta(\hat{\theta}(X)-\theta)^2 \geq \frac{1}{I(\theta)}$。 将$I(\theta)$的值代入上式，得到 $E_\theta(\hat{\theta}(X)-\theta)^2 \geq \theta^2$。 因此，Cramer-Rao下界为$\theta^2$。 所以答案是(b) $\theta^2$。 |
(b) |
| zhipu | To find the Cramer-Rao lower bound on \( E_\theta(\hat{\theta}(X) - \theta)^2 \) for an unbiased estimator \(\hat{\theta}(X)\) of \(\theta\) for the distribution family \( f_\theta(x) = \theta e^{-\theta x} \) where \( x \geq 0 \), we need to follow these steps: 1. **Determine the Fisher Information \(I(\theta)\)**: The Fisher information \(I(\theta)\) is given by: \[ I(\theta) = E_\theta \left[ \left( \frac{\partial \ln f_\theta(x)}{\partial \theta} \right)^2 \right] \] 2. **Compute the log-likelihood function**: The likelihood function is \( L(\theta; x) = f_\theta(x) = \theta e^{-\theta x} \). The log-likelihood function is: \[ \ln L(\theta; x) = \ln(\theta) - \theta x \] 3. **Differentiate the log-likelihood function with respect to \(\theta\)**: \[ \frac{\partial \ln L(\theta; x)}{\partial \theta} = \frac{1}{\theta} - x \] 4. **Square the derivative and take the expectation**: \[ \left( \frac{\partial \ln L(\theta; x)}{\partial \theta} \right)^2 = \left( \frac{1}{\theta} - x \right)^2 \] Now, we take the expectation: \[ I(\theta) = E_\theta \left[ \left( \frac{1}{\theta} - x \right)^2 \right] \] Since \( x \sim \text{Exp}(\theta) \) (Exponential distribution with rate \(\theta\)), the expectation \( E_\theta[x] = \frac{1}{\theta} \) and \( \text{Var}_\theta[x] = \frac{1}{\theta^2} \): \[ I(\theta) = E_\theta \left[ \left( \frac{1}{\theta} - x \right)^2 \right] = \text{Var}_\theta \left( \frac{1}{\theta} - x \right) + \left( E_\theta \left( \frac{1}{\theta} - x \right) \right)^2 \] Since \( E_\theta \left( \frac{1}{\theta} - x \right) = 0 \): \[ I(\theta) = \text{Var}_\theta \left( \frac{1}{\theta} - x \right) = \text{Var}_\theta[x] = \frac{1}{\theta^2} \] 5. **Apply the Cramer-Rao inequality**: The Cramer-Rao lower bound for the variance of any unbiased estimator \(\hat{\theta}(X)\) of \(\theta\) is given by: \[ \text{Var}_\theta(\hat{\theta}(X)) \geq \frac{1}{I(\theta)} \] Substituting \( I(\theta) = \frac{1}{\theta^2} \): \[ \text{Var}_\theta(\hat{\theta}(X)) \geq \theta^2 \] 6. **Interpret the result**: The Cramer-Rao lower bound on \( E_\theta(\hat{\theta}(X) - \theta)^2 \) is the same as the lower bound on the variance of \(\hat{\theta}(X)\), which is \(\theta^2\). Therefore, the correct answer is: (b) \(\theta^2\). |
(b) |
| gpto1 | To solve this problem, we need to find the Cramér-Rao lower bound (CRLB) for the variance of any unbiased estimator \(\hat{\theta}(X)\) of \(\theta\) for the given exponential distribution \(f_\theta(x) = \theta e^{-\theta x}\), where \(x \geq 0\). **Step 1: Recall the Cramér-Rao Lower Bound** For an unbiased estimator \(\hat{\theta}\) of \(\theta\), the CRLB is given by: \[ \operatorname{Var}(\hat{\theta}) \geq \frac{1}{I(\theta)} \] where \(I(\theta)\) is the Fisher Information. **Step 2: Compute the Fisher Information \(I(\theta)\)** First, write the log-likelihood function for a single observation \(X\): \[ \ell(\theta) = \ln f_\theta(x) = \ln \theta - \theta x \] Compute the first derivative with respect to \(\theta\): \[ \frac{\partial}{\partial \theta} \ell(\theta) = \frac{1}{\theta} - x \] Compute the second derivative: \[ \frac{\partial^2}{\partial \theta^2} \ell(\theta) = -\frac{1}{\theta^2} \] The Fisher Information is: \[ I(\theta) = -E_\theta\left[\frac{\partial^2}{\partial \theta^2} \ell(\theta)\right] = -\left(-\frac{1}{\theta^2}\right) = \frac{1}{\theta^2} \] **Step 3: Apply the CRLB** Using the Fisher Information, we find: \[ \operatorname{Var}(\hat{\theta}) \geq \frac{1}{I(\theta)} = \frac{1}{1/\theta^2} = \theta^2 \] So, the CRLB on the variance of any unbiased estimator of \(\theta\) is \(\theta^2\). **Answer: (b) θ²** |
(b) |
| deepseekv3 | To find the Cramer-Rao lower bound (CRLB) on the variance of an unbiased estimator \(\hat{\theta}(X)\) for the parameter \(\theta\) in the given distribution family \(f_\theta(x) = \theta e^{-\theta x}\) for \(x \geq 0\), we follow these steps: ### Step 1: Compute the Fisher Information \(I(\theta)\) The Fisher Information \(I(\theta)\) is given by: \[ I(\theta) = E\left[ \left( \frac{\partial}{\partial \theta} \ln f_\theta(X) \right)^2 \right] \] First, compute the log-likelihood function: \[ \ln f_\theta(x) = \ln \theta - \theta x \] Next, compute the derivative with respect to \(\theta\): \[ \frac{\partial}{\partial \theta} \ln f_\theta(x) = \frac{1}{\theta} - x \] Now, compute the Fisher Information: \[ I(\theta) = E\left[ \left( \frac{1}{\theta} - X \right)^2 \right] \] Expand the square: \[ I(\theta) = E\left[ \frac{1}{\theta^2} - \frac{2X}{\theta} + X^2 \right] \] Compute the expectations term by term. Recall that for the exponential distribution with parameter \(\theta\): \[ E[X] = \frac{1}{\theta}, \quad E[X^2] = \frac{2}{\theta^2} \] Thus, \[ I(\theta) = \frac{1}{\theta^2} - \frac{2}{\theta} \cdot \frac{1}{\theta} + \frac{2}{\theta^2} = \frac{1}{\theta^2} - \frac{2}{\theta^2} + \frac{2}{\theta^2} = \frac{1}{\theta^2} \] ### Step 2: Compute the Cramer-Rao Lower Bound The CRLB for the variance of an unbiased estimator \(\hat{\theta}(X)\) is: \[ \text{Var}(\hat{\theta}(X)) \geq \frac{1}{I(\theta)} = \theta^2 \] ### Final Answer The Cramer-Rao lower bound on \(E_\theta(\hat{\theta}(X) - \theta)^2\) is \(\theta^2\). **Answer:** (b) \(\theta^2\). |
(b) |