| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The Fisher information for a distribution family is given by the formula: \[ I(\theta) = E \left[ \left( \frac{\partial}{\partial\theta} \log f_\theta(X) \right)^2 \right] \] where $f_\theta(x)$ is the probability density function of the distribution, and $X$ is a random variable following that distribution. Given the distribution family $f_\theta(x) = \theta e^{-\theta x}$, we first find the log-likelihood function: \[ \log f_\theta(x) = \log(\theta e^{-\theta x}) = \log(\theta) - \theta x \] Next, we take the derivative of the log-likelihood with respect to $\theta$: \[ \frac{\partial}{\partial\theta} \log f_\theta(x) = \frac{\partial}{\partial\theta} (\log(\theta) - \theta x) = \frac{1}{\theta} - x \] Then, we square this derivative: \[ \left( \frac{\partial}{\partial\theta} \log f_\theta(x) \right)^2 = \left( \frac{1}{\theta} - x \right)^2 = \frac{1}{\theta^2} - \frac{2x}{\theta} + x^2 \] Now, we need to find the expected value of this squared derivative. Since $X$ follows the given distribution, we can use the properties of the distribution to find the expected values of the terms in the squared derivative. The expected value of $X$ for this distribution is $E[X] = \frac{1}{\theta}$, and the expected value of $X^2$ can be found using the second moment of the distribution, which is $E[X^2] = \frac{2}{\theta^2}$. Thus, the expected value of the squared derivative is: \[ E \left[ \left( \frac{\partial}{\partial\theta} \log f_\theta(X) \right)^2 \right] = E \left[ \frac{1}{\theta^2} - \frac{2X}{\theta} + X^2 \right] = \frac{1}{\theta^2} - \frac{2}{\theta}E[X] + E[X^2] \] \[ = \frac{1}{\theta^2} - \frac{2}{\theta} \cdot \frac{1}{\theta} + \frac{2}{\theta^2} = \frac{1}{\theta^2} \] Therefore, the Fisher information for the distribution family $f_\theta(x) = \theta e^{-\theta x}$ is $\theta^{-2}$. The correct answer is (d) $\theta^{-2}$. |
(d) |
| deepseekr1 | The Fisher information for the distribution family \( f_\theta(x) = \theta e^{-\theta x} \) (exponential distribution) is calculated as follows: 1. **Log-likelihood**: \( \log f_\theta(x) = \log \theta - \theta x \). 2. **First derivative with respect to \( \theta \)**: \[ \frac{\partial}{\partial \theta} \log f_\theta(x) = \frac{1}{\theta} - x \] 3. **Second derivative**: \[ \frac{\partial^2}{\partial \theta^2} \log f_\theta(x) = -\frac{1}{\theta^2} \] 4. **Fisher information via the second derivative**: \[ I(\theta) = -\mathbb{E}\left[-\frac{1}{\theta^2}\right] = \frac{1}{\theta^2} \] Alternatively, using the expectation of the squared first derivative: \[ I(\theta) = \mathbb{E}\left[\left(\frac{1}{\theta} - x\right)^2\right] = \text{Var}(X) = \frac{1}{\theta^2} \] (since \( \text{Var}(X) = \frac{1}{\theta^2} \) for an exponential distribution). **Answer**: \(\boxed{D}\) (Option d: \(\theta^{-2}\)). |
(d) |
| gpto1 | To determine the Fisher information \( I(\theta) \) for the given family of probability density functions: \[ f_\theta(x) = \theta e^{-\theta x}, \quad x \geq 0, \] we can follow the standard procedure involving the likelihood function. **Step 1: Compute the Log-Likelihood Function** First, compute the natural logarithm of the density function: \[ \ln f_\theta(x) = \ln \theta - \theta x. \] **Step 2: Compute the Derivative of the Log-Likelihood Function** Compute the derivative of \( \ln f_\theta(x) \) with respect to \( \theta \): \[ \frac{\partial}{\partial \theta} \ln f_\theta(x) = \frac{1}{\theta} - x. \] **Step 3: Compute the Square of the Derivative** Square the derivative obtained: \[ \left( \frac{\partial}{\partial \theta} \ln f_\theta(x) \right)^2 = \left( \frac{1}{\theta} - x \right)^2. \] **Step 4: Compute the Expected Value** Compute the expected value of the squared derivative with respect to the distribution of \( X \): \[ I(\theta) = E\left[ \left( \frac{1}{\theta} - X \right)^2 \right]. \] **Step 5: Simplify Using Known Moments** For an exponential distribution with parameter \( \theta \): - The mean is \( E[X] = \frac{1}{\theta} \). - The second moment is \( E[X^2] = \frac{2}{\theta^2} \). Compute the expected value: \[ I(\theta) = E\left[ \left( \frac{1}{\theta} - X \right)^2 \right] = E\left[ \left( X - \frac{1}{\theta} \right)^2 \right] = \operatorname{Var}(X). \] The variance of \( X \) is: \[ \operatorname{Var}(X) = E[X^2] - (E[X])^2 = \frac{2}{\theta^2} - \left( \frac{1}{\theta} \right)^2 = \frac{1}{\theta^2}. \] **Step 6: Conclude the Fisher Information** Therefore, the Fisher information is: \[ I(\theta) = \frac{1}{\theta^2}. \] **Answer: (d) \( \theta^{-2} \)** |
(d) |
| GPT-4o | To find the Fisher information for the distribution family \( f_\theta(x) = \theta e^{-\theta x} \), where \( x \geq 0 \), we need to follow several steps. Here, \( f_\theta(x) \) represents the probability density function of an exponential distribution with parameter \(\theta\). 1. **Compute the log-likelihood function**: The log-likelihood function \( \ell(\theta; x) \) for a single observation \( x \) is given by: \[ \ell(\theta; x) = \log f_\theta(x) = \log (\theta e^{-\theta x}) = \log \theta + \log e^{-\theta x} = \log \theta - \theta x. \] 2. **Compute the first derivative of the log-likelihood function with respect to \(\theta\)**: \[ \frac{\partial \ell(\theta; x)}{\partial \theta} = \frac{\partial}{\partial \theta} (\log \theta - \theta x) = \frac{1}{\theta} - x. \] 3. **Compute the second derivative of the log-likelihood function with respect to \(\theta\)**: \[ \frac{\partial^2 \ell(\theta; x)}{\partial \theta^2} = \frac{\partial}{\partial \theta} \left( \frac{1}{\theta} - x \right) = -\frac{1}{\theta^2}. \] 4. **Compute the expected value of the negative of the second derivative (the Fisher information)**: The Fisher information \( I(\theta) \) is defined as: \[ I(\theta) = \mathbb{E} \left[ -\frac{\partial^2 \ell(\theta; x)}{\partial \theta^2} \right]. \] Substitute the second derivative: \[ I(\theta) = \mathbb{E} \left[ -\left( -\frac{1}{\theta^2} \right) \right] = \mathbb{E} \left[ \frac{1}{\theta^2} \right]. \] Since \(\frac{1}{\theta^2}\) is constant with respect to \(x\), the expectation does not change its value: \[ I(\theta) = \frac{1}{\theta^2}. \] Thus, the Fisher information for the given distribution family \( f_\theta(x) = \theta e^{-\theta x} \) is \(\frac{1}{\theta^2}\). Therefore, the correct option is: (d) \(\theta^{-2}\). |
(d) |
| llama3.1-70B-Instruct | To find the Fisher information for the given distribution family \(f_\theta(x) = \theta e^{-\theta x}\), we first need to understand what Fisher information is. The Fisher information is a way of measuring the amount of information that a random variable (or a sample of random variables) carries about an unknown parameter of the distribution from which it is drawn. It is defined as the variance of the score, where the score is the derivative of the log-likelihood function with respect to the parameter. Given \(f_\theta(x) = \theta e^{-\theta x}\), let's find the Fisher information step by step. 1. **Log-likelihood function**: The first step is to find the log-likelihood function. Given a single observation \(x\), the likelihood function is \(L(\theta) = f_\theta(x) = \theta e^{-\theta x}\). The log-likelihood function is thus: \[l(\theta) = \ln(L(\theta)) = \ln(\theta e^{-\theta x}) = \ln(\theta) - \theta x\] 2. **Score**: The score is the derivative of the log-likelihood function with respect to \(\theta\). So, we differentiate \(l(\theta)\) with respect to \(\theta\): \[s(\theta) = \frac{d}{d\theta}l(\theta) = \frac{1}{\theta} - x\] 3. **Fisher Information**: The Fisher information \(I(\theta)\) is the variance of the score. However, since the score is a function of the random variable \(x\), we take the expectation of the square of the score and then subtract the square of the expectation of the score. But, given the nature of the score here, we directly compute the variance of the score by taking the expectation of the square of the score, noting that the expectation of the score itself is zero (a property of the score function in regular cases). Thus, we compute: \[I(\theta) = E[s(\theta)^2] = E\left[\left(\frac{1}{\theta} - x\right)^2\right]\] Given \(E[x] = \frac{1}{\theta}\) for an exponential distribution with rate \(\theta\), and \(Var(x) = \frac{1}{\theta^2}\), we can simplify the expression for \(I(\theta)\) by recognizing that the variance of \(x\) directly relates to the expectation of the square of the deviations from the mean. However, the direct computation of \(I(\theta)\) from the score's variance involves recognizing that the score's expectation is zero, and thus we directly compute the variance as the expectation of the square of the score. \[I(\theta) = E\left[\left(\frac{1}{\theta} - x\right)^2\right] = Var(x) = \frac{1}{\theta^2}\] Therefore, the correct answer is (d) \(\theta^{-2}\). |
(d) |
| llama3.1-405B-Instruct | To find the Fisher information for the distribution family $f_\theta(x) = \theta e^{-\theta x}$, we first need to recall the definition of Fisher information. The Fisher information $I(\theta)$ is defined as the variance of the score function, which in turn is defined as the partial derivative of the log-likelihood function with respect to the parameter $\theta$. Mathematically, this can be expressed as: \[I(\theta) = \text{Var}\left(\frac{\partial}{\partial \theta} \ln f_\theta(x)\right)\] Given $f_\theta(x) = \theta e^{-\theta x}$, let's first find the log-likelihood function: \[\ln f_\theta(x) = \ln(\theta) - \theta x\] Next, we find the partial derivative of the log-likelihood with respect to $\theta$, which is the score function: \[\frac{\partial}{\partial \theta} \ln f_\theta(x) = \frac{1}{\theta} - x\] Now, we need to find the variance of this score function. The variance of a random variable $Y$ is given by $\text{Var}(Y) = E(Y^2) - [E(Y)]^2$. In our case, $Y = \frac{1}{\theta} - x$. First, let's find $E(Y)$: \[E(Y) = E\left(\frac{1}{\theta} - x\right) = \frac{1}{\theta} - E(x)\] For the exponential distribution $f_\theta(x) = \theta e^{-\theta x}$, the expected value $E(x) = \frac{1}{\theta}$. Therefore: \[E(Y) = \frac{1}{\theta} - \frac{1}{\theta} = 0\] Next, we need to find $E(Y^2)$: \[E(Y^2) = E\left(\left(\frac{1}{\theta} - x\right)^2\right) = E\left(\frac{1}{\theta^2} - \frac{2x}{\theta} + x^2\right)\] \[E(Y^2) = \frac{1}{\theta^2} - \frac{2}{\theta}E(x) + E(x^2)\] For the exponential distribution, $E(x^2) = \frac{2}{\theta^2}$. Substituting $E(x) = \frac{1}{\theta}$ and $E(x^2) = \frac{2}{\theta^2}$: \[E(Y^2) = \frac{1}{\theta^2} - \frac{2}{\theta^2} + \frac{2}{\theta^2} = \frac{1}{\theta^2}\] Since $E(Y) = 0$, the variance of $Y$ (which is the Fisher information $I(\theta)$) is simply $E(Y^2)$: \[I(\theta) = \frac{1}{\theta^2}\] Therefore, the correct answer is (d) $\theta^{-2}$. |
(d) |
| DeepSeek-Coder-V2 | To find the Fisher information for the distribution family \( f_\theta(x) = \theta e^{-\theta x} \), \( x \geq 0 \), we need to follow these steps: 1. **Compute the log-likelihood function**: The probability density function (pdf) is given by: \[ f_\theta(x) = \theta e^{-\theta x} \] The log-likelihood function for a single observation \( x \) is: \[ \log f_\theta(x) = \log(\theta) + \log(e^{-\theta x}) = \log(\theta) - \theta x \] 2. **Compute the derivative of the log-likelihood function with respect to \( \theta \)**: \[ \frac{d}{d\theta} \log f_\theta(x) = \frac{d}{d\theta} (\log(\theta) - \theta x) = \frac{1}{\theta} - x \] 3. **Compute the second derivative of the log-likelihood function with respect to \( \theta \)**: \[ \frac{d^2}{d\theta^2} \log f_\theta(x) = \frac{d}{d\theta} \left( \frac{1}{\theta} - x \right) = -\frac{1}{\theta^2} \] 4. **Compute the Fisher information**: The Fisher information \( I(\theta) \) is given by the expected value of the negative of the second derivative of the log-likelihood function: \[ I(\theta) = -E\left[ \frac{d^2}{d\theta^2} \log f_\theta(x) \right] = -E\left[ -\frac{1}{\theta^2} \right] = \frac{1}{\theta^2} \] Thus, the Fisher information for the given distribution family is \( \frac{1}{\theta^2} \). Therefore, the correct option is: \[ \boxed{D} \] |
(d) |
| Qwen2-72B-Instruct | The Fisher information is a measure of the amount of information that an observable random variable X carries about an unknown parameter θ upon which the probability of X depends. For a distribution family with a probability density function (pdf) $f_\theta(x)$, the Fisher information $I(\theta)$ is given by: \[ I(\theta) = \mathbb{E}\left[\left(\frac{\partial}{\partial\theta}\log f_\theta(X)\right)^2\right] \] Given the distribution family $f_\theta(x) = \theta e^{-\theta x}$, we first find the log-likelihood function: \[ \log f_\theta(x) = \log(\theta) - \theta x \] Next, we take the derivative of the log-likelihood with respect to $\theta$: \[ \frac{\partial}{\partial\theta}\log f_\theta(x) = \frac{1}{\theta} - x \] Then, we square this derivative: \[ \left(\frac{\partial}{\partial\theta}\log f_\theta(x)\right)^2 = \left(\frac{1}{\theta} - x\right)^2 = \frac{1}{\theta^2} - 2\frac{x}{\theta} + x^2 \] The Fisher information is the expected value of this squared derivative. We can find this by integrating the squared derivative times the pdf over the domain of $x$: \[ I(\theta) = \mathbb{E}\left[\left(\frac{\partial}{\partial\theta}\log f_\theta(X)\right)^2\right] = \int_0^\infty \left(\frac{1}{\theta^2} - 2\frac{x}{\theta} + x^2\right) \theta e^{-\theta x} dx \] \[ I(\theta) = \int_0^\infty \left(\frac{1}{\theta} e^{-\theta x} - 2x e^{-\theta x} + \theta x^2 e^{-\theta x}\right) dx \] The first term integrates to 1 (the integral of the pdf over its domain is 1 by definition), the second term integrates to 0 (because it is the expected value of $x$ for an exponential distribution, which is $1/\theta$ times $-\theta$), and the third term integrates to 2 (because the expected value of $x^2$ for an exponential distribution is $2/\theta^2$ times $\theta$). Therefore: \[ I(\theta) = \frac{1}{\theta} + 0 + 2 = \frac{1}{\theta} + 2 \] However, the correct calculation for the Fisher information does not include the constant term, as it is the expected value of the squared score. Therefore, the correct Fisher information is: \[ I(\theta) = \frac{1}{\theta} \] Thus, the correct answer is (c) $\theta^{-1}$. |
(d) |
| baichuan4 | To find the Fisher information for the distribution family $f_\theta(x)=\theta e^{-\theta x}$ for $x \geq 0$, we first need to calculate the score function, which is the derivative of the log-likelihood with respect to $\theta$. The log-likelihood of $x$ given $\theta$ is: $$\log f_\theta(x) = \log(\theta) + \log(e^{-\theta x}) = \log(\theta) - \theta x$$ Now, we differentiate this with respect to $\theta$ to get the score function: $$\frac{d}{d\theta} \log f_\theta(x) = \frac{1}{\theta} - x$$ Next, we need to calculate the expectation of the square of the score function, which will give us the Fisher information: $$I(\theta) = \mathbb{E}\left[\left(\frac{1}{\theta} - x\right)^2\right]$$ To calculate this expectation, we need to integrate the square of the score function times the original density function: $$I(\theta) = \int_0^\infty (\theta e^{-\theta x})(\frac{1}{\theta} - x)^2 dx$$ Expanding the square and integrating term by term, we get: $$I(\theta) = \int_0^\infty (\theta e^{-\theta x})(\frac{1}{\theta^2} - \frac{2x}{\theta} + x^2) dx$$ $$I(\theta) = \frac{1}{\theta} \int_0^\infty e^{-\theta x} dx - 2 \int_0^\infty x e^{-\theta x} dx + \theta \int_0^\infty x^2 e^{-\theta x} dx$$ The first integral evaluates to $\frac{1}{\theta}(-\frac{e^{-\theta x}}{\theta})|_0^\infty = \frac{1}{\theta^2}$. The second integral evaluates to $2(-\frac{xe^{-\theta x}}{\theta} - \frac{e^{-\theta x}}{\theta^2})|_0^\infty = 2(\frac{1}{\theta^2})$. The third integral evaluates to $\theta(\frac{x^2e^{-\theta x}}{\theta} + \frac{2xe^{-\theta x}}{\theta^2} + \frac{2e^{-\theta x}}{\theta^3})|_0^\infty = \theta(\frac{2}{\theta^3})$. Putting these together, we get: $$I(\theta) = \frac{1}{\theta^2} - \frac{4}{\theta^2} + \frac{2}{\theta^2} = \frac{1}{\theta^2} - \frac{2}{\theta^2} = \frac{1}{\theta^2}$$ So the Fisher information is $\theta^{-2}$, which matches option (d). Therefore, the correct answer is (d) $\theta^{-2}$. |
(d) |
| doubao | The Fisher information is given by $I(\theta)=E[(\frac{\partial}{\partial\theta}\ln f_\theta(x))^2]$. For $f_\theta(x)=\theta e^{-\theta x}$, $\ln f_\theta(x)=\ln\theta-\theta x$. $\frac{\partial}{\partial\theta}\ln f_\theta(x)=\frac{1}{\theta}-x$. $(\frac{\partial}{\partial\theta}\ln f_\theta(x))^2=(\frac{1}{\theta}-x)^2$. To find the expected value, we need to multiply by $f_\theta(x)$ and integrate over $x$: \[ \begin{align*} I(\theta)&=\int_0^\infty(\frac{1}{\theta}-x)^2\theta e^{-\theta x}dx\\ &=\int_0^\infty(\frac{1}{\theta^2}-\frac{2x}{\theta}+x^2)\theta e^{-\theta x}dx\\ &=\frac{1}{\theta^2}\int_0^\infty\theta e^{-\theta x}dx-2\int_0^\infty x e^{-\theta x}dx+\int_0^\infty x^2\theta e^{-\theta x}dx\\ &=\frac{1}{\theta}-2(\frac{1}{\theta^2})+\frac{2}{\theta^3}\\ &=\frac{\theta^2-2\theta+2}{\theta^3} \end{align*} \] So none of the given options are correct. |
(d) |
| minimax | The Fisher information $I(\theta)$ for a parameter $\theta$ is defined as the variance of the score function, which is the derivative of the log-likelihood with respect to $\theta$. For a continuous distribution, the Fisher information can be calculated using the following formula: $$ I(\theta) = -E\left[\frac{\partial^2}{\partial \theta^2} \ln f_\theta(X)\right] $$ where $E[\cdot]$ denotes the expectation and $f_\theta(X)$ is the probability density function (PDF) of the distribution. Given the PDF $f_\theta(x)=\theta e^{-\theta x}$ for $x \geq 0$, let's first find the log-likelihood function: $$ \ln f_\theta(x) = \ln(\theta) + \ln(e^{-\theta x}) = \ln(\theta) - \theta x $$ Now, we take the derivative of the log-likelihood with respect to $\theta$: $$ \frac{\partial}{\partial \theta} \ln f_\theta(x) = \frac{1}{\theta} - x $$ Next, we take the second derivative with respect to $\theta$: $$ \frac{\partial^2}{\partial \theta^2} \ln f_\theta(x) = -\frac{1}{\theta^2} $$ The Fisher information is the negative expected value of this second derivative: $$ I(\theta) = -E\left[-\frac{1}{\theta^2}\right] = \frac{1}{\theta^2} E[1] $$ Since the expectation of a constant is just the constant itself, and the constant here is 1, we have: $$ I(\theta) = \frac{1}{\theta^2} $$ Therefore, the correct answer is (d) $\theta^{-2}$. |
(d) |
| kimi | The Fisher information is a measure of how much information a random variable carries about an unknown parameter upon which the probability of the variable depends. It is defined as the variance of the score function, which is the derivative of the log-likelihood function with respect to the parameter. Given the distribution family \( f_\theta(x) = \theta e^{-\theta x} \) for \( x \geq 0 \), we first find the log-likelihood function \( \ell(\theta) \) for a single observation \( x \): \[ \ell(\theta) = \log f_\theta(x) = \log(\theta e^{-\theta x}) = \log(\theta) - \theta x \] Now, we find the score function \( U(\theta) \), which is the derivative of the log-likelihood function with respect to \( \theta \): \[ U(\theta) = \frac{d}{d\theta} \ell(\theta) = \frac{d}{d\theta} (\log(\theta) - \theta x) = \frac{1}{\theta} - x \] The Fisher information \( I(\theta) \) is the variance of the score function: \[ I(\theta) = \text{Var}(U(\theta)) \] Since \( U(\theta) = \frac{1}{\theta} - x \), and \( x \) is a random variable with the given distribution, we need to find the variance of \( U(\theta) \). However, \( x \) is not independent of \( \theta \), so we cannot directly calculate the variance in the usual way. Instead, we use the fact that the variance of the score function is given by the negative expected value of the second derivative of the log-likelihood function: \[ I(\theta) = -E\left(\frac{d^2}{d\theta^2} \ell(\theta)\right) \] Let's find the second derivative of \( \ell(\theta) \): \[ \frac{d^2}{d\theta^2} \ell(\theta) = \frac{d}{d\theta} \left(\frac{1}{\theta} - x\right) = -\frac{1}{\theta^2} \] Now, we take the expectation: \[ I(\theta) = -E\left(-\frac{1}{\theta^2}\right) = \frac{1}{\theta^2} \] So the Fisher information for the distribution family \( f_\theta(x) = \theta e^{-\theta x} \) is \( \frac{1}{\theta^2} \), which corresponds to option (d). |
(d) |
| wenxinyiyan | The Fisher information for a distribution family $\{f_\theta: \theta \in \Theta\}$ is defined as $$ I(\theta)=\int_{\mathbb{R}} \left(\frac{\partial}{\partial \theta} \log f_\theta(x)\right)^2 f_\theta(x) d x $$Given the distribution family $f_\theta(x)=\theta e^{-\theta x}, x \geq 0$, we have $$ \log f_\theta(x)=\log \theta-\theta x $$and $$ \frac{\partial}{\partial \theta} \log f_\theta(x)=\frac{1}{\theta}-x $$Therefore, \begin{aligned} I(\theta) & =\int_0^{\infty}\left(\frac{1}{\theta}-x\right)^2 \theta e^{-\theta x} d x \\ & =\int_0^{\infty}\left(\frac{1}{\theta^2}-2 \frac{x}{\theta}+x^2\right) \theta e^{-\theta x} d x \\ & =\frac{1}{\theta^2} \int_0^{\infty} \theta e^{-\theta x} d x-\frac{2}{\theta} \int_0^{\infty} x \theta e^{-\theta x} d x+\int_0^{\infty} x^2 \theta e^{-\theta x} d x \\ & =\frac{1}{\theta^2}-\frac{2}{\theta} \cdot \frac{1}{\theta}+\frac{2}{\theta^2} \\ & =\frac{1}{\theta^2} \end{aligned} Hence, the correct option is (d) $\theta^{-2}$. |
(d) |
| zhipu | To find the Fisher information for the given distribution family, we need to use the definition of Fisher information, which for a single parameter \(\theta\) is given by: \[ I(\theta) = -E\left[\frac{\partial^2}{\partial \theta^2} \log f_\theta(X)\right] \] First, we need to compute the log-likelihood function for the distribution \(f_\theta(x) = \theta e^{-\theta x}\). The log-likelihood function \( \ell(\theta; x) \) is: \[ \ell(\theta; x) = \log f_\theta(x) = \log(\theta) - \theta x \] Next, we compute the first derivative of the log-likelihood function with respect to \(\theta\): \[ \frac{\partial \ell(\theta; x)}{\partial \theta} = \frac{1}{\theta} - x \] Now, we compute the second derivative of the log-likelihood function with respect to \(\theta\): \[ \frac{\partial^2 \ell(\theta; x)}{\partial \theta^2} = -\frac{1}{\theta^2} \] The Fisher information is the negative expected value of this second derivative: \[ I(\theta) = -E\left[\frac{\partial^2 \ell(\theta; x)}{\partial \theta^2}\right] = -E\left[-\frac{1}{\theta^2}\right] = \frac{1}{\theta^2} \] Thus, the Fisher information for the distribution family \(f_\theta(x) = \theta e^{-\theta x}\) is \(\theta^{-2}\). So, the correct option is (d) \(\theta^{-2}\). |
(d) |
| gpto1 | To determine the Fisher information \( I(\theta) \) for the given family of probability density functions: \[ f_\theta(x) = \theta e^{-\theta x}, \quad x \geq 0, \] we can follow the standard procedure involving the likelihood function. **Step 1: Compute the Log-Likelihood Function** First, compute the natural logarithm of the density function: \[ \ln f_\theta(x) = \ln \theta - \theta x. \] **Step 2: Compute the Derivative of the Log-Likelihood Function** Compute the derivative of \( \ln f_\theta(x) \) with respect to \( \theta \): \[ \frac{\partial}{\partial \theta} \ln f_\theta(x) = \frac{1}{\theta} - x. \] **Step 3: Compute the Square of the Derivative** Square the derivative obtained: \[ \left( \frac{\partial}{\partial \theta} \ln f_\theta(x) \right)^2 = \left( \frac{1}{\theta} - x \right)^2. \] **Step 4: Compute the Expected Value** Compute the expected value of the squared derivative with respect to the distribution of \( X \): \[ I(\theta) = E\left[ \left( \frac{1}{\theta} - X \right)^2 \right]. \] **Step 5: Simplify Using Known Moments** For an exponential distribution with parameter \( \theta \): - The mean is \( E[X] = \frac{1}{\theta} \). - The second moment is \( E[X^2] = \frac{2}{\theta^2} \). Compute the expected value: \[ I(\theta) = E\left[ \left( \frac{1}{\theta} - X \right)^2 \right] = E\left[ \left( X - \frac{1}{\theta} \right)^2 \right] = \operatorname{Var}(X). \] The variance of \( X \) is: \[ \operatorname{Var}(X) = E[X^2] - (E[X])^2 = \frac{2}{\theta^2} - \left( \frac{1}{\theta} \right)^2 = \frac{1}{\theta^2}. \] **Step 6: Conclude the Fisher Information** Therefore, the Fisher information is: \[ I(\theta) = \frac{1}{\theta^2}. \] **Answer: (d) \( \theta^{-2} \)** |
(d) |
| deepseekv3 | To find the Fisher information for the distribution family \( f_\theta(x) = \theta e^{-\theta x} \) where \( x \geq 0 \), follow these steps: ### Step 1: Compute the log-likelihood function The log-likelihood function is given by: \[ \ell(\theta; x) = \log f_\theta(x) = \log \theta - \theta x \] ### Step 2: Compute the first derivative of the log-likelihood with respect to \(\theta\) \[ \frac{\partial \ell(\theta; x)}{\partial \theta} = \frac{1}{\theta} - x \] ### Step 3: Compute the second derivative of the log-likelihood with respect to \(\theta\) \[ \frac{\partial^2 \ell(\theta; x)}{\partial \theta^2} = -\frac{1}{\theta^2} \] ### Step 4: Compute the Fisher information The Fisher information \( I(\theta) \) is the negative expectation of the second derivative: \[ I(\theta) = -\mathbb{E}\left[\frac{\partial^2 \ell(\theta; x)}{\partial \theta^2}\right] = -\mathbb{E}\left[-\frac{1}{\theta^2}\right] = \frac{1}{\theta^2} \] Thus, the Fisher information for the given distribution family is \( \theta^{-2} \). ### Final Answer: The correct option is **(d) \(\theta^{-2}\)**. |
(d) |