| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The Fisher information for a parametric family of distributions is defined as the expected value of the negative second derivative of the log-likelihood function with respect to the parameter. For the Gaussian distribution family $f_\theta(x) = N(0, \theta)$, we can find the Fisher information as follows: The probability density function of a Gaussian distribution with mean 0 and variance $\theta$ is given by: \[f_\theta(x) = \frac{1}{\sqrt{2\pi\theta}}\exp\left(-\frac{x^2}{2\theta}\right)\] The log-likelihood function is: \[\ell(\theta; x) = \log f_\theta(x) = -\frac{1}{2}\log(2\pi\theta) - \frac{x^2}{2\theta}\] Taking the derivative with respect to $\theta$: \[\frac{\partial\ell}{\partial\theta} = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2}\] Taking the second derivative with respect to $\theta$: \[\frac{\partial^2\ell}{\partial\theta^2} = \frac{1}{2\theta^2} - \frac{x^2}{\theta^3}\] The Fisher information is the expected value of the negative second derivative of the log-likelihood function: \[I(\theta) = E\left[-\frac{\partial^2\ell}{\partial\theta^2}\right] = E\left[\frac{x^2}{\theta^3} - \frac{1}{2\theta^2}\right]\] Since $x \sim N(0, \theta)$, $E[x^2] = \theta$. Therefore: \[I(\theta) = \frac{\theta}{\theta^3} - \frac{1}{2\theta^2} = \frac{1}{\theta^2} - \frac{1}{2\theta^2} = \frac{1}{2\theta^2}\] Thus, the Fisher information for the Gaussian distribution family $f_\theta(x) = N(0, \theta)$ is $\boxed{\text{(d) } 0.5\theta^{-2}}$. |
(d) |
| deepseekr1 | To find the Fisher information for the Gaussian distribution family \( f_\theta(x) = N(0, \theta) \), we start with the log-likelihood function for the normal distribution with mean 0 and variance \(\theta\): \[ \log f(x; \theta) = -\frac{1}{2} \log(2\pi\theta) - \frac{x^2}{2\theta} \] First, we compute the first derivative of the log-likelihood with respect to \(\theta\): \[ \frac{d}{d\theta} \log f(x; \theta) = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2} \] Next, we square this derivative: \[ \left( -\frac{1}{2\theta} + \frac{x^2}{2\theta^2} \right)^2 = \frac{(x^2 - \theta)^2}{4\theta^4} \] We then take the expectation of this expression: \[ I(\theta) = \frac{1}{4\theta^4} E[(X^2 - \theta)^2] \] Expanding the expectation: \[ E[(X^2 - \theta)^2] = E[X^4 - 2\theta X^2 + \theta^2] \] For \(X \sim N(0, \theta)\), we know \(E[X^2] = \theta\) and \(E[X^4] = 3\theta^2\). Substituting these values: \[ E[X^4 - 2\theta X^2 + \theta^2] = 3\theta^2 - 2\theta^2 + \theta^2 = 2\theta^2 \] Thus, the Fisher information is: \[ I(\theta) = \frac{1}{4\theta^4} \cdot 2\theta^2 = \frac{1}{2\theta^2} \] We verify this result using the second derivative method. The second derivative of the log-likelihood is: \[ \frac{d^2}{d\theta^2} \log f(x; \theta) = \frac{1}{2\theta^2} - \frac{x^2}{\theta^3} \] Taking the negative expectation: \[ - E\left[ \frac{1}{2\theta^2} - \frac{X^2}{\theta^3} \right] = - \left( \frac{1}{2\theta^2} - \frac{\theta}{\theta^3} \right) = - \left( \frac{1}{2\theta^2} - \frac{1}{\theta^2} \right) = \frac{1}{2\theta^2} \] Both methods confirm that the Fisher information is \(\frac{1}{2\theta^2}\), which corresponds to option (d). \[ \boxed{D} \] |
(d) |
| gpto1 | To find the Fisher information \(I(\theta)\) for the Gaussian distribution family \(f_\theta(x) = N(0,\theta)\), we start by recalling the definition of Fisher information for a parameter \(\theta\): \[ I(\theta) = E_\theta\left[ \left( \frac{\partial}{\partial \theta} \ln f_\theta(X) \right)^2 \right] \] **Step 1: Write the probability density function (pdf)** For a normal distribution \(N(0,\theta)\), the pdf is: \[ f_\theta(x) = \frac{1}{\sqrt{2\pi\theta}} \exp\left( -\frac{x^2}{2\theta} \right) \] **Step 2: Compute the log-likelihood function** \[ \ln f_\theta(x) = -\frac{1}{2} \ln(2\pi\theta) - \frac{x^2}{2\theta} \] **Step 3: Compute the derivative with respect to \(\theta\)** \[ \frac{\partial}{\partial \theta} \ln f_\theta(x) = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2} \] **Step 4: Compute the square of the derivative** \[ \left( \frac{\partial}{\partial \theta} \ln f_\theta(x) \right)^2 = \left( -\frac{1}{2\theta} + \frac{x^2}{2\theta^2} \right)^2 \] Expanding the square: \[ \left( \frac{\partial}{\partial \theta} \ln f_\theta(x) \right)^2 = \frac{1}{4\theta^2} - \frac{x^2}{2\theta^3} + \frac{x^4}{4\theta^4} \] **Step 5: Compute the expected value** We need to calculate \(I(\theta) = E\left[ \left( \frac{\partial}{\partial \theta} \ln f_\theta(X) \right)^2 \right]\), where \(X \sim N(0,\theta)\). First, find the moments of \(X\): - \(E[X^2] = \theta\) - \(E[X^4] = 3\theta^2\) Now, compute the expectation: \[ I(\theta) = \frac{1}{4\theta^2} - \frac{1}{2\theta^3} E[X^2] + \frac{1}{4\theta^4} E[X^4] \] Substitute the moments: \[ I(\theta) = \frac{1}{4\theta^2} - \frac{1}{2\theta^3} (\theta) + \frac{1}{4\theta^4} (3\theta^2) \] Simplify: \[ I(\theta) = \frac{1}{4\theta^2} - \frac{1}{2\theta^2} + \frac{3}{4\theta^2} \] Combine like terms: \[ I(\theta) = \left( \frac{1}{4} - \frac{1}{2} + \frac{3}{4} \right) \frac{1}{\theta^2} = \frac{1}{2\theta^2} \] **Conclusion:** The Fisher information for the Gaussian distribution \(N(0,\theta)\) is: \[ I(\theta) = \frac{1}{2\theta^2} \] Comparing this with the given options, we see that option (d) matches: **Option (d): \(0.5\theta^{-2}\)** **Answer:** **Option (d): 0.5 × θ⁻²** |
(d) |
| GPT-4o | To find the Fisher information for the Gaussian distribution family \( f_\theta(x) = N(0, \theta) \), we need to calculate the Fisher information \( I(\theta) \). The Fisher information for a parameter \(\theta\) is given by: \[ I(\theta) = \mathbb{E} \left[ \left( \frac{\partial}{\partial \theta} \ln f_\theta(X) \right)^2 \right], \] where \( X \sim N(0, \theta) \). First, let's write down the probability density function of the Gaussian distribution \( N(0, \theta) \): \[ f_\theta(x) = \frac{1}{\sqrt{2\pi\theta}} \exp\left( -\frac{x^2}{2\theta} \right). \] Next, we compute the log-likelihood function: \[ \ln f_\theta(x) = \ln \left( \frac{1}{\sqrt{2\pi\theta}} \right) + \ln \left( \exp\left( -\frac{x^2}{2\theta} \right) \right). \] Simplifying, we get: \[ \ln f_\theta(x) = -\frac{1}{2} \ln (2\pi\theta) - \frac{x^2}{2\theta}. \] Now, we need to find the derivative of the log-likelihood function with respect to \(\theta\): \[ \frac{\partial}{\partial \theta} \ln f_\theta(x) = \frac{\partial}{\partial \theta} \left( -\frac{1}{2} \ln (2\pi\theta) - \frac{x^2}{2\theta} \right). \] Computing the derivatives, we get: \[ \frac{\partial}{\partial \theta} \left( -\frac{1}{2} \ln (2\pi\theta) \right) = -\frac{1}{2} \cdot \frac{1}{\theta}, \] and \[ \frac{\partial}{\partial \theta} \left( -\frac{x^2}{2\theta} \right) = \frac{x^2}{2\theta^2}. \] Thus, \[ \frac{\partial}{\partial \theta} \ln f_\theta(x) = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2}. \] Next, we square this derivative: \[ \left( \frac{\partial}{\partial \theta} \ln f_\theta(x) \right)^2 = \left( -\frac{1}{2\theta} + \frac{x^2}{2\theta^2} \right)^2. \] Expanding the square: \[ \left( -\frac{1}{2\theta} + \frac{x^2}{2\theta^2} \right)^2 = \left( \frac{1}{4\theta^2} - \frac{x^2}{2\theta^3} + \frac{x^4}{4\theta^4} \right). \] Now, take the expectation with respect to \( X \sim N(0, \theta) \). Note that for a Gaussian distribution \( X \sim N(0, \theta) \), we have \(\mathbb{E}[X^2] = \theta\) and \(\mathbb{E}[X^4] = 3\theta^2\). So, we compute: \[ \mathbb{E} \left[ \left( \frac{\partial}{\partial \theta} \ln f_\theta(x) \right)^2 \right] = \mathbb{E} \left[ \frac{1}{4\theta^2} - \frac{X^2}{2\theta^3} + \frac{X^4}{4\theta^4} \right]. \] Taking the expectation: \[ \mathbb{E} \left[ \frac{1}{4\theta^2} \right] = \frac{1}{4\theta^2}, \] \[ \mathbb{E} \left[ -\frac{X^2}{2\theta^3} \right] = -\frac{1}{2\theta^3} \mathbb{E}[X^2] = -\frac{1}{2\theta^3} \cdot \theta = -\frac{1}{2\theta^2}, \] \[ \mathbb{E} \left[ \frac{X^4}{4\theta^4} \right] = \frac{1}{4\theta^4} \math |
(d) |
| llama3.1-70B-Instruct | To find the Fisher information for the Gaussian distribution family \(f_\theta(x) = N(0, \theta)\), we first need to understand what the 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. For a parameter \(\theta\), the Fisher information is defined as: \[I(\theta) = E\left[\left(\frac{\partial \ln f_\theta(X)}{\partial \theta}\right)^2\right]\] where \(E\) denotes the expected value, and \(f_\theta(X)\) is the probability density function of the random variable \(X\) given the parameter \(\theta\). For the Gaussian distribution \(N(0, \theta)\), the probability density function is: \[f_\theta(x) = \frac{1}{\sqrt{2\pi\theta}} \exp\left(-\frac{x^2}{2\theta}\right)\] Taking the natural logarithm of \(f_\theta(x)\), we get: \[\ln f_\theta(x) = -\frac{1}{2}\ln(2\pi\theta) - \frac{x^2}{2\theta}\] Now, we differentiate \(\ln f_\theta(x)\) with respect to \(\theta\): \[\frac{\partial \ln f_\theta(x)}{\partial \theta} = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2}\] To find the Fisher information, we square this derivative and take the expected value. However, we can simplify the process by recognizing that the expected value of \(X^2\) for a Gaussian distribution \(N(0, \theta)\) is \(\theta\). Thus, we can directly compute the expected value of the squared derivative: \[E\left[\left(\frac{\partial \ln f_\theta(X)}{\partial \theta}\right)^2\right] = E\left[\left(-\frac{1}{2\theta} + \frac{X^2}{2\theta^2}\right)^2\right]\] \[= E\left[\frac{1}{4\theta^2} - \frac{X^2}{2\theta^3} + \frac{X^4}{4\theta^4}\right]\] Given \(E[X^2] = \theta\) and \(E[X^4] = 3\theta^2\) for a Gaussian distribution, we substitute these values: \[= \frac{1}{4\theta^2} - \frac{\theta}{2\theta^3} + \frac{3\theta^2}{4\theta^4}\] \[= \frac{1}{4\theta^2} - \frac{1}{2\theta^2} + \frac{3}{4\theta^2}\] \[= \frac{1}{2\theta^2}\] Therefore, the Fisher information for the Gaussian distribution family \(f_\theta(x) = N(0, \theta)\) is \(\frac{1}{2\theta^2}\), which corresponds to option (d). |
(d) |
| llama3.1-405B-Instruct | The Fisher information for a parameter $\theta$ is defined as: $$I(\theta) = E\left[\left(\frac{\partial \log f_\theta(X)}{\partial \theta}\right)^2\right]$$ For the Gaussian distribution family $f_\theta(x) = N(0, \theta)$, we have: $$f_\theta(x) = \frac{1}{\sqrt{2\pi\theta}} \exp\left(-\frac{x^2}{2\theta}\right)$$ Taking the logarithm of the density function, we get: $$\log f_\theta(x) = -\frac{1}{2}\log(2\pi\theta) - \frac{x^2}{2\theta}$$ Taking the derivative of the log-density with respect to $\theta$, we get: $$\frac{\partial \log f_\theta(x)}{\partial \theta} = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2}$$ Now, we need to compute the expected value of the squared derivative: $$I(\theta) = E\left[\left(-\frac{1}{2\theta} + \frac{x^2}{2\theta^2}\right)^2\right]$$ Using the fact that $E[X^2] = \theta$ for a Gaussian distribution with mean 0 and variance $\theta$, we get: $$I(\theta) = E\left[\frac{1}{4\theta^2} - \frac{x^2}{2\theta^3} + \frac{x^4}{4\theta^4}\right]$$ $$I(\theta) = \frac{1}{4\theta^2} - \frac{\theta}{2\theta^3} + \frac{3\theta^2}{4\theta^4}$$ Simplifying, we get: $$I(\theta) = \frac{1}{4\theta^2} - \frac{1}{2\theta^2} + \frac{3}{4\theta^2}$$ $$I(\theta) = \frac{1}{2\theta^2}$$ Therefore, the correct answer is: (d) $0.5\theta^{-2}$ |
(d) |
| DeepSeek-Coder-V2 | To find the Fisher information for the Gaussian distribution family \( f_\theta(x) = N(0, \theta) \), we need to follow these steps: 1. **Identify the probability density function (PDF):** The PDF of the Gaussian distribution with mean 0 and variance \(\theta\) is: \[ f_\theta(x) = \frac{1}{\sqrt{2\pi\theta}} \exp\left(-\frac{x^2}{2\theta}\right) \] 2. **Compute the log-likelihood function:** The log-likelihood function for a single observation \(x\) is: \[ \log f_\theta(x) = \log\left(\frac{1}{\sqrt{2\pi\theta}}\right) - \frac{x^2}{2\theta} \] Simplifying, we get: \[ \log f_\theta(x) = -\frac{1}{2}\log(2\pi\theta) - \frac{x^2}{2\theta} \] 3. **Compute the derivative of the log-likelihood function with respect to \(\theta\):** \[ \frac{\partial}{\partial\theta} \log f_\theta(x) = \frac{\partial}{\partial\theta} \left( -\frac{1}{2}\log(2\pi\theta) - \frac{x^2}{2\theta} \right) \] Breaking it down: \[ \frac{\partial}{\partial\theta} \left( -\frac{1}{2}\log(2\pi\theta) \right) = -\frac{1}{2} \cdot \frac{1}{2\pi\theta} \cdot 2\pi = -\frac{1}{2\theta} \] \[ \frac{\partial}{\partial\theta} \left( -\frac{x^2}{2\theta} \right) = \frac{x^2}{2\theta^2} \] Combining these, we get: \[ \frac{\partial}{\partial\theta} \log f_\theta(x) = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2} \] 4. **Compute the second derivative of the log-likelihood function with respect to \(\theta\):** \[ \frac{\partial^2}{\partial\theta^2} \log f_\theta(x) = \frac{\partial}{\partial\theta} \left( -\frac{1}{2\theta} + \frac{x^2}{2\theta^2} \right) \] Breaking it down: \[ \frac{\partial}{\partial\theta} \left( -\frac{1}{2\theta} \right) = \frac{1}{2\theta^2} \] \[ \frac{\partial}{\partial\theta} \left( \frac{x^2}{2\theta^2} \right) = -\frac{x^2}{2\theta^3} \] Combining these, we get: \[ \frac{\partial^2}{\partial\theta^2} \log f_\theta(x) = \frac{1}{2\theta^2} - \frac{x^2}{2\theta^3} \] 5. **Compute the Fisher information:** The Fisher information is the expected value of the second derivative of the log-likelihood function: \[ I(\theta) = -E\left[ \frac{\partial^2}{\partial\theta^2} \log f_\theta(x) \right] \] Substituting the second derivative: \[ I(\theta) = -E\left[ \frac{1}{2\theta^2} - \frac{x^2}{2\theta^3} \right] \] Since \(E[x^2] = \theta\) for the Gaussian distribution \(N(0, \theta)\): \[ I(\theta) = -\left( \frac{1}{2\theta^2} - \frac{\theta}{2\theta^3} \right) \] Simplifying: \[ I(\theta) = -\left( \frac{1}{2\theta^2} - \frac{1}{2\theta^2} \right) = 0 \] This is incorrect. Let's re-evaluate the expectation: \[ I(\theta) = E\left[ \left( \frac{x^2}{2\theta^2} - \frac{1}{2\theta} \right)^2 \right] \] \[ I(\theta) = E\left[ \frac{x^4}{4\theta^4} - \frac{x^2}{2\theta^3} + \frac{1}{4\theta^2} \right] \] Since \(E[x^4] = 3\theta^2\) and \(E[x^2] = \theta\): \[ I(\theta) = \frac{3\theta^2}{4\theta^4} - \frac{\theta}{2\theta^3} + \frac{1}{4\theta^2} \] Simplifying: \[ I(\theta) = \frac{3}{4\theta^2} - \frac{1}{2\theta^2} + \frac{1}{4\theta^2} = \frac{1}{2\theta^2} \] Thus, the correct Fisher information for the Gaussian distribution family \( f_\theta(x) = N(0, \theta) \) is: \[ \boxed{D} \] |
(d) |
| Qwen2-72B-Instruct | The Fisher information for a parametric family of distributions is defined as the expected value of the negative second derivative of the log-likelihood function with respect to the parameter. For the Gaussian distribution family $f_\theta(x) = N(0, \theta)$, the log-likelihood function is given by: $$ \log L(\theta | x) = -\frac{n}{2} \log(2\pi) - \frac{n}{2} \log(\theta) - \frac{1}{2\theta} \sum_{i=1}^n x_i^2 $$ where $n$ is the sample size and $x_i$ are the observations. Taking the second derivative of the log-likelihood function with respect to $\theta$ gives: $$ \frac{\partial^2}{\partial \theta^2} \log L(\theta | x) = \frac{n}{2\theta^2} - \frac{1}{\theta^3} \sum_{i=1}^n x_i^2 $$ Taking the expected value of the negative second derivative gives the Fisher information: $$ I(\theta) = E\left[ -\frac{\partial^2}{\partial \theta^2} \log L(\theta | x) \right] = \frac{n}{2\theta^2} - \frac{1}{\theta^3} E\left[ \sum_{i=1}^n x_i^2 \right] $$ Since $x_i \sim N(0, \theta)$, we have $E[x_i^2] = \theta$, and therefore: $$ I(\theta) = \frac{n}{2\theta^2} - \frac{n}{\theta^2} = \frac{n}{2\theta^2} $$ For a single observation ($n=1$), the Fisher information is: $$ I(\theta) = \frac{1}{2\theta^2} $$ Therefore, the correct answer is (d) $0.5\theta^{-2}$. |
(d) |
| baichuan4 | The Fisher information for a Gaussian distribution with mean $\mu$ and variance $\theta$ is given by: $$I(\theta) = \mathbb{E}\left[\left(\frac{\partial}{\partial \theta} \log f_\theta(X)\right)^2\right]$$ For the given problem, we have a Gaussian distribution with mean $0$ and variance $\theta$, i.e., $f_\theta(x) = N(0, \theta)$. The log-likelihood of this distribution is: $$\log f_\theta(x) = -\frac{1}{2}\log(2\pi\theta) - \frac{x^2}{2\theta}$$ Taking the derivative with respect to $\theta$, we get: $$\frac{\partial}{\partial \theta} \log f_\theta(x) = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2}$$ Squaring this derivative and taking the expectation, we have: $$I(\theta) = \mathbb{E}\left[\left(-\frac{1}{2\theta} + \frac{X^2}{2\theta^2}\right)^2\right]$$ Since $X \sim N(0, \theta)$, we know that $\mathbb{E}[X^2] = \theta$. Therefore, $$I(\theta) = \mathbb{E}\left[\left(-\frac{1}{2\theta} + \frac{\theta}{2\theta^2}\right)^2\right] = \mathbb{E}\left[\left(-\frac{1}{2\theta} + \frac{1}{2\theta}\right)^2\right] = 0$$ However, this is not one of the given options. Let's calculate it again more carefully. $$I(\theta) = \mathbb{E}\left[\left(-\frac{1}{2\theta} + \frac{X^2}{2\theta^2}\right)^2\right] = \mathbb{E}\left[\frac{1}{4\theta^2} - \frac{X^2}{2\theta^3} + \frac{X^4}{4\theta^4}\right]$$ Now, using the fact that $\mathbb{E}[X^2] = \theta$ and $\mathbb{E}[X^4] = 3\theta^2$ for a standard normal distribution, we get: $$I(\theta) = \frac{1}{4\theta^2} - \frac{\theta}{2\theta^3} + \frac{3\theta^2}{4\theta^4} = \frac{1}{4\theta^2} - \frac{1}{2\theta^2} + \frac{3}{4\theta^2} = \frac{1}{2\theta^2}$$ Therefore, the correct option is (d) $0.5\theta^{-2}$. |
(d) |
| doubao | The Fisher information for a Gaussian distribution $N(0,\theta)$ is $\frac{1}{2\theta}$. So the answer is (c). | (d) |
| minimax | The Fisher information $I(\theta)$ for a parameter $\theta$ of a probability distribution is given by the expected value of the squared derivative of the log-likelihood with respect to the parameter $\theta$. For a Gaussian distribution $N(0,\theta)$, the probability density function (pdf) is: $$ f_\theta(x) = \frac{1}{\sqrt{2\pi\theta}} \exp\left(-\frac{x^2}{2\theta}\right) $$ where $\theta$ is the variance of the distribution. To find the Fisher information, we first need to compute the log-likelihood function $\ell(\theta)$: $$ \ell(\theta) = \log(f_\theta(x)) = \log\left(\frac{1}{\sqrt{2\pi\theta}}\right) - \frac{x^2}{2\theta} $$ Now, we take the derivative of the log-likelihood with respect to $\theta$: $$ \frac{d\ell(\theta)}{d\theta} = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2} $$ The Fisher information is the expected value of the square of this derivative: $$ I(\theta) = E\left[\left(-\frac{1}{2\theta} + \frac{x^2}{2\theta^2}\right)^2\right] $$ Since $x$ is normally distributed with mean $0$ and variance $\theta$, the expected value of $x^2$ is $\theta$ (because the variance of a normal distribution is equal to the expected value of $x^2$ minus the square of the expected value of $x$, and since the mean is zero, the variance is just the expected value of $x^2$). Now, we can simplify the expression for the Fisher information: $$ I(\theta) = E\left[\left(-\frac{1}{2\theta} + \frac{\theta}{2\theta^2}\right)^2\right] $$ $$ I(\theta) = E\left[\left(-\frac{1}{2\theta} + \frac{1}{2\theta}\right)^2\right] $$ $$ I(\theta) = E\left[0^2\right] $$ $$ I(\theta) = 0 $$ However, this result seems counterintuitive because we know that the Fisher information should not be zero for a parameter that we are estimating. The mistake here is that we have not properly accounted for the fact that the variance of $x^2$ is not just $\theta^2$; it also includes the term $2\theta^2$ due to the fact that the variance of a squared normal random variable is not simply the square of the variance of the original variable. Let's correct this by computing the Fisher information properly: $$ I(\theta) = E\left[\left(-\frac{1}{2\theta} + \frac{x^2}{2\theta^2}\right)^2\right] $$ $$ I(\theta) = E\left[\frac{1}{4\theta^2} - \frac{x^2}{2\theta^3} + \frac{x^4}{4\theta^4}\right] $$ We know that $E[x^2] = \theta$ and $E[x^4] = 3\theta^2$ for a standard normal distribution (since the fourth central moment of a standard normal distribution is $3\sigma^4$ where $\sigma^2$ is the variance). Substituting these values in, we get: $$ I(\theta) = \frac{1}{4\theta^2} - \frac{\theta}{2\theta^3} + \frac{3\theta^2}{4\theta^4} $$ $$ I(\theta) = \frac{1}{4\theta^2} - \frac{1}{2\theta^2} + \frac{3}{4\theta^2} $$ $$ I(\theta) = \frac{1 + 3 - 2}{4\theta^2} $$ $$ I(\theta) = \frac{2}{4\theta^2} $$ $$ I(\theta) = \frac{1}{2\theta^2} $$ Therefore, the correct answer is (d) $0.5\theta^{-2}$. |
(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. For a family of probability distributions $f_\theta(x)$, the Fisher information $I(\theta)$ is defined as: $$I(\theta) = \mathbb{E}\left[\left(\frac{\partial}{\partial\theta}\log f_\theta(x)\right)^2\right]$$ where $\mathbb{E}$ denotes the expectation. For the Gaussian distribution family $f_\theta(x) = N(0, \theta)$, where $\theta$ is the variance, the probability density function is given by: $$f_\theta(x) = \frac{1}{\sqrt{2\pi\theta}}e^{-\frac{x^2}{2\theta}}$$ To find the Fisher information, we first need to compute the log-likelihood function $\log f_\theta(x)$: $$\log f_\theta(x) = \log\left(\frac{1}{\sqrt{2\pi\theta}}\right) - \frac{x^2}{2\theta}$$ Now, we take the derivative of the log-likelihood with respect to $\theta$: $$\frac{\partial}{\partial\theta}\log f_\theta(x) = \frac{\partial}{\partial\theta}\left(-\frac{1}{2}\log(2\pi\theta) - \frac{x^2}{2\theta}\right) = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2}$$ Next, we square this derivative: $$\left(\frac{\partial}{\partial\theta}\log f_\theta(x)\right)^2 = \left(-\frac{1}{2\theta} + \frac{x^2}{2\theta^2}\right)^2$$ Now we take the expectation of this squared derivative: $$I(\theta) = \mathbb{E}\left[\left(-\frac{1}{2\theta} + \frac{x^2}{2\theta^2}\right)^2\right]$$ Since $x \sim N(0, \theta)$, the expected value of $x^2$ is $\theta$. We can compute the expectation as follows: $$I(\theta) = \mathbb{E}\left[\left(-\frac{1}{2\theta} + \frac{x^2}{2\theta^2}\right)^2\right] = \mathbb{E}\left[\frac{1}{4\theta^2} - \frac{x^2}{2\theta^3} + \frac{x^4}{4\theta^4}\right]$$ $$= \frac{1}{4\theta^2} - \frac{\theta}{2\theta^3} + \frac{\mathbb{E}[x^4]}{4\theta^4}$$ The fourth moment of a normal distribution $x^4$ is $3\theta^2$ for $x \sim N(0, \theta)$, so: $$I(\theta) = \frac{1}{4\theta^2} - \frac{1}{2\theta} + \frac{3\theta^2}{4\theta^4}$$ $$= \frac{1}{4\theta^2} - \frac{1}{2\theta} + \frac{3}{4\theta^2}$$ $$= \frac{1}{2\theta^2}$$ Therefore, the correct option is (d) $\boxed{0.5\theta^{-2}}$. |
(d) |
| wenxinyiyan | The Fisher information for a parametric family of distributions is defined as $$I(\theta) = E\left[\left(\frac{\partial}{\partial\theta} \log f_\theta(X)\right)^2\right]$$ where $f_\theta(x)$ is the probability density function of the distribution. For the Gaussian distribution $N(0, \theta)$ with mean 0 and variance $\theta$, the probability density function is $$f_\theta(x) = \frac{1}{\sqrt{2\pi\theta}} e^{-\frac{x^2}{2\theta}}$$ The log-likelihood function is $$\log f_\theta(x) = -\frac{1}{2} \log(2\pi\theta) - \frac{x^2}{2\theta}$$ Differentiating with respect to $\theta$, we get $$\frac{\partial}{\partial\theta} \log f_\theta(x) = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2}$$ Squaring this, we get $$\left(\frac{\partial}{\partial\theta} \log f_\theta(x)\right)^2 = \frac{1}{4\theta^2} - \frac{x^2}{2\theta^3} + \frac{x^4}{4\theta^4}$$ To find the Fisher information, we need to take the expectation of this with respect to $X$. Since $E[X^2] = \theta$ and $E[X^4] = 3\theta^2$ for a Gaussian distribution with mean 0 and variance $\theta$, we have $$I(\theta) = E\left[\frac{1}{4\theta^2} - \frac{X^2}{2\theta^3} + \frac{X^4}{4\theta^4}\right] = \frac{1}{4\theta^2} - \frac{1}{2\theta^2} + \frac{3\theta^2}{4\theta^4} = \frac{1}{2\theta^2}$$ Therefore, the correct option is (d) $0.5\theta^{-2}$. |
(d) |
| zhipu | To determine the Fisher information for the Gaussian distribution family \( f_\theta(x) = N(0, \theta) \), where \( N(0, \theta) \) denotes a normal distribution with mean 0 and variance \( \theta \), we need to follow these steps: 1. **Write down the probability density function (PDF) of the Gaussian distribution:** \[ f_\theta(x) = \frac{1}{\sqrt{2\pi\theta}} \exp\left(-\frac{x^2}{2\theta}\right) \] 2. **Compute the natural logarithm of the PDF:** \[ \ell_\theta(x) = \log f_\theta(x) = -\frac{1}{2} \log(2\pi\theta) - \frac{x^2}{2\theta} \] 3. **Differentiate the log-likelihood function with respect to \( \theta \):** \[ \frac{\partial \ell_\theta(x)}{\partial \theta} = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2} \] 4. **Compute the expected value of the square of the derivative of the log-likelihood function:** \[ \mathbb{E}\left[\left(\frac{\partial \ell_\theta(x)}{\partial \theta}\right)^2\right] = \mathbb{E}\left[\left(-\frac{1}{2\theta} + \frac{x^2}{2\theta^2}\right)^2\right] \] Expanding the square: \[ \mathbb{E}\left[\left(-\frac{1}{2\theta} + \frac{x^2}{2\theta^2}\right)^2\right] = \mathbb{E}\left[\frac{1}{4\theta^2} - \frac{x^2}{2\theta^3} + \frac{x^4}{4\theta^4}\right] \] 5. **Evaluate each expectation term separately:** - \(\mathbb{E}\left[\frac{1}{4\theta^2}\right] = \frac{1}{4\theta^2}\) - \(\mathbb{E}\left[-\frac{x^2}{2\theta^3}\right] = -\frac{1}{2\theta^3} \mathbb{E}[x^2]\) - \(\mathbb{E}\left[\frac{x^4}{4\theta^4}\right] = \frac{1}{4\theta^4} \mathbb{E}[x^4]\) For a Gaussian distribution \( N(0, \theta) \): - \(\mathbb{E}[x^2] = \theta\) - \(\mathbb{E}[x^4] = 3\theta^2\) (using the fourth moment of a normal distribution) Substituting these expectations: \[ \mathbb{E}\left[\frac{1}{4\theta^2}\right] = \frac{1}{4\theta^2} \] \[ \mathbb{E}\left[-\frac{x^2}{2\theta^3}\right] = -\frac{\theta}{2\theta^3} = -\frac{1}{2\theta^2} \] \[ \mathbb{E}\left[\frac{x^4}{4\theta^4}\right] = \frac{3\theta^2}{4\theta^4} = \frac{3}{4\theta^2} \] 6. **Sum these terms to get the Fisher information:** \[ \mathbb{E}\left[\left(\frac{\partial \ell_\theta(x)}{\partial \theta}\right)^2\right] = \frac{1}{4\theta^2} - \frac{1}{2\theta^2} + \frac{3}{4\theta^2} = \frac{1 - 2 + 3}{4\theta^2} = \frac{2}{4\theta^2} = \frac{1}{2\theta^2} \] Thus, the Fisher information for the Gaussian distribution family \( f_\theta(x) = N(0, \theta) \) is \( \frac{1}{2\theta^2} \). The correct option is: (d) \( 0.5\theta^{-2} \). |
(d) |
| gpto1 | To find the Fisher information \(I(\theta)\) for the Gaussian distribution family \(f_\theta(x) = N(0,\theta)\), we start by recalling the definition of Fisher information for a parameter \(\theta\): \[ I(\theta) = E_\theta\left[ \left( \frac{\partial}{\partial \theta} \ln f_\theta(X) \right)^2 \right] \] **Step 1: Write the probability density function (pdf)** For a normal distribution \(N(0,\theta)\), the pdf is: \[ f_\theta(x) = \frac{1}{\sqrt{2\pi\theta}} \exp\left( -\frac{x^2}{2\theta} \right) \] **Step 2: Compute the log-likelihood function** \[ \ln f_\theta(x) = -\frac{1}{2} \ln(2\pi\theta) - \frac{x^2}{2\theta} \] **Step 3: Compute the derivative with respect to \(\theta\)** \[ \frac{\partial}{\partial \theta} \ln f_\theta(x) = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2} \] **Step 4: Compute the square of the derivative** \[ \left( \frac{\partial}{\partial \theta} \ln f_\theta(x) \right)^2 = \left( -\frac{1}{2\theta} + \frac{x^2}{2\theta^2} \right)^2 \] Expanding the square: \[ \left( \frac{\partial}{\partial \theta} \ln f_\theta(x) \right)^2 = \frac{1}{4\theta^2} - \frac{x^2}{2\theta^3} + \frac{x^4}{4\theta^4} \] **Step 5: Compute the expected value** We need to calculate \(I(\theta) = E\left[ \left( \frac{\partial}{\partial \theta} \ln f_\theta(X) \right)^2 \right]\), where \(X \sim N(0,\theta)\). First, find the moments of \(X\): - \(E[X^2] = \theta\) - \(E[X^4] = 3\theta^2\) Now, compute the expectation: \[ I(\theta) = \frac{1}{4\theta^2} - \frac{1}{2\theta^3} E[X^2] + \frac{1}{4\theta^4} E[X^4] \] Substitute the moments: \[ I(\theta) = \frac{1}{4\theta^2} - \frac{1}{2\theta^3} (\theta) + \frac{1}{4\theta^4} (3\theta^2) \] Simplify: \[ I(\theta) = \frac{1}{4\theta^2} - \frac{1}{2\theta^2} + \frac{3}{4\theta^2} \] Combine like terms: \[ I(\theta) = \left( \frac{1}{4} - \frac{1}{2} + \frac{3}{4} \right) \frac{1}{\theta^2} = \frac{1}{2\theta^2} \] **Conclusion:** The Fisher information for the Gaussian distribution \(N(0,\theta)\) is: \[ I(\theta) = \frac{1}{2\theta^2} \] Comparing this with the given options, we see that option (d) matches: **Option (d): \(0.5\theta^{-2}\)** **Answer:** **Option (d): 0.5 × θ⁻²** |
(d) |
| deepseekv3 | To find the Fisher information for the Gaussian distribution family \( f_\theta(x) = N(0, \theta) \), we follow these steps: ### Step 1: Write the probability density function (PDF) The PDF of a Gaussian distribution \( N(0, \theta) \) is: \[ f_\theta(x) = \frac{1}{\sqrt{2\pi\theta}} \exp\left(-\frac{x^2}{2\theta}\right). \] ### Step 2: Compute the log-likelihood function The log-likelihood function is: \[ \ln f_\theta(x) = -\frac{1}{2} \ln(2\pi\theta) - \frac{x^2}{2\theta}. \] ### Step 3: Compute the first derivative with respect to \( \theta \) Differentiate \( \ln f_\theta(x) \) with respect to \( \theta \): \[ \frac{\partial}{\partial \theta} \ln f_\theta(x) = -\frac{1}{2\theta} + \frac{x^2}{2\theta^2}. \] ### Step 4: Compute the second derivative with respect to \( \theta \) Differentiate the first derivative with respect to \( \theta \): \[ \frac{\partial^2}{\partial \theta^2} \ln f_\theta(x) = \frac{1}{2\theta^2} - \frac{x^2}{\theta^3}. \] ### Step 5: Compute the expected value of the second derivative The Fisher information is the negative of the expected value of the second derivative: \[ I(\theta) = -E\left[\frac{\partial^2}{\partial \theta^2} \ln f_\theta(x)\right] = -E\left[\frac{1}{2\theta^2} - \frac{x^2}{\theta^3}\right]. \] Since \( E[x^2] = \theta \) for \( N(0, \theta) \), we have: \[ I(\theta) = -\left(\frac{1}{2\theta^2} - \frac{\theta}{\theta^3}\right) = -\left(\frac{1}{2\theta^2} - \frac{1}{\theta^2}\right) = \frac{1}{2\theta^2}. \] ### Step 6: Match the result with the given options The Fisher information is \( I(\theta) = \frac{1}{2\theta^2} \), which corresponds to option (d). ### Final Answer: The correct option is **(d)** \( 0.5\theta^{-2} \). |
(d) |