| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The conditional distribution of a standard Brownian motion \(W(s)\) given \(W(t) = b\) for \(0 < s < t\) is normally distributed with mean \(\frac{s}{t}b\) and variance \((s/t)(1-s/t)\). Given \(W(1) = 2\) and we want to find the conditional distribution of \(W(0.5)\): - Mean: \(\frac{0.5}{1} \cdot 2 = 1\) - Variance: \((0.5/1)(1-0.5/1) = 0.25\) So, the conditional PDF of \(W(0.5)\) given \(W(1) = 2\) is normally distributed with mean 1 and variance 0.25. Therefore, the answer is \([1, 0.25]\). |
[1.0, 0.25] |
| deepseekr1 | The conditional PDF of \( W\left(\frac{1}{2}\right) \) given \( W(1) = 2 \) is a normal distribution. Using properties of Brownian motion and conditional expectations for jointly normal variables: - **Mean**: The conditional expectation is derived as \( \mathbb{E}[W(s) \mid W(t) = 2] = \frac{s}{t} \cdot 2 = \frac{1/2}{1} \cdot 2 = 1 \). - **Variance**: The conditional variance is \( \text{Var}(W(s) \mid W(t) = 2) = s - \frac{s^2}{t} = \frac{1}{2} - \frac{(1/2)^2}{1} = \frac{1}{4} \). Thus, the mean is **1** and the variance is **1/4**. \boxed{[1, \frac{1}{4}]} |
[1.0, 0.25] |
| gpto1 | To solve this problem, we need to find the conditional probability density function (PDF) of \( W(s) \) given that \( W(t) = 2 \), where \( W(t) \) is a standard Brownian motion, and \( 0 < s < t \). Specifically, we need to compute the mean and variance of this conditional distribution. **Step 1: Understand the Properties of Brownian Motion** - **Standard Brownian Motion** \( W(t) \) is a continuous-time stochastic process with: - \( W(0) = 0 \) - Independent increments: \( W(t) - W(s) \) is independent of \( W(u) \) for all \( u \leq s \) - Normally distributed increments: \( W(t) - W(s) \sim N(0, t - s) \) - Continuous paths **Step 2: Consider the Joint Distribution of \( W(s) \) and \( W(t) \)** Because \( W(s) \) and \( W(t) \) are jointly normally distributed (since any linear combination of normally distributed random variables is also normally distributed), we can use properties of the multivariate normal distribution to find the conditional distribution. **Step 3: Compute the Covariance and Variances** - **Mean Values:** - \( E[W(s)] = 0 \) - \( E[W(t)] = 0 \) - **Variances:** - \( \operatorname{Var}(W(s)) = s \) - \( \operatorname{Var}(W(t)) = t \) - **Covariance:** - Since \( W(s) \leq W(t) \), the covariance is \( \operatorname{Cov}(W(s), W(t)) = s \) - This is because \( \operatorname{Cov}(W(s), W(t)) = E[W(s)W(t)] - E[W(s)]E[W(t)] = E[W(s)W(t)] \) - And \( E[W(s)W(t)] = E[W(s)W(s) + W(s)(W(t) - W(s))] = s + 0 = s \) **Step 4: Use the Conditional Mean and Variance Formulas** For jointly normally distributed variables \( X \) and \( Y \), the conditional distribution \( X | Y = y \) is normal with: - **Conditional Mean:** \[ E[X | Y = y] = E[X] + \frac{\operatorname{Cov}(X, Y)}{\operatorname{Var}(Y)} (y - E[Y]) \] - **Conditional Variance:** \[ \operatorname{Var}(X | Y) = \operatorname{Var}(X) - \frac{[\operatorname{Cov}(X, Y)]^2}{\operatorname{Var}(Y)} \] **Step 5: Plug in the Values \( s = \frac{1}{2} \) and \( t = 1 \)** - **Conditional Mean:** \[ E[W(1/2) | W(1) = 2] = 0 + \frac{(1/2)}{1} \times (2 - 0) = 1 \] - **Conditional Variance:** \[ \operatorname{Var}(W(1/2) | W(1) = 2) = \frac{1}{2} - \frac{(1/2)^2}{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \] **Step 6: State the Result** The conditional distribution of \( W(1/2) \) given \( W(1) = 2 \) is normal with mean \( 1 \) and variance \( \frac{1}{4} \). **Answer: [1, ¼] |
[1.0, 0.25] |
| GPT-4o | To find the conditional probability density function (PDF) of \( W(s = 1/2) \) given that \( W(t = 1) = 2 \) for a standard Brownian motion \( W(t) \), we can use the properties of Brownian motion and conditional distributions. ### Solution: 1. **Brownian Motion Properties**: - \( W(t) \) is a standard Brownian motion which means \( W(t) \sim \mathcal{N}(0, t) \). - For \( 0 < s < t \), the conditional distribution of \( W(s) \) given \( W(t) \) is normally distributed. 2. **Conditional Distribution**: The conditional distribution of \( W(s) \) given \( W(t) = x \) is: \[ W(s) | W(t) = x \sim \mathcal{N}\left( \frac{s}{t} x, \left(1 - \frac{s}{t}\right)s \right). \] Here, \( s = \frac{1}{2} \) and \( t = 1 \), and \( x = 2 \). 3. **Calculate Mean and Variance**: - Mean of \( W(1/2) \) given \( W(1) = 2 \): \[ \mathbb{E}[W(1/2) | W(1) = 2] = \frac{1/2}{1} \cdot 2 = 1. \] - Variance of \( W(1/2) \) given \( W(1) = 2 \): \[ \text{Var}(W(1/2) | W(1) = 2) = \left(1 - \frac{1/2}{1}\right) \cdot \frac{1}{2} = \left(1 - \frac{1}{2}\right) \cdot \frac{1}{2} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}. \] 4. **Conditional PDF**: Given that \( W(1/2) | W(1) = 2 \) follows a normal distribution with mean 1 and variance \( \frac{1}{4} \), the conditional PDF is: \[ f_{W(1/2) | W(1) = 2}(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\left( -\frac{(x - \mu)^2}{2 \sigma^2} \right), \] where \( \mu = 1 \) and \( \sigma^2 = \frac{1}{4} \). Simplified: \[ f_{W(1/2) | W(1) = 2}(x) = \frac{1}{\sqrt{2 \pi \cdot \frac{1}{4}}} \exp\left( -\frac{(x - 1)^2}{2 \cdot \frac{1}{4}} \right). \] \[ f_{W(1/2) | W(1) = 2}(x) = \frac{1}{\sqrt{\pi / 2}} \exp\left( -2(x - 1)^2 \right). \] ### Result: Thus, the mean and variance of \( W(s = 1/2) \) given that \( W(t = 1) = 2 \) are: \[ \text{Mean} = 1 \] \[ \text{Variance} = \frac{1}{4} \] Returning the list as requested: \[ [1, \frac{1}{4}] \] |
[1.0, 0.25] |
| llama3.1-70B-Instruct | The conditional PDF of W(s) given W(t) = w is a normal distribution with mean (s/t)w and variance s(1 - s/t). In this case, s = 1/2, t = 1, and w = 2. So, the conditional PDF of W(1/2) given W(1) = 2 is a normal distribution with mean (1/2/1)2 = 1 and variance 1/2(1 - 1/2/1) = 1/4. Therefore, the mean is 1 and the variance is 1/4. The list of [mean, variance] is [1, 1/4]. |
[1.0, 0.25] |
| llama3.1-405B-Instruct | The conditional PDF of W(s = 1/2) given that W(t = 1) = 2 is a normal distribution with mean 1 and variance 1/2. To see this, we can use the fact that the conditional distribution of a multivariate normal distribution is also normal. Specifically, if X and Y are jointly normal with mean vector μ and covariance matrix Σ, then the conditional distribution of X given Y = y is normal with mean μ_X + Σ_XY Σ_YY^(-1) (y - μ_Y) and variance Σ_XX - Σ_XY Σ_YY^(-1) Σ_YX. In this case, we have X = W(1/2) and Y = W(1), and we know that W(t) is a standard Brownian motion, so μ = [0, 0] and Σ = [[1/2, 1/2], [1/2, 1]]. Plugging in these values, we get: μ_X = 0 Σ_XX = 1/2 Σ_XY = 1/2 Σ_YY = 1 Σ_YX = 1/2 So the conditional mean is: μ_X + Σ_XY Σ_YY^(-1) (y - μ_Y) = 0 + (1/2) (1)^(-1) (2 - 0) = 1 And the conditional variance is: Σ_XX - Σ_XY Σ_YY^(-1) Σ_YX = 1/2 - (1/2) (1)^(-1) (1/2) = 1/2 Therefore, the conditional PDF of W(s = 1/2) given that W(t = 1) = 2 is a normal distribution with mean 1 and variance 1/2. The list of [mean, variance] is [1, 1/2]. |
[1.0, 0.25] |
| DeepSeek-Coder-V2 | To solve the problem, we need to find the conditional probability density function (PDF) of \( W(s = 1/2) \) given that \( W(t = 1) = 2 \). We will use the properties of Brownian motion and the conditional distribution formula. Given: - \( W(t) \) is the standard Brownian motion. - \( 0 < s < t \) - \( s = 1/2 \) - \( t = 1 \) - \( W(t = 1) = 2 \) The conditional distribution of \( W(s) \) given \( W(t) \) for \( 0 < s < t \) is normally distributed with: - Mean: \( W(s) + \frac{t - s}{t} (W(t) - W(s)) \) - Variance: \( \frac{s(t - s)}{t} \) Let's calculate the mean and variance step by step. 1. **Mean Calculation:** \[ \text{Mean} = W(s) + \frac{t - s}{t} (W(t) - W(s)) \] Substituting \( s = 1/2 \) and \( t = 1 \): \[ \text{Mean} = W(1/2) + \frac{1 - 1/2}{1} (2 - W(1/2)) \] Simplifying: \[ \text{Mean} = W(1/2) + \frac{1/2}{1} (2 - W(1/2)) \] \[ \text{Mean} = W(1/2) + \frac{1}{2} (2 - W(1/2)) \] \[ \text{Mean} = W(1/2) + 1 - \frac{1}{2} W(1/2) \] \[ \text{Mean} = \frac{1}{2} W(1/2) + 1 \] 2. **Variance Calculation:** \[ \text{Variance} = \frac{s(t - s)}{t} \] Substituting \( s = 1/2 \) and \( t = 1 \): \[ \text{Variance} = \frac{(1/2)(1 - 1/2)}{1} \] Simplifying: \[ \text{Variance} = \frac{(1/2)(1/2)}{1} \] \[ \text{Variance} = \frac{1/4}{1} \] \[ \text{Variance} = \frac{1}{4} \] Therefore, the mean and variance of the conditional distribution of \( W(s = 1/2) \) given \( W(t = 1) = 2 \) are: \[ \text{Mean} = 1 \] \[ \text{Variance} = \frac{1}{4} \] The final answer is: \[ \boxed{[1, \frac{1}{4}]} \] |
[1.0, 0.25] |
| Qwen2-72B-Instruct | The conditional distribution of a standard Brownian motion \(W(s)\) given \(W(t) = b\) for \(0 < s < t\) is normally distributed with mean \(\frac{s}{t}b\) and variance \((s/t)(1-s/t)\). Given \(W(1) = 2\) and we want to find the conditional distribution of \(W(0.5)\): - Mean: \(\frac{0.5}{1} \cdot 2 = 1\) - Variance: \((0.5/1)(1-0.5/1) = 0.25\) So, the conditional PDF of \(W(0.5)\) given \(W(1) = 2\) is normally distributed with mean 1 and variance 0.25. Therefore, the answer is \([1, 0.25]\). |
[1.0, 0.25] |
| baichuan4 | To solve this problem, we need to use the properties of Brownian motion, specifically the fact that a Brownian motion has independent increments and that the increment \(W(t) - W(s)\) for \(t > s\) is normally distributed with mean \(0\) and variance \(t - s\). Given that \(W(1) = 2\), we want to find the conditional probability density function (PDF) of \(W(1/2)\). By the properties of Brownian motion, we know that \(W(1) - W(1/2)\) is normally distributed with mean \(0\) and variance \(1 - 1/2 = 1/2\). The conditional PDF of \(W(1/2)\) given \(W(1) = 2\) is normally distributed with mean \(E[W(1/2) | W(1) = 2]\) and variance \(Var[W(1/2) | W(1) = 2]\). We can calculate the mean as follows: \[E[W(1/2) | W(1) = 2] = E[W(1) - (W(1) - W(1/2)) | W(1) = 2] = 2 - E[W(1) - W(1/2) | W(1) = 2] = 2 - 0 = 2\] For the variance, we have: \[Var[W(1/2) | W(1) = 2] = Var[W(1) - (W(1) - W(1/2)) | W(1) = 2] = Var[W(1) - W(1/2) | W(1) = 2] = 1/2\] So, the mean is \(2\) and the variance is \(1/2\). The list [mean, variance] would be \([2, 1/2]\). |
[1.0, 0.25] |
| doubao | The conditional distribution of W(s) given W(t) is a normal distribution. The mean of this conditional distribution is W(s) = s/t * W(t). In this case, s = 1/2 and t = 1, so the mean is (1/2) * 2 = 1. The variance is s * (t - s) / t = (1/2) * (1 - 1/2) / 1 = 1/4. So the list is [1, 1/4]. | [1.0, 0.25] |
| minimax | To find the conditional probability density function (PDF) of a standard Brownian motion \( W(s) \) at time \( s \) given \( W(t) \) at time \( t \), we can use the properties of Brownian motion. The standard Brownian motion \( W(t) \) has the following properties: 1. \( W(0) = 0 \). 2. \( W(t) \) has independent increments. 3. \( W(t) \) has Gaussian increments with mean 0 and variance \( t \). Given \( W(t = 1) = 2 \), we want to find the conditional PDF of \( W(s) \) at \( s = 1/2 \). The increment \( W(1) - W(1/2) \) is independent of \( W(1/2) \), and since \( W(1) \) is given, we can use the fact that the difference between two normally distributed variables is also normally distributed. Let's denote \( X = W(1/2) \) and \( Y = W(1) \). We know that \( Y = 2 \). The increment \( Y - X \) is normally distributed with mean \( 0 \) and variance \( 1 - 1/2 = 1/2 \), because the variance of the Brownian motion at time \( t \) is \( t \). The joint distribution of \( X \) and \( Y \) is bivariate normal, and the conditional distribution of \( X \) given \( Y \) is also normal. The mean and variance of \( X \) given \( Y \) can be found using the following formulas: Mean of \( X \) given \( Y \): \[ \mu_{X|Y} = \mu_X + \frac{\text{Cov}(X,Y)}{\text{Var}(Y)}(Y - \mu_Y) \] Variance of \( X \) given \( Y \): \[ \text{Var}(X|Y) = \text{Var}(X) - \frac{[\text{Cov}(X,Y)]^2}{\text{Var}(Y)} \] Since \( X \) and \( Y \) are both normally distributed with mean 0 and variances \( 1/2 \) and \( 1 \) respectively, and they are increments of Brownian motion, their covariance is equal to the variance of \( X \), which is \( 1/2 \). So, we have: \[ \mu_{X|Y} = 0 + \frac{1/2}{1}(2 - 0) = 1 \] \[ \text{Var}(X|Y) = 1/2 - \frac{(1/2)^2}{1} = 1/2 - 1/4 = 1/4 \] Therefore, the conditional PDF of \( W(s = 1/2) \) given \( W(t = 1) = 2 \) is a normal distribution with mean \( \mu_{X|Y} = 1 \) and variance \( \text{Var}(X|Y) = 1/4 \). The list of [mean, variance] is: \[ [1, 1/4] \] |
[1.0, 0.25] |
| kimi | The problem involves finding the conditional probability density function (PDF) of a standard Brownian motion at time \( s = \frac{1}{2} \) given that the Brownian motion at time \( t = 1 \) is equal to 2. We are also asked to find the mean and variance of this conditional distribution. Since \( W(t) \) is a standard Brownian motion, it has the following properties: 1. \( W(0) = 0 \) with probability 1. 2. The process has independent increments. 3. For any \( 0 \leq s < t \), the increment \( W(t) - W(s) \) is normally distributed with mean 0 and variance \( t - s \). Given \( W(1) = 2 \), we want to find the conditional distribution of \( W(\frac{1}{2}) \). Let's denote \( W(\frac{1}{2}) \) by \( X \) and \( W(1) \) by \( Y \). We know that \( Y = 2 \), and we want to find the distribution of \( X \) given \( Y = 2 \). The joint distribution of \( X \) and \( Y \) can be found using the fact that the increments of Brownian motion are normally distributed. Since \( W(\frac{1}{2}) \) and \( W(1) - W(\frac{1}{2}) \) are independent and normally distributed, we can write: \( X \sim N(0, \frac{1}{4}) \) (because \( \text{Var}(W(\frac{1}{2})) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \)) \( Y - X \sim N(0, \frac{1}{2}) \) (because \( \text{Var}(W(1) - W(\frac{1}{2})) = 1 - \frac{1}{2} = \frac{1}{2} \)) Now, \( Y = X + (Y - X) \), and given \( Y = 2 \), we can write: \( 2 = X + (Y - X) \) Since \( X \) and \( Y - X \) are independent, we can find the conditional distribution of \( X \) given \( Y = 2 \) by using the fact that the sum of two independent normal random variables is also normally distributed. Let \( Z = Y - X \), then \( Z \sim N(0, \frac{1}{2}) \). Given \( Y = 2 \), we have \( X = Y - Z \). Since \( Z \) is normally distributed with mean 0 and variance \( \frac{1}{2} \), the conditional distribution of \( X \) given \( Y = 2 \) is also normal. The mean of \( X \) given \( Y = 2 \) is the same as the mean of \( X \), which is 0, because the mean of \( Z \) is 0. The variance of \( X \) given \( Y = 2 \) is the same as the variance of \( X \), which is \( \frac{1}{4} \), because the variance of \( Z \) is \( \frac{1}{2} \), and the variance of \( X \) given \( Y \) is not affected by the value of \( Y \) (since \( X \) and \( Z \) are independent). Therefore, the mean and variance of the conditional distribution of \( W(\frac{1}{2}) \) given \( W(1) = 2 \) are both \( \frac{1}{4} \). The solution is: \[ \boxed{[0, \frac{1}{4}]} \] |
[1.0, 0.25] |
| wenxinyiyan | 首先，我们知道布朗运动具有几个关键性质： 1. 对于任意$t > s$，$W(t) - W(s)$与$W(s)$是独立的。 2. $W(t) - W(s)$服从均值为0，方差为$t - s$的正态分布。 根据这些性质，我们可以推导出： * $W(1) - W(1/2)$与$W(1/2)$是独立的。 * $W(1) - W(1/2)$服从$N(0, 1/2)$。 由于$W(1) = 2$，我们可以得出： $W(1/2) = 2 - [W(1) - W(1/2)]$ 由于$W(1) - W(1/2)$与$W(1/2)$独立，并且$W(1) - W(1/2)$服从$N(0, 1/2)$，我们可以得出$W(1/2)$在给定$W(1) = 2$的条件下的条件分布。 条件分布的均值和方差可以通过以下方式计算： * 均值：$E[W(1/2) | W(1) = 2] = E[2 - (W(1) - W(1/2))] = 2 - E[W(1) - W(1/2)] = 2 - 0 = 2$ * 方差：$Var[W(1/2) | W(1) = 2] = Var[2 - (W(1) - W(1/2))] = Var[W(1) - W(1/2)] = 1/2$ 因此，给定$W(t = 1) = 2$时，$W(s = 1/2)$的条件PDF的均值和方差分别为2和1/2。 返回列表：[2, 1/2] |
[1.0, 0.25] |
| zhipu | To find the conditional probability density function (PDF) of \( W(s) \) given \( W(t) \), and to determine the mean and variance of this conditional distribution, we can use properties of the standard Brownian motion. ### Step-by-Step Solution: 1. **Properties of Brownian Motion:** - \( W(t) \) is a standard Brownian motion, meaning it has independent increments and \( W(t) \sim \mathcal{N}(0, t) \). - For \( 0 < s < t \), the increment \( W(t) - W(s) \) is independent of \( W(s) \) and \( W(t) - W(s) \sim \mathcal{N}(0, t-s) \). 2. **Conditional Distribution:** - Given \( W(t) = w_t \), the conditional distribution of \( W(s) \) is normal with: \[ W(s) | W(t) = w_t \sim \mathcal{N}\left( \frac{s}{t} w_t, s \left(1 - \frac{s}{t}\right) \right) \] 3. **Specific Values:** - Given \( s = \frac{1}{2} \) and \( t = 1 \), and \( W(1) = 2 \): - The mean of the conditional distribution: \[ \text{Mean} = \frac{s}{t} w_t = \frac{\frac{1}{2}}{1} \cdot 2 = 1 \] - The variance of the conditional distribution: \[ \text{Variance} = s \left(1 - \frac{s}{t}\right) = \frac{1}{2} \left(1 - \frac{\frac{1}{2}}{1}\right) = \frac{1}{2} \left(1 - \frac{1}{2}\right) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \] ### Final Results: - **Mean:** 1 - **Variance:** \( \frac{1}{4} \) Thus, the list of [mean, variance] is: \[ [1, \frac{1}{4}] \] ### Summary: The conditional PDF of \( W(s = \frac{1}{2}) \) given that \( W(t = 1) = 2 \) is a normal distribution with mean 1 and variance \( \frac{1}{4} \). Therefore, the list of [mean, variance] is: \[ [1, \frac{1}{4}] \] |
[1.0, 0.25] |
| gpto1 | To solve this problem, we need to find the conditional probability density function (PDF) of \( W(s) \) given that \( W(t) = 2 \), where \( W(t) \) is a standard Brownian motion, and \( 0 < s < t \). Specifically, we need to compute the mean and variance of this conditional distribution. **Step 1: Understand the Properties of Brownian Motion** - **Standard Brownian Motion** \( W(t) \) is a continuous-time stochastic process with: - \( W(0) = 0 \) - Independent increments: \( W(t) - W(s) \) is independent of \( W(u) \) for all \( u \leq s \) - Normally distributed increments: \( W(t) - W(s) \sim N(0, t - s) \) - Continuous paths **Step 2: Consider the Joint Distribution of \( W(s) \) and \( W(t) \)** Because \( W(s) \) and \( W(t) \) are jointly normally distributed (since any linear combination of normally distributed random variables is also normally distributed), we can use properties of the multivariate normal distribution to find the conditional distribution. **Step 3: Compute the Covariance and Variances** - **Mean Values:** - \( E[W(s)] = 0 \) - \( E[W(t)] = 0 \) - **Variances:** - \( \operatorname{Var}(W(s)) = s \) - \( \operatorname{Var}(W(t)) = t \) - **Covariance:** - Since \( W(s) \leq W(t) \), the covariance is \( \operatorname{Cov}(W(s), W(t)) = s \) - This is because \( \operatorname{Cov}(W(s), W(t)) = E[W(s)W(t)] - E[W(s)]E[W(t)] = E[W(s)W(t)] \) - And \( E[W(s)W(t)] = E[W(s)W(s) + W(s)(W(t) - W(s))] = s + 0 = s \) **Step 4: Use the Conditional Mean and Variance Formulas** For jointly normally distributed variables \( X \) and \( Y \), the conditional distribution \( X | Y = y \) is normal with: - **Conditional Mean:** \[ E[X | Y = y] = E[X] + \frac{\operatorname{Cov}(X, Y)}{\operatorname{Var}(Y)} (y - E[Y]) \] - **Conditional Variance:** \[ \operatorname{Var}(X | Y) = \operatorname{Var}(X) - \frac{[\operatorname{Cov}(X, Y)]^2}{\operatorname{Var}(Y)} \] **Step 5: Plug in the Values \( s = \frac{1}{2} \) and \( t = 1 \)** - **Conditional Mean:** \[ E[W(1/2) | W(1) = 2] = 0 + \frac{(1/2)}{1} \times (2 - 0) = 1 \] - **Conditional Variance:** \[ \operatorname{Var}(W(1/2) | W(1) = 2) = \frac{1}{2} - \frac{(1/2)^2}{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \] **Step 6: State the Result** The conditional distribution of \( W(1/2) \) given \( W(1) = 2 \) is normal with mean \( 1 \) and variance \( \frac{1}{4} \). **Answer: [1, ¼] |
[1.0, 0.25] |
| deepseekv3 | To solve the problem, we need to find the conditional probability density function (PDF) of \( W(s) \) given \( W(t) = 2 \), where \( W(t) \) is a standard Brownian motion, and \( 0 < s < t \). Specifically, we are tasked with finding the mean and variance of \( W(s = 1/2) \) given \( W(t = 1) = 2 \). --- ### Step 1: Understand the properties of Brownian motion For a standard Brownian motion \( W(t) \): - \( W(0) = 0 \). - \( W(t) \) has independent increments. - \( W(t) \) is normally distributed with mean 0 and variance \( t \), i.e., \( W(t) \sim N(0, t) \). --- ### Step 2: Conditional distribution of \( W(s) \) given \( W(t) \) Given \( W(t) = x \), the conditional distribution of \( W(s) \) is normal with: - Mean: \( \mu = \frac{s}{t} x \). - Variance: \( \sigma^2 = s \left(1 - \frac{s}{t}\right) \). --- ### Step 3: Apply the given values Here, \( s = 1/2 \), \( t = 1 \), and \( W(t) = 2 \). Substituting these values: - Mean: \( \mu = \frac{1/2}{1} \cdot 2 = 1 \). - Variance: \( \sigma^2 = \frac{1}{2} \left(1 - \frac{1/2}{1}\right) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \). --- ### Final Answer The mean and variance of \( W(s = 1/2) \) given \( W(t = 1) = 2 \) are: \[ [1, \frac{1}{4}] \] |
[1.0, 0.25] |