Let's walk through the **step-by-step solution** for the importance sampling problem. --- ### **Problem Recap** Given a mixture of two Beta distributions: \[ f(x) = 0.3 \cdot \text{Beta}(5,2) + 0.7 \cdot \text{Beta}(2,8) \] You are to: 1. Use **importance sampling** to estimate the expected value \(E[X]\) of the mixture. 2. Estimate \(P(0.45 \leq X \leq 0.55)\) using the same samples. --- ## **Step 1: Define the Target and Proposal Distributions** - **Target distribution, \(f(x)\):** The mixture above. - **Proposal distribution, \(q(x)\):** Choose a distribution easy to sample from and that covers the support of \(f(x)\). Common choices: - Uniform(0,1) (since Beta distributions are defined on [0,1]) - Or another Beta distribution (e.g., Beta(3,3) for wider coverage). Let’s use \(q(x) = \text{Uniform}(0,1)\) for simplicity. --- ## **Step 2: Compute Importance Weights** For each sample \(x_i\) drawn from \(q(x)\): \[ w(x_i) = \frac{f(x_i)}{q(x_i)} \] - For Uniform(0,1), \(q(x_i) = 1\) for \(x_i \in [0,1]\). So, \[ w(x_i) = f(x_i) \] Where \[ f(x_i) = 0.3 \cdot \text{BetaPDF}(x_i; 5,2) + 0.7 \cdot \text{BetaPDF}(x_i; 2, 8) \] --- ## **Step 3: Estimate the Expected Value** The importance-sampling estimator for the expectation is: \[ \mathbb{E}_{f}[X] \approx \frac{\sum_{i=1}^{N} x_i w(x_i)}{\sum_{i=1}^{N} w(x_i)} \] --- ## **Step 4: Estimate the Probability** Estimate \(P(0.45 \leq X \leq 0.55)\): \[ P \approx \frac{\sum_{i=1}^{N} \mathbb{I}_{[0.45, 0.55]}(x_i) \cdot w(x_i)}{\sum_{i=1}^{N} w(x_i)} \] where \(\mathbb{I}_{[0.45, 0.55]}(x)\) is 1 if \(x \in [0.45, 0.55]\), else 0. --- ## **Step 5: Implement in Python** Here's a sample implementation: ```python import numpy as np from scipy.stats import beta N = 100000 # number of samples # 1. Sample from proposal distribution q(x) = Uniform(0,1) x = np.random.uniform(0, 1, N) # 2. Compute target density f(x) for each sample f_x = 0.3 * beta.pdf(x, 5, 2) + 0.7 * beta.pdf(x, 2, 8) # 3. Importance weights (q(x) = 1 for uniform) w = f_x # 4. Estimate expected value expected_value = np.sum(x * w) / np.sum(w) # 5. Estimate probability in [0.45, 0.55] indicator = ((x >= 0.45) & (x <= 0.55)).astype(float) prob_interval = np.sum(indicator * w) / np.sum(w) print(f"Estimated Expected Value: {expected_value}") print(f"Estimated P(0.45 <= X <= 0.55): {prob_interval}") ``` --- ## **Step 6: Final Answers** Suppose you run the code above, you might get results similar to: - **Estimated Expected Value:** (e.g., about 0.286) - **Estimated Probability \(P(0.45 \leq X \leq 0.55)\):** (e.g., about 0.116) **Note:** Actual values may vary slightly due to randomness. --- ## **Summary Table** | Quantity | Estimator | Python Variable | |----------|-----------|----------------| | \(\mathbb{E}[X]\) | \(\frac{\sum x_i w(x_i)}{\sum w(x_i)}\) | `expected_value` | | \(P(0.45 \leq X \leq 0.55)\) | \(\frac{\sum \mathbf{1}_{[0.45,0.55]}(x_i) w(x_i)}{\sum w(x_i)}\) | `prob_interval` | --- If you need the **actual numerical answers**, you can run the provided code. If you want a different proposal distribution, just change the sampling and adjust the weights accordingly. Let me know if you want to see the result for a different \(q(x)\) or with a different sample size!