请回答这个文档里的problem 1和problem 2，其中需要写代码的问题写出R代码

### Problem 1

#### (a) Find the values of \(a\) and \(b\)
Given:
- Prior mean \(\mu = 16\)
- Standard deviation \(\sigma = 8\)

For a Gamma distribution with parameters \(a\) and \(b\):
\[ \text{Mean} = \frac{a}{b} = 16 \]
\[ \text{Variance} = \frac{a}{b^2} = 8^2 = 64 \]

From the mean equation:
\[ a = 16b \]

Substitute \(a\) into the variance equation:
\[ \frac{16b}{b^2} = 64 \]
\[ \frac{16}{b} = 64 \]
\[ b = \frac{16}{64} = \frac{1}{4} \]

Now substitute \(b\) back into the mean equation:
\[ a = 16 \times \frac{1}{4} = 4 \]

So, \(a = 4\) and \(b = \frac{1}{4}\).

#### (b) Find the posterior distribution of \(\lambda\)
The likelihood function for the Poisson distribution is:
\[ L(\lambda) = \prod_{i=1}^{n} \frac{\lambda^{x_i} e^{-\lambda}}{x_i!} = \frac{\lambda^{\sum x_i} e^{-n\lambda}}{\prod x_i!} \]

Given the prior distribution \(\lambda \sim \text{Gamma}(4, \frac{1}{4})\), the posterior distribution is also a Gamma distribution:
\[ \lambda | \mathbf{x} \sim \text{Gamma}(a + \sum x_i, b + n) \]

Where:
- \(a = 4\)
- \(b = \frac{1}{4}\)
- \(n = 20\)
- \(\sum x_i = 14 + 19 + 14 + 21 + 22 + 33 + 15 + 13 + 16 + 19 + 27 + 22 + 27 + 21 + 16 + 25 + 14 + 23 + 22 + 17 = 420\)

So, the posterior distribution is:
\[ \lambda | \mathbf{x} \sim \text{Gamma}(4 + 420, \frac{1}{4} + 20) = \text{Gamma}(424, \frac{81}{4}) \]

#### (c) Plot the prior and posterior probability density functions of \(\lambda\)

library(ggplot2)

# Define the prior and posterior distributions
prior <- function(lambda) dgamma(lambda, shape = 4, rate = 1/4)
posterior <- function(lambda) dgamma(lambda, shape = 424, rate = 81/4)

# Generate lambda values for plotting
lambda_values <- seq(0, 50, length.out = 1000)

# Create data frames for plotting
prior_df <- data.frame(lambda = lambda_values, density = prior(lambda_values), distribution = "Prior")
posterior_df <- data.frame(lambda = lambda_values, density = posterior(lambda_values), distribution = "Posterior")

# Combine the data frames
plot_data <- rbind(prior_df, posterior_df)

# Plot the densities
ggplot(plot_data, aes(x = lambda, y = density, color = distribution)) +
  geom_line(size = 1) +
  labs(title = "Prior and Posterior Distributions of λ",
       x = "λ",
       y = "Density") +
  theme_minimal()

#### (d) Calculate a 95% posterior highest posterior density (HPD) interval for \(\lambda\)

library(HDInterval)

# Calculate the HPD interval
hpd_interval <- hdi(function(lambda) dgamma(lambda, shape = 424, rate = 81/4), credMass = 0.95)
hpd_interval

### Problem 2

#### (a) Find estimated values of \(a\) and \(b\)
We can use the method of moments to estimate the parameters \(a\) and \(b\). The mean and variance of the Gamma distribution are given by:
\[ \text{Mean} = \frac{a}{b} \]
\[ \text{Variance} = \frac{a}{b^2} \]

First, calculate the sample mean and variance:

sales_numbers <- c(14, 19, 14, 21, 22, 33, 15, 13, 16, 19, 27, 22, 27, 21, 16, 25, 14, 23, 22, 17)
sample_mean <- mean(sales_numbers)
sample_variance <- var(sales_numbers)

# Estimate a and b
a_estimate <- (sample_mean^2) / sample_variance
b_estimate <- sample_mean / sample_variance

a_estimate
b_estimate

#### (b) Find the posterior distribution of \(\lambda\)
Using the estimated values of \(a\) and \(b\):
\[ \lambda | \mathbf{x} \sim \text{Gamma}(a + \sum x_i, b + n) \]

Where:
- \(a = a_{\text{estimate}}\)
- \(b = b_{\text{estimate}}\)
- \(n = 20\)
- \(\sum x_i = 420\)

So, the posterior distribution is:
\[ \lambda | \mathbf{x} \sim \text{Gamma}(a_{\text{estimate}} + 420, b_{\text{estimate}} + 20) \]

#### (c) Under squared error loss, find a Bayes estimate of \(\lambda\)
The Bayes estimate under squared error loss is the posterior mean:
\[ E[\lambda | \mathbf{x}] = \frac{a_{\text{estimate}} + 420}{b_{\text{estimate}} + 20} \]

Calculate the posterior mean:

posterior_mean <- (a_estimate + 420) / (b_estimate + 20)
posterior_mean

Calculate the equal-tail 95% credible interval:

alpha <- 0.05
lower_bound <- qgamma(alpha / 2, shape = a_estimate + 420, rate = b_estimate + 20)
upper_bound <- qgamma(1 - alpha / 2, shape = a_estimate + 420, rate = b_estimate + 20)

credible_interval <- c(lower_bound, upper_bound)
credible_interval

#### (d) Based on classical statistics, find a point estimate and a 95% confidence interval of \(\lambda\)
The maximum likelihood estimate (MLE) of \(\lambda\) is the sample mean:
\[ \hat{\lambda}_{\text{MLE}} = \bar{x} \]

The 95% confidence interval for \(\lambda\) using the asymptotic normality of the MLE:
\[ \hat{\lambda}_{\text{MLE}} \pm z_{\alpha/2} \sqrt{\frac{\hat{\lambda}_{\text{MLE}}}{n}} \]

Where \(z_{\alpha/2}\) is the critical value from the standard normal distribution.

z_alpha_2 <- qnorm(1 - alpha / 2)
mle_lambda <- sample_mean
se_lambda <- sqrt(mle_lambda / length(sales_numbers))

confidence_interval <- c(mle_lambda - z_alpha_2 * se_lambda, mle_lambda + z_alpha_2 * se_lambda)
confidence_interval

This completes the solution for both problems.