The distribution of wind speeds in a city follows a mixture of log-normal distributions: If we denote the wind speed (in miles per hour) by \(X\), we have \[\begin{equation*} \log(X) \sim \mathcal{N}(\mu_K, \sigma_K^2), \end{equation*}\] where \(K\) is random with \[\begin{equation*} P(K = 1) = 0.1, \quad P(K = 2) = 0.4, \quad P(K = 3) = 0.5, \end{equation*}\] and the means and standard deviations are given by \[\begin{equation*} \mu = \begin{pmatrix} 2.1 \\ 1.6 \\ -0.5 \end{pmatrix}, \quad \sigma = \begin{pmatrix} 0.55 \\ 0.7 \\ 0.6 \end{pmatrix}. \end{equation*}\] The density of the log-normal distribution with parameters \(\mu\in\mathbb{R}\) and \(\sigma > 0\) is \[ \varphi(x; \mu, \sigma) = \frac{1}{x\sqrt{2\pi\sigma^2}} \exp\left(-\frac{\left(\log(x)-\mu\right)^2}{2\sigma^2}\right) \] for all \(x \geq 0\). This function is available in R as dlnorm(). You can generate samples from the wind speed distribution as follows:

wind_speed <- function(n, w, mu, sigma) {
    k <- length(w)
    i <- sample(k, n, replace = TRUE, prob = w)
    rlnorm(n, mu[i], sigma[i])

w <- c(0.1, 0.4, 0.5)
mu <- c(2.1, 1.6, -0.5)
sigma <- c(0.55, 0.7, 0.6)
wind_speed(5, w, mu, sigma)
## [1]  0.4719161  0.1363222  1.4373320  6.0327490 10.6146743

Task 1

We are interested in the probability that the wind speed is larger than 40 miles per hour.

Task 2

Our aim here is to find a more efficient estimator for the probability from task 1.

Task 3

Now, for simplicity, assume that the wind speed follows a single log-normal distribution with parameters \(\mu\) and \(\sigma\): \[\begin{equation*} \log(X) \sim \mathcal{N}(\mu, \sigma^2). \end{equation*}\] Furthermore, we assume that \(\mu=0\) is known and \(\sigma\) is unknown, random and exponentially distributed: \(\sigma \sim \mathrm{Exp}(1)\).

On five different days we have observed the wind speeds \(x_1=0.50\), \(x_2=1.67\), \(x_3=2.22\), \(x_4=0.22\) and \(x_5=4.36\). Using these data we want to learn about the unknown value \(\sigma\). From Bayes rule we know that the conditional distribution of \(\sigma\), given the data \((x_1, \ldots, x_5)\) has density \[\begin{equation*} f(s) = \frac{1}{Z} \tilde f(s), \quad \tilde f(s) = \pi(s) \prod_{i=1}^5 \varphi(x_i; 0, s), \end{equation*}\] where \(Z\) is the normalising constant, \(\pi(s)\) is the density of the \(\mathrm{Exp}(1)\) distribution, and \(\varphi(x_i; 0, s)\) is the density of the log-normal distribution with parameters \(\mu=0\) and \(\sigma=s\). In Bayesian statistics, \(\pi\) is called the prior density and \(f\) is called the posterior density.

We want to sample from the posterior density \(f\) using envelope rejection sampling.