如何在r中生成概率密度函数和期望值？

Question

The task:

Eric the fly has a friend, Ernie. Assume that the two flies sit at independent locations, uniformly distributed on the globe's surface. Let D denote the Euclidean distance between Eric and Ernie (ie, on a straight line through the interior of the globe).

Make a conjecture about the probability density function of D and give an estimate of its expected value, E(D).

So far I have made a function to generate two points on the globe's surface, but I am unsure what to do next:

sample3d <- function(2)
  {
  df <- data.frame()
  while(n > 0){
    x <- runif(1,-1,1)
    y <- runif(1,-1,1)
    z <- runif(1,-1,1)
    r <- x^2 + y^2 + z^2
    if (r < 1){
      u <- sqrt(x^2+y^2+z^2)
      vector = data.frame(x = x/u,y = y/u, z = z/u)
      df <- rbind(vector,df)
      n = n- 1
    }
  }
  df
}
E <- sample3d(2)

Answer 1

This is an interesting problem. I'll outline a computational approach; I'll leave the math up to you.

1. First we fix a random seed for reproducibility.

    set.seed(2018); 

2. We sample 10^4 points from the unit sphere surface.

    sample3d <- function(n = 100) { df <- data.frame(); while(n > 0) { x <- runif(1,-1,1) y <- runif(1,-1,1) z <- runif(1,-1,1) r <- x^2 + y^2 + z^2 if (r < 1) { u <- sqrt(x^2 + y^2 + z^2) vector = data.frame(x = x/u,y = y/u, z = z/u) df <- rbind(vector,df) n = n- 1 } } df } df <- sample3d(10^4); 

   Note that sample3d is not very efficient, but that's a different issue.

3. We now randomly sample 2 points from df , calculate the Euclidean distance between those two points (using dist ), and repeat this procedure N = 10^4 times.

    # Sample 2 points randomly from df, repeat N times N <- 10^4; dist <- replicate(N, dist(df[sample(1:nrow(df), 2), ])); 

   As pointed out by @JosephWood, the number N = 10^4 is somewhat arbitrary. We are using a bootstrap to derive the empirical distribution. For N -> infinity one can show that the empirical bootstrap distribution is the same as the (unknown) population distribution (Bootstrap theorem). The error term between empirical and population distribution is of the order 1/sqrt(N) , so N = 10^4 should lead to an error around 1%.

4. We can plot the resulting probability distribution as a histogram:

    # Let's plot the distribution ggplot(data.frame(x = dist), aes(x)) + geom_histogram(bins = 50); 

4. Finally, we can get empirical estimates for the mean and median.

    # Mean mean(dist); #[1] 1.333021 # Median median(dist); #[1] 1.41602 

   These values are close to the theoretical values:

    mean.th = 4/3 median.th = sqrt(2)