问题
I'm giving part of a presentation on numerical integration. While the talk itself will go into better forms of numerical integration (mainly importance sampling and stratified sampling), I'm mentioning during part of my section Monte Carlo integration sampling from the uniform distribution.
I've found that:
mean(sin(runif(1e8, 0, pi)))
is giving an answer of 0.636597, rather than 1 that is expected. This answer seems pretty consistent with increasing sample size, and I'm unsure why there's so much error. Other computations such as:
mean(sin(runif(1e6, 0, 2 * pi)))
give 0.0005398996, much closer to the expected answer of 0.
Can someone help me see why
mean(sin(runif(1e8, 0, pi)))
is giving such an inaccurate answer? Is this user error, or is it to be expected when sampling from the uniform distribution?
回答1:
I came back to make my answer complete, in case future readers need to know the logic. Note, the true value is 2 not 1, as stated in your question.
So, you just computed the mean function values at samples, but forgot to multiply interval length.
set.seed(0); pi * mean(sin(runif(1000, 0, pi)))
# [1] 2.001918
is what you need.
A deterministic view of this result is mean value theorem for integral, or Riemann sum approximation of integral.
So we can also do
pi * mean(sin(seq(0, pi, length = 1000)))
# [1] 1.997998
Monte Carlo integration is more useful via importance sampling. Read Monte Carlo integration using importance sampling given a proposal function for a good example.
来源：https://stackoverflow.com/questions/40971079/wrong-result-when-doing-simple-monte-carlo-integration-in-r