Solved – What are the advantages of a simulation over an exact binomial calculation

binomial distributionintuitionsimulation

Let's say that I want to calculate the probability of achieving more than 8 heads in 10 coin tosses. The easiest and most accurate way will be to simply use the binomial distribution and sum the probability of getting 9 an 10 heads.

However, it is also possible to simulate numerous coin tosses and then see which proportion of those answers our question. In which general scenario (doesn't have to be restricted to coin tosses) would this method of simulation be better than using the binomial and why?

Best Answer

Some advantages of simulation:

you can use it when you don't know how to do the exact calculation
you can use it to check you got the exact calculation correct
you can generalize the situation and examine sensitivity of the exact solution to changes in the circumstances (what if there's heterogeneity or dependence, for example?); often the simple situation can be done exactly but the more general cases may be somewhat trickier.

Related Solutions

Solved – Binomial distribution where probability of success is dependent on another binomial distribution

This is known as a compound distribution. Some compound distributions simplify, and can be recognized as another well-known distribution, but I don't think this binomial-binomial compound is one of those. (There is another type of binomial-binomial compound which does simplify to just a binomial, where you toss all coins, and then reflip the heads.)

I think the simplest way to handle the distribution is as a mixture of $11$ different binomial distributions parametrized by the number of heads in the initial batch, from $0$ to $10$. There are other possibilities based on recognizing this as a compound of a multinomial distribution.

Solved – Generate binomial sample with (pretty) exact probability

I'm not sure whether this is actually advisable, but it should be straightforward to generate. What you are asking for, essentially, is an underdispersed binomial distribution. You can get this by sampling (with replacement, if you want more than 1 value) from a vector of the integers 0:size, where you specify a set of underdispersed probabilities. You just have to figure out what the probabilities are that you want. First, consider this figure:

enter image description here

From this you can see that the probabilities of each possible value from a binomial distribution can be matched by a normal distribution with mean $n\pi$ and variance $n\pi(1-\pi)$. Thus, you can make an underdispersed version by using appropriately scaled densities from a normal distribution with the same mean but a smaller SD. Imagine that you want the SD to be cut in half, then:

set.seed(1773)
hSD.norm  = dnorm(0:18, mean=9, sd=sqrt(18*.25)*.5)
ud.probs  = hSD.norm/sum(hSD.norm)
N         = 10000
vals      = sample(0:18, size=N, replace=TRUE, prob=ud.probs)

enter image description here

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Related Solutions

Solved – Binomial distribution where probability of success is dependent on another binomial distribution

Solved – Generate binomial sample with (pretty) exact probability

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