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Asked by Derek O'Connor on 3 Mar 2011

A **Bernoulli Process** is a sequence of iid (independent,
identically-distributed) random variables `x1,x2,...,xi,...`, where
the probability mass function of `xi` is

xi = S(success) with probability p, xi = F(failure) with probability q, with p+q = 1.

Depending on the application, (S,F) may be (1,0), (H,T), (true, false), etc.

Here are two obvious ways to generate a (1,0) Bernoulli process in Matlab:

- Vectorized Form:
`bernp1 = rand(n,1) <= p; Process vector bernp1` - Component Form:
`for k = 1:n, bernp2 = rand <= p; Process scalar bernp2; end;`

The "best" answer will depend on the application and computer at hand.

If the computer has lots of memory then the vectorized version is the simplest, and may be the fastest.

If "Process vector or scalar" takes a lot of time then the time to
generate `bernp` by either method may be negligible.

If a fixed amount of memory is available for `bernp` then we must use
the component form or a mini-vectorized form.

Using `rand` to generate 1 random bit seems wasteful, given that `rand`
returns a double precision floating point number which has about 53
random bits. Can we use `rand` more efficiently?

My interest in the Bernoulli process was sparked by re-reading Feller Volume 1. At the bottom of page 18 Feller says: "In this volume we shall consider only discrete sample spaces". A closer examination shows that Feller derives (nearly) everything in this book from the Bernoulli process.

Derek O'Connor

*No products are associated with this question.*

Answer by Andrew Newell on 3 Mar 2011

Interesting question. The number of wasted bits depends on the number of bits you need to define p. If p = 1/sqrt(3), no wasted bits! Any algorithm trying to beat your algorithm 1 would require a lot of special cases, and would be competing with a built-in algorithm.

The Statistics Toolbox uses a variation on your first algorithm.

( *edited to expand on answer* )

Derek O'Connor on 4 Mar 2011

Andrew,

I don't understand the first part of your answer.

The value of p would be represented by a double precision f.p. number, i.e., a 64-bit string. This same string is used repeatedly for each rand generated.

Derek O'Connor

Andrew Newell on 4 Mar 2011

Each time you generate a new random number, you need to compare it with p, and you don't know in advance how many bits you'll need to compare. It might be all 53.

Answer by Walter Roberson on 4 Mar 2011

Yes, you can make better use of rand.

If the probability is expressible in lowest terms with a denominator of 2^n, n <= 26. With the 2^n denominator, you take n consecutive bits at a time out of the float representation.

For other probabilities, you use Arithmetic Coding

Andrew Newell on 4 Mar 2011

I don't know what you mean. Float representation of what? And what do you do with the bits?

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