Re: Sequence of Bernoulli RVs
- To: mathgroup at smc.vnet.net
- Subject: [mg72998] Re: Sequence of Bernoulli RVs
- From: "Ray Koopman" <koopman at sfu.ca>
- Date: Sun, 28 Jan 2007 00:59:52 -0500 (EST)
- References: <epf9es$c1o$1@smc.vnet.net>
Virgil Stokes wrote:
> The following code will get the expected number of "runs" (sequence of
> equal values) for a sequence of Bernoulli RVs in terms of p, which is
> the probability of the value 1, and of course, (1-p), the probability of 0.
>
> n=3
> l1 = Tuples[{(1-p),p},n]; (* generate Bernoulli sequence of length n *)
> xx = Table[Split[l1[[k]]],{k,1,Length[l1]}]; (* list of runs *)
> l2 = Map[Length,Map[Split,xx]]; (* lengths of each run *)
> tt = Table[Times@@l1[[k]],{k,1,Length[l2]}]; (* probabilities for each
> run *)
> expectedruns = Inner[Times,tt,l2]//FullSimplify (* mean of a function
> of a Discrete RV *)
>
> It may not be very obvious that this is correct. Please try it over a
> range of values of n (n=1,2,...) if you are in doubt. The expected
> value for such a sequence is given by
>
> E[number of runs] = 1 + 2(n-1)(p-1)p
>
> Finally, my question --- Is it possible to derive this equation (in
> terms of p and n) using Mathematica?
>
> --V. Stokes
I don't see a role for Mathematica to play here. If an (n+1)st bit
is appended to an n-bit sequence, the probability that it creates
a new run is 2p(1-p), and the probability that it extends the old
terminal run is 1 - 2p(1-p). Hence
ExpectedRuns[n+1] = (ExpectedRuns[n] + 1) * 2p(1-p) +
ExpectedRuns[n] * (1 - 2p(1-p))
= ExpectedRuns[n] + 2p(1-p).
Since
ExpectedRuns[1] = 1,
it follows immediately that
ExpectedRuns[n] = 1 + 2(n-1)p(1-p).
What would you have Mathematica do?
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