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Sequence of Bernoulli RVs
*To*: mathgroup at smc.vnet.net
*Subject*: [mg72974] Sequence of Bernoulli RVs
*From*: Virgil Stokes <vs at it.uu.se>
*Date*: Sat, 27 Jan 2007 05:45:39 -0500 (EST)
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
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