Correction to "Sequence of Bernoulli RVs"

*To*: mathgroup at smc.vnet.net*Subject*: [mg73004] Correction to "Sequence of Bernoulli RVs"*From*: Virgil Stokes <vs at it.uu.se>*Date*: Sun, 28 Jan 2007 01:23:57 -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 all possible sequences 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)(1-p)p Finally, my question --- Is it possible to derive this equation (in terms of p and n) using Mathematica? --V. Stokes PS. The error was in the equation above: (p-1) has been replaced by (1-p) and the is equation is correct!