Re: Restting Probabilities

*To*: mathgroup at smc.vnet.net*Subject*: [mg29316] Re: [mg29284] Restting Probabilities*From*: Tomas Garza <tgarza01 at prodigy.net.mx>*Date*: Wed, 13 Jun 2001 03:10:47 -0400 (EDT)*References*: <200106110838.EAA06919@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

I'm not altogether sure I grasped your problem (why talk of a matrix, when for example your x is a vector?). However, if I get you right, you want to construct a sequence of (in your example) 5 Bernoulli random variables with varying probabilities according to x, except that, whenever you get a 1, you go back to the first value of x and proceed from there on (i.e., reset your probabilities) until you get another 1 (or finish). I find the problem not too trivial, so that if you are having only a "little" difficulty, then you're OK. A possible approach, which I'm sure can be enormously improved by many of the distinguished members of this group, uses the function defined below "genVect" to generate the vector as follows: In[1]:= << Statistics`DiscreteDistributions` In[2]:= b[u_, k_] := Random[BernoulliDistribution[u[[k]]]]; In[3]:= genVect[x_] := Module[{r = x}, y = {b[r, 1]}; (r = If[Last[y] == 1, RotateRight[r, Length[y]], r]; y = Flatten[{y, b[r, #]}]) & /@ {2, 3, 4, 5}] Example: In[4]:= genVect[x] // Last Out[4]= {0, 1, 0, 0, 0} The last list (i.e., {0,1,0,0,0}) is the result you are looking for. Let me explain (if you're really new to Mathematica, I'm afraid you'll have to think hard). Starting from a length-1 vector with a variable produced with b (the Bernoulli random generator defined above) using the first probability (0.05 in the example), you append a new variable depending on the value of y: if 0, just take the next probability in the original list; if 0, then shift the original list one position to the right, so that the first value of x is now in the second position, and produce the second variable. And so on. The procedure constructs, a list of vectors until it gets to the last one, which is the one you want. I'm sure this can be improved, but it seems to work. I checked it with a simulation running the procedure 10,000 times and looking at the frequencies. They correspond closely to the probabilities in x, taking into account that every time there occurs a 1, the next value must have a frequency close to 0.05, and so on. The simulation process is performed with In[5]:= sim = {}; Do[AppendTo[sim, genVect[x] // Last], {7000}] // Timing Out[5]= {13.02 Second, Null} and the 10,000 results are kept in "sim". Tomas Garza Mexico City ----- Original Message ----- From: "Michael Loop" <loopm at yahoo.com> To: mathgroup at smc.vnet.net Subject: [mg29316] [mg29284] Restting Probabilities > I am a new user of Mathematica, and I am having a little difficulty grasping > the syntax. I am trying to generate a matrix of 0's and 1's given by a > Bernoulli probability distribution. If we look at a case for a matrix with > five different probabilities, x={0.05,0.09,.15,.23,.47}. I have been able > to construct a matrix with these probabilities, but what I cannot do is > reset the probabilites back to the start of the list when a 1 was the last > result in the table. In other words, each time a random Bernoulli generates > a 1, I want the probability for the next xi in the row to be 0.05 again and > continue to preoceed down the list again until another 1 is reached. If > anyone could provide me with some help on the proper syntax for resetting > the probability at the start of the list {x}after a 1 is reached in the row, > I would be grateful. >

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