From: m1@m1.com   
      
   Hello...   
      
      
   More of my philosophy of if i can do math..   
      
   Archimedes Plutonium just written:   
    >In fact we do not know if Amine can do any math   
      
      
   So you are asking if i can do math?   
      
   Yes i can do math, and here is my logical proof of it, you can   
   read my following thoughts of how i can do math:   
      
   About Markov chains in mathematics and more..   
      
   In mathematics, many Markov chains automatically find their own way to   
   an equilibrium distribution as the chain wanders through time. This   
   happens for many Markov chains, but not all. I will talk about the   
   conditions required for the chain to find its way to an equilibrium   
   distribution.   
      
   If in mathematics we give a Markov chain on a finite state space and   
   asks if it converges to an equilibrium distribution as t goes to   
   infinity. An equilibrium distribution will always exist for a finite   
   state space. But you need to check whether the chain is irreducible and   
   aperiodic. If so, it will converge to equilibrium. If the chain is   
   irreducible but periodic, it cannot converge to an equilibrium   
   distribution that is independent of start state. If the chain is   
   reducible, it may or may not converge.   
      
   So i will give an example:   
      
   Suppose that for the course you are currently taking there are two   
   volumes on the market and represent them by A and B. Suppose further   
   that the probability that a teacher using volume A keeps the same volume   
   next year is 0.4 and the probability that it will change for volume B   
   is 0.6. Furthermore the probability that a professor using B this   
   year changes to next year for A is 0.2 and the probability that it   
   again uses volume B is 0.8. We notice that the matrix of transition is:   
      
   |0.4 0.6|   
   |       |   
   |0.2 0.8|   
      
   The interesting question for any businessman is whether his   
   market share will stabilize over time. In other words, does it exist   
   a probability vector (t1, t2) such that:   
      
   (t1, t2) * (transition matrix above) = (t1, t2) [1]   
      
   So notice that the transition matrix above is irreducible and aperiodic,   
   so it will converge to an equilibrum distribution that is (t1, t2) that   
   i will mathematically find, so the system of equations of [1] above is:   
      
   0.4 * t1 + 0.2 * t2 = t1   
   0.6 * t1 + 0.8 * t2 = t2   
      
   this gives:   
      
   -0.6 * t1 + 0.2 * t2 = 0   
   0.6 * t1 - 0.2 * t2 = 0   
      
   But we know that (t1, t2) is a vector of probability, so we have:   
      
   t1 + t2 = 1   
      
   So we have to solve the following system of equations:   
      
   t1 + t2 = 1   
   0.6 * t1 - 0.2 * t2 = 0   
      
   So i have just solved it with R, and this gives the vector:   
      
   (0.25,0.75)   
      
   Which means that in the long term, volume A will grab 25% of the market   
   while volume B will grab 75% of the market unless the advertising   
   campaign does change the probabilities of transition.   
      
      
   Thank you,   
   Amine Moulay Ramdane.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)