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Math Help - Difficult Combinatorics Problem, Help PLZ

  1. #1
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    Difficult Combinatorics Problem, Help PLZ

    i have this assignment due tmrw where i have to answer this question, plzzzzzzzz help i really dont understand it, im being marked for the process of thinking so steps will be very helpful

    in a sequence of p zeroes and q ones, the ith term ti, is called a change point if ti does not equal ti-1, for i= 2,3,4....p+q. For example, the sequence of 0,1,1,0,1,0,1,0 has p=q=4, and five change points t2, t4, t6, t7, t8. For all possible sequences of p zeroes and q ones with 1 less then or equal to p which is less then or equal to q, determine the average number of change points.

    i know its a difficult problem but plz jus try to help i really dont understand it!

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  2. #2
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    Quote Originally Posted by ChiragDesai01 View Post
    i have this assignment due tmrw where i have to answer this question, plzzzzzzzz help i really dont understand it, im being marked for the process of thinking so steps will be very helpful

    in a sequence of p zeroes and q ones, the ith term ti, is called a change point if ti does not equal ti-1, for i= 2,3,4....p+q. For example, the sequence of 0,1,1,0,1,0,1,0 has p=q=4, and five change points t2, t4, t6, t7, t8. For all possible sequences of p zeroes and q ones with 1 less then or equal to p which is less then or equal to q, determine the average number of change points.

    i know its a difficult problem but plz jus try to help i really dont understand it!

    Define X_i = \begin{cases}<br />
1 \text{ if } t_i \neq t_{i-1}\\<br />
0 \text{ otherwise}<br />
\end{cases}

    For any i from 1 to p+q-1,
    Pr(t_{i-1} = 0) = p / (p+q) and Pr(t_i=1 | t_{i-1} = 0) = q / (p+q-1)
    Pr(t_{i-1} = 1) = q / (p+q) and Pr(t_i=0| t_{i-1} = 1) = p / (p+q-1)
    so
    Pr(X_i = 1) = \frac{p}{p+q} \frac{q}{p+q-1} + \frac{q}{p+q} \frac{p}{p+q-1} = \frac{2pq}{(p+q)(p+q-1)}
    This is also E(X_i).

    The number we want, the average number of change points, is
    E(X_1 + X_2 + \cdots + X_{p+q-1}). There is an extremely useful theorem saying that if X and Y are random variables then E(X+Y) = E(X) + E(Y). It's important to note that it is not necessary that X and Y be independent in order to apply this theorem. This good for us because the X_i's are not independent. So let's apply it as follows:

    E(X_1 + X_2 + \cdots + X_{p+q-1})
     = E(X_1) + E(X_2) + \cdots + E(X_{p+q-1})
     = (p+q-1) E(X_i)
     = (p+q-1) \frac{2pq}{(p+q)(p+q-1)}
     = \frac{2pq}{p+q}

    Hope this helps-- I'm in a rush, but then so are you.

    jw
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  3. #3
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    broooooo thanks alot! this dus help im jus guna copy this into my notebook cuz im getting marked on process and thinking thanks boss
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