Results 1 to 2 of 2

Math Help - Probability of the next entry in a series of discrete trials.

  1. #1
    Senior Member
    Joined
    Feb 2008
    Posts
    410

    Probability of the next entry in a series of discrete trials.

    Hi, all.

    Suppose you have a series of events, all the same, which can result in X or ~X. The probability of each is unknown. As you watch the series go forward, every entry (to start) is X. So:

    S = \{X, X, X, X, X... \}

    At any point, however, it is possible that the pattern will cease, and that we will get a ~X. But let's say it hasn't happened yet.

    Let's also say that the number of X's we have observed so far is n, such that the probability of this occurring is:

    P(S_n)=[P(X)]^n

    So, how do we determine P(X) ?

    We could determine a confidence level, I think...

    [P(X)]^n>.05

    P(X)>.05^{\frac{1}{n}}

    So, there's a 95% chance that P(X)\in(.05^\frac{1}{n},1], yes? But how do we actually find P(X)?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    We could use the confidence interval to find a minimum probability. Let's say that the error is given as E. So:

    E^{\frac{1}{n}} is the lower bound of P(X) in the 1-E confidence level. Let's call Q(X) the probable expectation that the next entry in the sequence is X. So, Q(X)\geq(1-E)E^{\frac{1}{n}}. We can take the derivative with respect to E to find the value of E which maximizes Q(X):

    \frac{d}{dE}(1-E)E^{\frac{1}{n}}=0

    \frac{d}{dE}(E^{\frac{1}{n}}-E^{\frac{1}{n}+1})=0

    \frac{1}{n}E^{\frac{1}{n}-1}-[\frac{1}{n}+1]E^{\frac{1}{n}})=0

    \frac{1}{En}E^{\frac{1}{n}}-[\frac{1}{n}+1]E^{\frac{1}{n}})=0

    [\frac{1}{En}-\frac{1}{n}-1]E^{\frac{1}{n}}=0

    \frac{1}{En}-\frac{1}{n}-1=0

    \frac{1}{En}=\frac{n+1}{n}

    En=\frac{n}{n+1}

    E=\frac{1}{n+1}

    Now, recall our Q(X) relationship:

    Q(X)\geq(1-E)E^{\frac{1}{n}}

    And substitute for E:

    Q(X)\geq(1-\frac{1}{n+1})[\frac{1}{n+1}]^{\frac{1}{n}}

    Q(X)\geq\frac{n}{(n+1)^{\frac{1}{n}}(n+1)}

    Q(X)\geq\frac{n}{(n+1)^{\frac{n+1}{n}}}

    It seems like we should be able to do better than this, though.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Probability- result of multiple trials
    Posted in the Statistics Forum
    Replies: 2
    Last Post: February 13th 2011, 06:24 PM
  2. Independent Trials & Discrete Random Variables
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: March 24th 2010, 03:39 PM
  3. Replies: 0
    Last Post: March 17th 2010, 01:49 PM
  4. Bernoulli trials probability
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 30th 2009, 02:15 PM
  5. [SOLVED] Binomial Trials, Probability, Please Help, I will love you.
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: April 25th 2006, 09:14 PM

Search Tags


/mathhelpforum @mathhelpforum