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Expected value of stationary independent increments.

  1. Apr 7, 2010 #1
    1. The problem statement, all variables and given/known data
    Let [tex]\{ X(t),t \ge 0\} [/tex] be a random process with stationary independent increments and assume that [tex] X(0)=0 [/tex]. Show that:

    [tex]E[X(t)] = {\mu _1}t[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I tried to work backwards by argueing that the mean between time intervals it equal to the product of the mean at t =1 and the time interval
    [tex]\mu \Delta {t_1} + \mu \Delta {t_2} + \cdot \cdot \cdot + \mu \Delta {t_n} = \mu (\Delta {t_1} + \Delta {t_2} + \cdot \cdot \cdot + \Delta {t_n}) = \mu {t_n}[/tex]

    Allthough this gives me the write ansewr i dont feel its a very solid proof. Its seems there should be a straight forward way to show this with the diffenition of the expected value but i cant seem to see it.
    Last edited: Apr 7, 2010
  2. jcsd
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