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Homework Help: Equality of distribution for random varibles

  1. Jun 21, 2010 #1
    If 2 random variables X, and Y have the same distribution, does that mean that for another random variables Z, X + Z and Y + Z also have the same distribution?

    From looking at the convolution formula, the answer should be yes, because the convolution of random variables depends only on the distribution of the 2 variables being added.

    However, if we consider the Poisson process N(t), then the stationary increment property says that N(5) - N(4) has the same distribution as N(1).

    Now consider the random variable (N(4) - N(5)) - N(1). If the above claim is true then this has the same distribution as N(1) - N(1), which has variance 0, i.e. Var((N(4) - N(5)) - N(1)) = 0. This would seem to imply that the count of random Poisson events between time 4 and 5 must always be exactly the same as between time 0 and 1 and there is actually no randomness at all, which can't possibly be true.
     
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  3. Jun 21, 2010 #2

    LCKurtz

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    But that isn't of the form N(r) - N(s).
     
  4. Jun 21, 2010 #3
    But it equals N(1) - N(1) which is in the form N(r) - N(s).

    What if we wrote it like this:

    1. Clearly Var(N(1) - N(1)) = 0.
    2. Now N(5) - N(4) has the same distribution as N(1) by the stationary increments property.
    3. Thus (N(5) - N(4)) - N(1) has the same distribution as N(1) - N(1) (by the claim that if X has the same distribution as Y, then X - Z has the same distribution as Y - Z)
    4. So, in particular, they have the same variance, i.e. Var((N(5) - N(4)) - N(1)) = Var(N(1) - N(1)) = 0.
     
    Last edited: Jun 21, 2010
  5. Jun 21, 2010 #4

    LCKurtz

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    It has been a long time since I looked at this stuff, but I think your problem is that

    N(5)-N(4) is itself not a member of your Poisson process N(t) and you can't claim the stationary increment property to (N(5)-N(4)) - N(1).
     
  6. Jun 21, 2010 #5
    Hmm, I don't think I did that. The only time I used stationarity was at 2) where I said N(5) - N(4) has the same distribution as N(1). Which is OK. I didn't use stationarity to get (N(5) - N(4)) - N(1) having the same distribution as N(1) - N(1).
     
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