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Stochastic process proof

  1. Feb 4, 2012 #1
    If a stoch. process Xt has independent and weak stationary increments. var(Xt) = σ^2 for all t, prove that Cov(xt,xs) = min(t,s)σ^2

    I'm not sure how to do this. I tried using the definition of covariance but that doesn't really lead me anywhere. If it's stationary that means the distribution doesn't change as time changes. I was thinking of setting s = t+k and showing the covariance being min(t, t+k)var(Xt) but I don't know how to get to there.
     
  2. jcsd
  3. Feb 4, 2012 #2
    is var(xt)=σ^2 or σ^2*t? if it's σ^2*t, then you can say.

    cov(xt,xs)=cov(xt-xs+xs,xs) and use independent + stationary increments to prove it.

    i'm not sure if it holds true if it's simply σ^2 though...
     
  4. Feb 4, 2012 #3
    I think that you mean X(t), find Cov(X(t),X(s)). Independent and stationary implies X is a Poisson process. Suppose t>s. Then X(t)=X(s)+X(t-s) and
    Cov(X(s)+X(t-s),X(s)) = Var(X(s))+Cov(X(t-s),X(s)) = Var(X(s)) by independence.
    If s>t then you find Var(X(t)). Recall that the variance is the rate times the length of the time interval.
     
  5. Feb 5, 2012 #4
    independent and stationary does NOT imply X is a poisson process.

    the opposite is true, but many stochastic processes that are independent and stationary are not necessarily poisson processes (brownian motion, for example has independent and stationary increments).
     
  6. Feb 5, 2012 #5
    var(xt) =σ^2 for all t. I think μ stays the same for all increments since its a stationary process. But that doesn't really get me anywhere. It's asking to prove that the covariance between increments is the variance x min{s,t}. So if t>s then it would be sσ^2 and vice versa. I really don't know how to get there though.
     
  7. Feb 5, 2012 #6
    i think given that information, cov(xs,xt)=min(s,t)σ^2 only if var(xt)=tσ^2. i'd double check that the problem isn't a typo or something.
     
  8. Feb 6, 2012 #7
    Thank you very much jimmypoopins. I could have said if it's a point process then it is Poisson, the context in which this result is usually first seen, but even that isn't necessary to say because the result is more general. Thanks.
     
  9. Feb 10, 2012 #8
    Kuma,

    How did this problem turn out for you? It's a very standard problem assigned but I think you had the problem written wrong. Let us know.
     
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