Proving Covariance for Stationary Stochastic Processes

[Kuma]

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.

 

[jimmypoopins]

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...

 

[alan2]

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.

 

[jimmypoopins]

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).

 

[Kuma]

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.

 

[jimmypoopins]

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.

 

[alan2]

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.

 

[alan2]

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.