Re: April 2007 IBM Ponder This Challenge
I don't think you can deduce S is stationary from what you've done here. If t were restricted to say [-10,10], then we could easily construct a suitable X and S process.
We can define a backwards process in this case by noting that the increments in S are IID, so we can add up their negatives in reverse order.
This isn't quite the same as conditioning S on going through 0 since it ignores the issue of a prior (but I'm sure we can show that no such S exists anyway since it'll need infinite mass for all time, so we need another way of justifying the construction).
Marv
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