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Jun 15, 2020 at 7:27 history edited CommunityBot
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Jun 29, 2016 at 14:44 comment added user94231 @michael ah, got it. Thanks for the suggestion.
Jun 29, 2016 at 14:04 comment added user83457 As you say , it has to be normalized. $\frac {X_i - \lambda t} {\sqrt { \lambda}}$ should be converging to a wiener process. As I said, I would think you could find this, maybe in Billingsley's book, but it is intuitive and not hard. If you believe that, then you have the difference of independent processes each of which is converging to a wiener process.
Jun 29, 2016 at 11:53 comment added user94231 @michael could you explain your proposal in a little more detail?
Jun 29, 2016 at 11:53 history edited user94231 CC BY-SA 3.0
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Jun 28, 2016 at 12:08 history edited user94231 CC BY-SA 3.0
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Jun 28, 2016 at 10:48 comment added user83457 You will have an easier time finding the limiting behaviour of $X_i - \lambda t$ and putting them together using independence. I would expect billingsley to discuss it, but I don't know for sure that he does.
Jun 28, 2016 at 3:35 history asked user94231 CC BY-SA 3.0