Talk:Hewitt-Savage zero-one law

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The example seems nearly vacuous, since the only way a random variable taking values in [0, ∞) could fail to have strictly positive expectation is for it to be almost surely zero. Michael Hardy 22:51, 21 August 2006 (UTC)

The emphasis of that example, though, isn't the point about the expectation but rather the application to the Hewitt-Savage zero-one law to see that the probability of the random series diverging is either zero or one. I agree that the example is very simple (and is, indeed, the same one as used for Kolmogorov's zero-one law), but that's not necessarily a bad thing. I would welcome a better showcase example for this result. Sullivan.t.j 23:12, 21 August 2006 (UTC)

I have added a comment on the example being rather simple and the distinction between being able to apply a zero-one law and being able to work out which of the two possible values is the correct one. I have also added a similar warning to Kolmogorov's zero-one law. Sullivan.t.j 23:32, 21 August 2006 (UTC)

The definition given here of exchangeable event collapses to that of event in the algebra generated by the X_i. In fact if we assume the X_i to be independent and identically distribuited then we have, taken B_i borel sets P{X_j \in B_i}P{X_i \in B_i}=P{X_i \in B_i,X_j \in B_j)=P{X_j \in B_i}P{X_i \in B_i}, for each i,j \in N; now we can choose the permutation (i,j) on the niddle event and the equality between the two sides still holds. What i mean is that the distribution of a sequence (X_1,X_2,...) of iid is ALWAYS independent of their order so every event genrated by these variables should be "exchangeable" according to the definition; but we don't want this. What we could do is either remove the iid hypotheses or define an exchangeable event to be a set in \sigma(X_1.....) that remains fixed under the action of a finite permutation of the variables.(Loeve does the first, Shiryaev the latter) The Hewitt-Savage can be proved by using both definitions.

Sorry the joint distribution was completely misprinted. Should be

P{X_j \in B_j}P{X_i \in B_i}=P{X_i \in B_i,X_j \in B_j)=P{X_j \in B_i}P{X_i \in B_j} --213.140.19.113 12:29, 18 April 2007 (UTC)