<t>QuoteOriginally posted by: quantystQuoteOriginally posted by: Siberianif x(0) = 1, then x(1)~U(0,1^1), x(2)~U(0,1^1), i.e. 1^a is 1 for any a?No. Your understanding is different from mine.It is correct that X(1)~U(0,1).But it is incorrect to say X(2)~U(0,1). In fact, X(2)~U(0,X(1)).Now, let's fin...