ARAB BANK LOSES APPEAL Peers Reject Suit for Funds Frozen Under Israeli. Robinh ( talk) 20:55, 25 April 2009 (UTC) Reply We dice up premium ingredients to create your very own savory bowls. NEW SEASON UNDER WAY Sowing of Lawns and Hardy Plants Has Already Begun in the. Algebraist 14:31, 25 April 2009 (UTC) Reply The digits of pi certainly are random in the Bayesian paradigm. The obvious value to choose is infinity, in which case the expectation is also infinite. Thus to get a meaningful expectation value for the point at which the first repetition occurs, you need to decide what value this variable should take if no such repetition occurs. If instead of considering pi, we consider a random number (between 0 and 1, with independent uniformly-distributed digits, say), then the probability that such a repetition occurs is not 1 (it's not 1/9, either your calculation is an overestimate due to some double counting). Furthermore, even if pi is normal, it's not obvious (at least to me) that it must contain such a repetition. Sp in ni ng Spark 14:19, 25 April 2009 (UTC) Reply Expectations are only meaningful for random events, and the digits of pi are not random, so you're never going to get a meaningful expectation here. I would still like to know how this can have a finite probability but not have an expectation value. C 13:48, 25 April 2009 (UTC) Reply Yes, I was making that assumption (and indeed I knew that it was an assumption - forgive my slopiness, I am an engineer, not a mathematician).Does this mean that expectation value is meaningless here? Sp in ni ng Spark 13:06, 25 April 2009 (UTC) Reply Your first assumption would have to be that pi is a normal number, which is not actually known.
Hi, I read the following surprising comment in a maths textbook: "Let f ( x )
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