Talk:Markov's inequality

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Last year the interval [0, ∞) was changed to [0, ∞]. I feel that this is not a good change.

The first form specifies the non-negative real numbers; the second form includes the possibility that the function h can map from a real number to ∞. If ∞ is in the range of h, and the preimage of ∞ has non-zero probability under the appropriate probability measure, then clearly E[h(X)] = ∞, which must be greater than any probability!

Most importantly, the latter version of the interval goes against the convention with which I am familiar (which is, of course, to use the former version.)

I have changed it back; if anyone would like to discuss this further please feel free! Ben Cairns 01:27, 24 Jan 2005 (UTC)

The proof provide is too terse for the level of most people with just an undergraduate level understanding of probability. To see an easier proof to understand check the following link: http://mathworld.wolfram.com/MarkovsInequality.html

Now I've reinstated a proof I put here months ago, which someone removed. Michael Hardy 19:50, 6 October 2006 (UTC)