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)

[edit] There's a small error in the article

In the first general measure theoretic statement, it says that the bound is given by an integral of f. It should be |f|.

No idea who should be told to fix it, but it should be fixed... —The preceding unsigned comment was added by 132.68.1.29 (talk) 13:19, 24 December 2006 (UTC).

Edit: never mind, fixed it.

[edit] Diagrams

The explanation of this inequality would benefit greatly from a diagram. Sanchom (talk) 08:03, 26 February 2007 (UTC)