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I'm studying randomized algorithms and I sometimes come across results like

(1) The algorithm has an expected $O(f(n))$ cost.

and

(2) With constant probability, the cost is bounded by $O(f(n))$.

I'm perfectly fine with statements like (2), but I'm puzzled to what extent a statement like (1) is useful: For certain probability distributions of a random variable, the expected value itself occurs with less than constant probability; for other distributions, it occurs with 1ドル-1/n$ probability. Of course, in many cases, (1) is extended via concentration bounds to show high probability, but in cases where this isn't done, it doesn't seem that a statement on the "expected cost" lets us derive any implications on the actual performance of the algorithm, right?

asked Jul 29, 2013 at 1:58
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1 Answer 1

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If you have a cost of 1ドル$ with probability one half and 0ドル$ with probability one half, you never reach the expected cost of $\frac{1}{2}$.

But what the expected cost gives you is the average of costs. If you repeat the algorithm and make the average of the cost (after an infinite number of times) you will get an average cost equal to the expected cost.

I hope it's clear ...

answered Jul 29, 2013 at 15:43
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