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#1




The expected value of continuous function
I don't really know how to express my question and I am trying to make it clear
There are two ways to define the expected value The first way is the total outcome divided by the number of experiments the second way is summing up the possible outcome times their probability In rolling a fair dice example using the first way: [(n/6)(1)+(n/6)(2)+...+(n/6)6]/n, n goes to infinity using the second way: (1/6)(1)+(1/6)(2)+...+(1/6)(6) which turns out to be the same result Now, what I am confused about is the continuous case. For example: p(x) = 2x, 0<x<1 using the second way, possible outcomes are x, and their respective probability is 2x summing up in continuous case becomes the integral integral from 0 to 1, (2x)(x) The question is, how can I do this in "first way definition"? since the continuous case cannot take a single point, so I can only take a subset, but it has infinitely many subsets, like 0<x<0.3 with probability 0.09 0<x<0.4 with probability 0.16 0.22<x<0.33 with probability 0.0121 and infinitely more and I just stuck here How can I do it in the "first definition", and more importantly, how can I prove that it is equivalent to the second definition? Thanks a lot! 
#2




I don’t follow your first method in the discrete case. For example suppose n=5, which I think corresponds to rolling 5 dice. One possible outcome of that is three 2s, one 4 and one 6. How does that outcome relate to your sum of all outcomes?

#3




An explanation from economics is that the expected value is the maximum price someone that is risk neutral who guesses the answers, would pay to take the test where the prize is the number of correct answers. Now that is somewhat circular as the definition of risk neutral is based upon making decisions according to expected value.

#4




Many people who define the mean as "sum the values and divide by the quantity" are confusing "how to find" with a definition.
Consider the following definition for the mean in a finite, discrete case: That constant value such that if you replace each value in the original set with this constant, you still get the same total sum. Illustration: Code:
5 6 8 9 => 28 7 7 7 7 => 28 Now, for the continuous function case, the goal is to find the constant function such that if you integrate both over the same interval, you get the same value. (I hope that the "how to" is also intuitively found here as above).
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#5




I don't understand what you are saying in the continuous cast.
Let's say I have the pdf x/2 0<x<2 The expected value is 4/3. The pdf integrates to one and the weighted pdf integrates to 4/3. Physically it would be the centroid or balancing point. The constant function that integrates to that is 2/3. 
#6




Quote:
I was intentionally not putting all of the pieces in as I've found that when one "struggles" a bit with the core concept, they remember it far longer than when their hand is held in the explanations. I may not always hit that right balance at the start, but the following conversations still serve the intended purpose. Note that centroid is another version of "average" that is not necessarily going to be the same as the mean. (I would say that the centroid would be the generalization of the median.)
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