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My question entails finding a continuous function equation that is the continuous function equivalent of a modified discrete probability calculation. This is in support of research that I have been doing for a number of years.

I'm looking for an equation that is based on a continuous Gaussian distribution. However, I find it easiest to explain using a discrete math example. Take the following distribution {70:.05, 80:.10, 90:.20, 100:.30, 110:.20, 120:.10, 130:.05} with a mean of 100. From this, if we then create a cumulative distribution that says, there is a 100% chance of obtaining at least 70 or more, a 95% chance of obtaining at least 10 or more in addition to 70 (i.e. 80-70), an 85% chance of obtaining at least 10 or more in addition to 80 (i.e. 90-80) and so forth we obtain a new set that looks like the following: {70:1, 10:.95, 10:.85, 10:.65, 10:.35, 10:.15, 10:.05}. The sum of the probability weighted values within this set are still equal to the mean from the first distribution (100).

Next, we take the set from above and then we square each probability to obtain a new set {70:1$^2$, 10:.95$^2$, 10:.85$^2$, 10:.65$^2$, 10:.35$^2$, 10:.15$^2$, 10:.05$^2$}. Weighting all of the values in this set by each respective squared probability and summing them yields a value of 91.95.

My question is this: What is the continuous function equivalent of the above discrete calculation that gives us the result of 91.95 using the following:

  1. A Gaussian distribution (normal distribution)
  2. Mean of μ
  3. Variance of σ2
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perhaps you are doing the following:

if $F$ is a cdf, $G = F^2$ and $H = 1 - (1 - F)^2$ are also cdfs. it looks as tho you are obtaining H from your [discrete] F.

btw - $G$ is the cdf of the larger of two independent observations from $F$ and $H$ is the cdf of the smaller of the two.

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Thanks for helping ronaf. I think you may be on the right track. However, I'm not so sure about H in your example. Rather, I may be looking for the sum of (1-F(x))^2 * (x-?) for all x's? Using my example, (x-?) would be (80-70), (90-80) and so forth. The result of 91.95 does not represent a probability. It represents a modified mean, if that makes sense? – Jason Barker Sep 9 '10 at 6:53

jason - you are correct: i did not address the 91.95 part of your question. so i'll try again: if $0\le X \sim F,\ {\rm E} X = \int_0^\infty xdF = \int_0^\infty (1-F)dx$. the 91.95 looks like $ \int (1-F)^2dx$ = $\int (1-H)dx = {\rm E} \min\{X_1,X_2\}$, where $X_1$ and $X_2$ are independent copies of $X \sim F$.

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