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This question might sound a little bit mystical, but it seemed like an interesting idea, so I am posting it here. Despite the title, I know it probably does not work miracles, but here goes anyway.

I have been thinking about predicting outcomes of random events, and had the following idea. Let $X$ be a random variable.

Then the average value we can expect $X$ to have is $\mu := E(X)$, its mathematical expectation.

This seems like a good start, but as we know from probability, random variables also have a cerain degree of dispersion. But here instead of the standard deviation, I'll use $\delta_1 := E(|X-E(X)|)$ instead, because what I'm really interested in is: "How far away from the expected value can I expect to find the value of my variable?" This also generalizes more faithfully, I think.

So, this would lead me to think that I'd most likely find the value of my variable to be $\mu\pm\delta_1$.

But here comes the next question: "How far from $\delta_1$ will the actual error $|X-E(X)|$ lie on average?" So we define $\delta_2:=E(\big||X-E(X)|-\delta_1\big|)$. So we have another correction: we predict that the value should be $\mu\pm\delta_1\pm\delta_2$, where the choices of $+$ and $-$ in the $\pm$ are independent (not necessarily in any statistical sense of the word).

We can of course continue this indefinitely and recursively define:

$$X_0:=X$$

$$\delta_0:=\mu = E(X)$$

$$X_{n+1}:=|X_n-\delta_n|$$

$$\delta_{n+1}:=E(X_{n+1})$$

And now we take the final prediction for where the value of $X$ is going to be found to be: $$\mu +\sum_{n=1}^\infty(\pm\delta_n)$$ where the choices of $+$ and $-$ in $\pm$ are again completely arbitrary and mutually independent.

My questions are:

What exactly does this sum describe?

To me, since I do not know much physics and ascribe mystical properties to it, this looks like some sort of quantum states for the variable.

On a more serious note: I'd expect (maybe after adding some appropriate conditions) the set of such sums to be dense in the interval $(\mu -\sum_{n=1}^\infty\delta_n,\mu +\sum_{n=1}^\infty\delta_n)$. [Edit: actually, after thinking about this some more, it doesn't seem so likely anymore. My intuition about it has completely abandoned me, to be honest.]

Has this been explored before? (And does it have a name?)

Also, if you prefer to work with standard deviations (or their squares), you can modify the above recursive definition to read $X_{n+1}=(X_n-\delta_n)^2$. Such a sequence may seem more standard, but I'm not sure it is so intuitively obvious what it describes anymore. So thoughts about this variant are also welcome.

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    Could you explain how does: $X_{n+1}:=|X_n-\delta_n|$?2012-01-27
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    @EmmadKareem: Well, that's simply a definition. I use it because I want to measure how far from the previous average value $E(|X_n-\delta_n|)$ the true value of $|X_n-\delta_n|$ will be found on average, so we can "apply this correction" then.2012-01-27
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    Interesting question. No literature that I know. Starting from $X$ uniform on an interval, the sum $S=\mu+\sum\limits_n\pm\delta_n$ is also uniform on the same interval. But in general, the distribution of $S$ is not the distribution of $X$ since it is always symmetric around $\mu$.2012-01-28
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    Did you make some progress on this question?2012-09-19
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    @did: I was quite busy doing other stuff, so I didn't really have time to think about it. I plan to return to it when my time admits. If I make some progress, I'll post it here and notify you.2012-09-19
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    Nice to know. $ $2012-09-19
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    but what about $\delta^{bis}_{2}:=E(\big|-|X-E(X)|-\delta_1\big|)$ it's the symmetric part of the situation?2013-02-22

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