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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:


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



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: $X_{n+1}:=|X_n-\delta_n|$? – NoChance Jan 27 '12 at 2:55
@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. – Dejan Govc Jan 27 '12 at 9:33
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 $\mu$. – Did Jan 28 '12 at 13:51
Did you make some progress on this question? – Did Sep 19 '12 at 18:36
@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. – Dejan Govc Sep 19 '12 at 19:05

Not an answer but.

As explained in the comments, the question is about symmetric distributions since the random variable defined by such a series is always symmetric. One can rephrase the construction as follows.

Consider the transformation $T$ defined by $T(Y)=|Y-\mathrm E(Y)|$, for every nonnegative random variable $Y$. The problem is to compare the distributions of the two symmetric random variables $\varepsilon_0 Y$ and $$ Z=\sum\limits_{n\geqslant1}\varepsilon_nt_n(Y),\qquad\text{with}\quad t_n(Y)=\mathrm E(T^n(Y)), $$ where $T^n$ is the transformation $T$ iterated $n$ times and $(\varepsilon_n)_{n\geqslant0}$ is an i.i.d. sequence of symmetric Bernoulli random variables.

Before asking for a comparison between the distributions of $\varepsilon_0Y$ and $Z$, one might wish to determine the distributions of $Y$ such that this procedure defines a proper random variable $Z$. This concerns the behaviour of the sequence $(t_n(Y))_{n\geqslant1}$ since, basically, one is asking whether the series $\sum\limits_nt_n(Y)^2$ converges or not.

Even before this, one can wonder whether the sequence $(t_n(Y))_{n\geqslant1}$ is always nonincreasing. Note that $t_1(Y)\gt\mathrm E(Y)$ may happen...

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Thanks for your thoughts. – Dejan Govc May 8 '12 at 11:27
You are welcome. Thanks for this interesting (and somewhat frustrating) question. – Did May 13 '12 at 10:14

Not an answer at all, but concern the sequence of confidence intervals (assumption $X$ has infinite support etc.) :
k=1. Confidence interval for $E(X)$ at 95%.
k=2. Confidence interval for confidence interval for $E(X)$ at 95%.
k=3. Confidence interval for confidence interval for confidence interval for...

Statement of level k collapse to statement of level k-1 but with different alfa.
So :

"Confidence interval for confidence interval for $E(X)$ at 95%." is identical with
"Confidence interval for $E(X)$ at 95%*95%."

Your construction could be used to make series of confidence intervals with different/unknown but fixed alfas.

I wonder if you see my post useful, please comment.

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