Might such a sequence of mathematical expectations be able to predict uncertain events? 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.
 A: 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...
A: 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.
