TITLE: Recreating an Integer Sequence After Convolution
QUESTION [5 upvotes]: ...and encoding it as a probability distribution.
Suppose we have a sequence of non-negative integers that is periodic with period $N$: 
\begin{equation*}
A_{1},A_{2},...,A_{N},A_{1}...
\end{equation*}
Each $A_{k}$ takes on a value no greater than some constant $B$:
\begin{equation*}
0 \leq A_{k} \leq B
\end{equation*}
We then take this sequence and do a simple convolution, for some constant $L > 0$ and $1 \leq n \leq N$:
\begin{equation*}
S_{L}(n) = A_{n} + A_{n+1} +...+ A_{n+L-1}.
\end{equation*}
From $S_{L}(n)$ we then form a probability distribution $P(n)$ which gives the frequency of each of its values. Let $e_{j}(k) = 1$ if $j = k$ and $0$ otherwise. Then:
\begin{equation*}
P(n) = (e_{n}(S_{L}(1)) + e_{n}(S_{L}(2)) +...+ e_{n}(S_{L}(N))) / N.
\end{equation*}
What I would like to find out is the extent to which this process can be reversed. I have two data points:
1) I know (pretty much) everything about the probability distribution $P(n)$: the distribution itself, its mean, range, variance, skewness, kurtosis, etc.
2) I can tell you the frequency of values of $A_{k}$ in one period, so that if the sequence is 1,0,2,3,1,0, I can tell you there are two 0's, two 1's, one 2, and one 3.
To what extent am I able to reconstruct the sequence $A_{k}$ from these two data points?

REPLY [2 votes]: No, you cannot recover the sequence $A_k$. As a trivial example, note that any cyclic permutation of $(A_1, A_2, \dots, A_N)$ would result in the same distribution (and frequencies). But since this is a periodic sequence, you probably don't care about distinguishing between cyclic permutations of the same sequence, so here's another example.
Consider L=1. Then your probability distribution is just equivalent to the frequencies, so any permutation of the $A_k$s would give the same distribution. If L=1 is too degenerate, here's another example with L=2.
Say $L=2$, and $N=10$. Then, both sequences $(1, 1, 0, 0, 0, 1, 1, 0, 0, 0)$ and $(1, 1, 0, 0, 0, 0, 1, 1, 0, 0)$ would have the same distribution of sums $S_L$: 2 twice, 1 four times, and 0 four times. You can easily extend this example to any $L$.