# Recreating an Integer Sequence After Convolution

... and also inscribing it as a probability circulation.

Intend we have a series of non-negative integers that is routine with duration $N$:

\begin{equation*} A_{1},A_{2},...,A_{N},A_{1}... \end{equation*}

Each $A_{k}$ tackles a value no more than some constant $B$:

\begin{equation*} 0 \leq A_{k} \leq B \end{equation*}

We after that take this series and also do a straightforward convolution, for some constant $L > 0$ and also $1 \leq n \leq N$:

\begin{equation*} S_{L}(n) = A_{n} + A_{n+1} +...+ A_{n+L-1}. \end{equation*}

From $S_{L}(n)$ we after that create a probability circulation $P(n)$ which offers the regularity of each of its values. Allow $e_{j}(k) = 1$ if $j = k$ and also $0$ or else. After that:

\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 love to figure out is the level to which this procedure can be turned around. I have 2 information factors:

1) I recognize (virtually) every little thing concerning the probability circulation $P(n)$: the circulation itself, its mean, array, difference, skewness, kurtosis, and so on

2) I can inform you the regularity of values of $A_{k}$ in one duration, to make sure that if the series is 1,0,2,3,1,0, I can inform you there are 2 0's, 2 1's, one 2, and also one 3.

To what level am I able to rebuild the series $A_{k}$ from these 2 information factors?

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2019-05-05 10:43:32
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No, you can not recoup the series $A_k$. As an unimportant instance, note that any kind of cyclic permutation of $(A_1, A_2, \dots, A_N)$ would certainly cause the very same circulation (and also regularities). Yet given that this is a routine series, you possibly uncommitted concerning comparing cyclic permutations of the very same series, so below's an additional instance.
Take into consideration L = 1. After that your chance circulation is simply equal to the regularities, so any kind of permutation of the $A_k$s would certainly offer the very same circulation. If L = 1 is also degenerate, below's an additional instance with L = 2.
Claim $L=2$, and also $N=10$. After that, both series $(1, 1, 0, 0, 0, 1, 1, 0, 0, 0)$ and also $(1, 1, 0, 0, 0, 0, 1, 1, 0, 0)$ would certainly have the very same circulation of amounts $S_L$ : 2 two times, 1 4 times, and also 0 4 times. You can conveniently expand this instance to any kind of $L$.