Suppose we have a sequence of non-negative integers that is periodic with period N: A 1 , A 2 , . . . , A N , A 1 . . .Each A k takes on a value no greater than some constant B: 0 ≤ A k ≤ BWe then take this sequence and do a simple convolution, for some constant L &gt; 0 and 1 ≤ n ≤ N: S L ( n ) = A n + A n + 1 + . . . + A n + L − 1 .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: P ( n ) = ( e n ( S L ( 1 ) ) + e n ( S L ( 2 ) ) + . . . + e n ( S L ( N ) ) ) / N .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.

Question

Suppose we have a sequence of non-negative integers that is periodic with period   N:      A      1    ,      A      2    ,  .  .  .  ,      A      N    ,      A      1    .  .  .Each       A      k   takes on a value no greater than some constant   B:  0  ≤      A      k    ≤  BWe then take this sequence and do a simple convolution, for some constant   L  &amp;gt;  0 and   1  ≤  n  ≤  N:      S      L    (  n  )  =      A      n    +      A          n      +      1        +  .  .  .  +      A          n      +      L      −      1        .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:  P  (  n  )  =  (      e      n    (      S      L    (  1  )  )  +      e      n    (      S      L    (  2  )  )  +  .  .  .  +      e      n    (      S      L    (  N  )  )  )  /  N  .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&#039;s, two 1&#039;s, one 2, and one 3.

Schetterai · Accepted Answer

No, you cannot recover the sequence       A    k  . As a trivial example, note that any cyclic permutation of   (      A    1    ,      A    2    ,  …  ,      A    N    ) would result in the same distribution (and frequencies). But since this is a periodic sequence, you probably don&#039;t care about distinguishing between cyclic permutations of the same sequence, so here&#039;s another example.Consider   L  =  1. Then your probability distribution is just equivalent to the frequencies, so any permutation of the Aks would give the same distribution. If   L  =  1 is too degenerate, here&#039;s another example with   L  =  2.Say   L  =  2, and   N  =  10. Then, both sequences   (  1  ,  1  ,  0  ,  0  ,  0  ,  0  ,  1  ,  1  ,  0  ,  0  ) and   (  1  ,  1  ,  0  ,  0  ,  0  ,  1  ,  1  ,  0  ,  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.

Suppose we have a sequence of non-negative integers that is periodic with period N :

Answered question

Answer & Explanation

New Questions in Precalculus