Binomial distribution with dependent trials?

Reynolds

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Sep 17, 2014
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I need your help with following problem:

String with n characters is given. For each character in string there is probability p that it is wrong. Now you take a sliding window of length k, k<= n, that slides over that string. For the given parameters p,k and n one must must determine the mean and variance of the number of the moving windows without any error.

For n = 5 and k = 2 we have sliding windows that contain letters of sting on positions 12, 23, 34 and 45.

I was thinking that I may define discrete random variable that counts how many windows are there with out any error, but very soon it becomes quite difficult to count. I was also trying to define some sort of generating function, but i did not get far. Thank you in advance!

 
Given a sequence (x1, x2,x3,...xn), my understanding of a set S of simple sliding window samples of size k is that
S = { (xj, xJ+1,xj+2,...xj+k-1), j = 1, 2, 3, n+1-k}

Thus if n=5 and k=2, the sample set would be
S = {(x1, x2), (x2, x3), (x3, x4), (x4, x5)}

This does not appear to be what you mean so could you please tell us what you mean by a sliding window of length k.
 
Given a sequence (x1, x2,x3,...xn), my understanding of a set S of simple sliding window samples of size k is that
S = { (xj, xJ+1,xj+2,...xj+k-1), j = 1, 2, 3, n+1-k}

Thus if n=5 and k=2, the sample set would be
S = {(x1, x2), (x2, x3), (x3, x4), (x4, x5)}

This does not appear to be what you mean so could you please tell us what you mean by a sliding window of length k.
Hi!
Thank you for the interest! :)
You are absolutely correct, but I have string here (so there is order), not only set.
For example let n = 5 and k = 2 and string is QWERT we would have QW, WE, ER and RT.
Hope I made it little more clear this time, sorry for the confusion. :)
 
Hi!
Thank you for the interest! :)
You are absolutely correct, but I have string here (so there is order), not only set.
For example let n = 5 and k = 2 and string is QWERT we would have QW, WE, ER and RT.
Hope I made it little more clear this time, sorry for the confusion. :)

Yes, I would assume there was order for the problem so that x1 preceded x2 preceded x3 ... in the example I gave.

Taking your example, suppose the word had one letter wrong. If that letter had been chosen randomly the probability would be 1/5 that any one letter in QWERT would be incorrect. What then would be the probability that one of the letters in the (ordered) set (Q, W) would be incorrect? How about the other sets (W, E), (E, R), and (R, T).
 
Yes, I would assume there was order for the problem so that x1 preceded x2 preceded x3 ... in the example I gave.

Taking your example, suppose the word had one letter wrong. If that letter had been chosen randomly the probability would be 1/5 that any one letter in QWERT would be incorrect. What then would be the probability that one of the letters in the (ordered) set (Q, W) would be incorrect? How about the other sets (W, E), (E, R), and (R, T).
I am not sure what r u trying to hint me :).
But I sense great dependance in your example. Because if the first letter is wrong i would have one more good window than i a case that second letter is wrong...
 
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