What Is the Probability of a Zero in the Sum of x + a*y in Z_[q]?

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SUMMARY

The discussion centers on calculating the probability of obtaining at least one zero in the expression x + a*y, where x and y are vectors in Zmq and a is an element in Zq. The initial calculations presented involve determining the probabilities P(A) and P(B) using basic probability principles, specifically focusing on the conditions when a = 0 or when x_i = -a*y_i for some indices. The final probability expression derived is 1/q + 1 - ((q-1)/q)^m - 1/q*(1 - (q-1)/q)^m, indicating a complex interaction between the variables and their distributions.

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  • Understanding of vector spaces over finite fields, specifically Zmq and Zq.
  • Familiarity with basic probability theory, including union and intersection of events.
  • Knowledge of combinatorial counting principles related to elements in finite sets.
  • Ability to manipulate algebraic expressions involving probabilities.
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  • Study the properties of finite fields, particularly Zq and their applications in probability.
  • Learn about combinatorial probability techniques to better understand counting elements with specific properties.
  • Explore advanced topics in probability theory, such as conditional probabilities and their applications in vector spaces.
  • Investigate the implications of zero elements in vector sums within the context of linear algebra.
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Mathematicians, statisticians, and computer scientists working with probability theory, particularly in the context of finite fields and vector spaces. This discussion is beneficial for anyone involved in theoretical research or applications requiring an understanding of probability distributions in algebraic structures.

petha
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Hi, I am given the following problem.

Given the vector

x+a*y x,yin Zmq, a in Zq. What is the probability that there will be at least one zero in the sum?
My reasoning so far.

x+a*y = 0 either if a=0 or x i = -a*yi for some (or all) 1≤ i ≤ m

So by basic probability P(A U B) = P(A) + P(B) -P(A and B).

1 P(A) = P(a=0) = 1/q
2 P(B) = 1-P(No zeros) = 1 - ((q-1)/q)m (qm elements in total, (q-1)m elements with no zeros.
P(A AND B) = P(A)*P(B) = 1/q(1-((q-1)/q) m)
So in total 1/q+1-((q-1)/q)m)-1/q*(1-(q-1)/q)m)

This looks like a total mess, but I am not certain what is wrong in my calculations.
 
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petha said:
Given the vector

x+a*y x,yin Zmq, a in Zq. What is the probability that there will be at least one zero in the sum?
My reasoning so far.

x+a*y = 0 ...
That says the entire vector is zero. I think you meant only that at least one dimension is zero.
... either if a=0 ...
How would that guarantee any zero terms in the sum? x might contain no zeroes.
 

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