If I pick two uniformly random vectors over integers $$(a_1,\dots,a_{n}),(b_1,\dots,b_{n})\in[-m,m]^n\cap\Bbb Z^n$$ what is the probability that $$\sum_{i=1}^na_ib_i=0$$ holds?
Probability of orthogonal vectors.
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probability
orthogonality
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0By $[-m,m]$, do you mean an interval in $\Bbb Z$ or an interval in $\Bbb R$ ? In the latter case, the probability is obviously zero. – 2017-02-04
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0@JohnBentin integers. – 2017-02-04
2 Answers
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The orthogonal space of a line given by a vector is a hyperplane, and such a hyperplane has measure $0$ over the ambient space, so I would say such a probability is $0$. Just notice that given a direction, the quantity of other directions that are orthogonal to the first is very tiny compared with all the possible directions.
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For $n=1$ the probability is zero if your vectors are independent, because:
\begin{align*} P(a_1b_1 = 0) &= P(\{a_1 = 0\} \cup \{b_1 = 0\})\\ &= P(a_1 = 0) + P(b_1 = 0) - P(\{a_1 = 0\}\cap \{b_1 = 0\})\\ &= P(a_1 = 0) + P(b_1 = 0) - P(a_1 = 0)P(b_1 = 0)\\ &= 0. \end{align*}
You could try to do the same with induction for any $n \in \mathbb{N}.$