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Assume a particle X in $d$-dimensional hypercube. Each dimension is independent of another and the particle's position is distributed uniformly in each.

A distance measure of $D = \frac{1}{2} \max\limits_{i\in 1\cdots d} \lvert \frac{1}{2} - X_i\rvert$ is computed

$P(D >a) = p(\frac{1}{2} \max\limits_{i\in 1\cdots d} \lvert \frac{1}{2} - X_i\rvert >a) = p(\max\limits_{i\in 1\cdots d} \lvert \frac{1}{2} - X_i\rvert > 2a)= 1- p(\max\limits_{i\in 1\cdots d} \lvert \frac{1}{2} - X_i\rvert \leq 2a) = 1 - \prod\limits_i p(\lvert \frac{1}{2} - X_i \rvert \leq 2a))$

$ = \prod\limits_i 2 p(X_i-\frac{1}{2} \leq 2a | X_i >\frac{1}{2}) = 1-(8a)^d$

Am I goin wrong somewhere?

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    Why is the probability that the maximum over dimensions of ${1 \over 2} - X_i$ equal to the product of the probabilities that *each* dimension is greater than $a$? An imagine that $d \to \infty$. Then surely the probability that the particle deviates from ${1 \over 2}$ on *at least one* dimension gets higher and higher (approaching 1.0). However, your formula states the probability is the product of a large (infinite) number of terms, each of which is less than 1.0, so that product gets *smaller* (approaching 0.). Rethink what is going on here. – 2017-02-01
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    Thanks for that insight. So it would be accurate to say $p(D>a) = p(\frac{1}{2} max |\frac{1}{2} - X_i| > a) = p(max |\frac{1}{2} - X_i| > 2a)$, but I dont know how to proceed from here. Any hints would be helpful – 2017-02-01

2 Answers 2

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The probability that the particle deviates from ${1 \over 2}$ on at least one dimension is 1 - probability that none of them do.

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The event $Da)=1-(4a)^d. $$ (of course if $a\geq1/4$ the probability is $0$).