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Let us have random vectors $X_1, \dots, X_N$ which are identically independently uniformly distributed in the $n$-dimensional unit hyperbox $[0; 1]^n$. Let $c = (0.5, \dots, 0.5)$ be the center of the hyperbox.

What is the conditional distribution of a vector $X_i$ given that the closest vector to the center has a fixed distance: $$ \min_{i = 1, \dots, N} d(X_i,c) = \delta $$

where $d$ is the maximum metric and $\delta$ is a fixed number? We can suppose that $\delta$ is sufficiently small.

My hypothesis is that the vectors have 2 possible distributions. This is illustrated (for the 2 dimensional case) in the following picture:

enter image description here

If the vector $X_i$ happens to be the one closest to $c$ (which happens with a probability of $\frac{1}{N}$) then it is uniformly distributed on the surface of the sphere $S = \{X\in \mathbb{R}^n\ \text{ such that } d(X,c) \leq \delta \}$ since it was previously uniformly distributed.

If it is not closest to $c$ (which happens with a probability of $\frac{N-1}{N}$) then all we know that it is not within $S$. So in order to get the conditional density function we need to divide by the original density function by the probability that $X_i$ does not fall into $S$ thus getting a uniform distribution in $[0; 1]^n / S$ which is the area shown in blue.

Is my hypothesis true? I have difficulties expressing myself precisely, especially in probability theory so I would like to know if this justification would pass as a proof.

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