I Computing the expectation of the minimum difference between the 0th i.i.d.r.v. and ith i.i.d.r.v.s where 1 ≤ i ≤ n

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The discussion focuses on calculating the expected minimum difference between an independent random variable \(X_0\) and \(n\) other independent random variables \(X_i\) uniformly distributed on [0,1]. The author presents a detailed solution involving the computation of an integral that incorporates the probability of the absolute difference between \(X_0\) and \(X_i\). The key formula derived shows that the expectation can be expressed as \(L = \frac{n+3}{2(n+1)(n+2)}\). The solution involves evaluating nested integrals and applying probability principles to derive the final result. This approach effectively demonstrates the relationship between uniform distributions and expected values in probability theory.
WMDhamnekar
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Problem :Let ##X_0,X_1,\dots,X_n## be independent random variables, each distributed uniformly on [0,1].Find ## E\left[ \min_{1\leq i\leq n}\vert X_0 -X_i\vert \right] ##.

Would any member of Physics Forum take efforts to explain with all details the following author's solution to this question?

Author's solution:
Let L be the expression in question. Then $$L=\displaystyle\int_0^1 E \left[ \min_{1 \leq i \leq n}\vert x- X_i \vert dx\right] =\displaystyle\int_0^1\displaystyle\int_0^1\left[P(\vert X_0 - x\vert\geq u )\right]^ndu dx $$
Since ## P(\vert X_0 -x \vert \geq u ) = \max(1-u-x,0) + \max(x-u ,0), x,u \in [0,1]## we have $$ P(\vert X_0 -x \vert \geq u )=\begin{cases} 1- 2u & 0 \leq u < x\\ 1-u -x & x \leq u < 1-x
& x \in[0,\frac12 ]\\ 0, & 1-x \leq u \leq 1\end{cases}$$
So,
$$L = 2\displaystyle\int_0^\frac12\left[\displaystyle\int_0^x (1-2u)^n du + \displaystyle\int_x^{1-x}(1-u-x)^n du\right]dx = \frac{n+3}{2(n+1)(n+2)}$$
 
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