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

In summary, the study addresses the computation of the expected minimum difference between the first independent and identically distributed random variable (i.i.d.r.v.) and subsequent i.i.d.r.v.s for a given range of indices. It explores the statistical properties and methodologies required to derive the expectation of these differences, providing insights into the behavior of the minimum differences as the number of variables increases. The findings have implications for understanding distribution characteristics and probabilistic modeling in various applications.
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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,
$$L = 2\displaystyle\int_0^\frac12\left[\displaystyle\int_0^x (1-2u)^n du + \displaystyle\int_x^{1-x}(1-u-x)^n du\right]dx = \frac{n+3}{2(n+1)(n+2)}$$
 

FAQ: 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

What does "i.i.d.r.v." stand for in the context of this problem?

"i.i.d.r.v." stands for "independent and identically distributed random variables." This means that each random variable in the set has the same probability distribution and is statistically independent of the others.

How do you define the 0th i.i.d.r.v. and the ith i.i.d.r.v.s in this context?

The 0th i.i.d.r.v. is a specific random variable selected as a reference point. The ith i.i.d.r.v.s (where 1 ≤ i ≤ n) are the other random variables in the set, which are compared to the 0th i.i.d.r.v. to compute differences.

What is the mathematical expectation in this context?

The mathematical expectation, or expected value, is the average value that a random variable takes on over numerous trials. In this context, it refers to the average value of the minimum difference between the 0th i.i.d.r.v. and the ith i.i.d.r.v.s over many instances.

How do you compute the minimum difference between the 0th i.i.d.r.v. and the ith i.i.d.r.v.s?

To compute the minimum difference, you calculate the absolute difference between the 0th i.i.d.r.v. and each of the ith i.i.d.r.v.s (for 1 ≤ i ≤ n) and then take the smallest of these differences. Mathematically, it is expressed as min(|X_0 - X_i|) for i from 1 to n, where X_0 is the 0th i.i.d.r.v. and X_i are the other i.i.d.r.v.s.

What are the practical applications of computing this expectation?

Computing the expectation of the minimum difference between i.i.d.r.v.s has applications in various fields such as risk management, quality control, and optimization problems. It helps in understanding the variability and reliability of systems where random variables are involved, such as in finance, engineering, and data science.

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