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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Thread starter WMDhamnekar
Start date Jan 22, 2024
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Expectation Integral calculus Probability Uniform distribution

Jan 22, 2024

WMDhamnekar

MHB

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] ##.

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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,
$$L = 2\displaystyle\int_0^\frac12\left[\displaystyle\int_0^x (1-2u)^n du + \displaystyle\int_x^{1-x}(1-u-x)^n du\right]dx = \frac{n+3}{2(n+1)(n+2)}$$

What is 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?

The expectation of the minimum difference between the 0th independent and identically distributed random variable (i.i.d.r.v.) and any of the ith i.i.d.r.v.s from 1 to n depends on the distribution of these variables. However, for uniformly distributed variables on [0,1], the expected minimum difference can be calculated and generally decreases as n increases, reflecting that more samples provide a higher chance of a closer value to the 0th variable.

How does the number of variables n affect the expected minimum difference?

As the number n of i.i.d.r.v.s increases, the expected minimum difference between the 0th variable and the closest among the ith variables typically decreases. This is because the probability of finding a variable closer to the 0th variable increases with more samples. In mathematical terms, for uniform distributions, this expectation tends to scale inversely with the number of variables.

Can this expectation be computed for any type of distribution?

Yes, the expectation can theoretically be computed for any distribution of the i.i.d.r.v.s, but the complexity and the form of the result will depend on the specifics of the distribution. For common distributions like normal, exponential, or uniform, specific formulas or numerical methods can be used. For others, especially those lacking simple closed forms, numerical simulation or advanced probabilistic techniques might be necessary.

What role does the distribution of the i.i.d.r.v.s play in determining this expectation?

The distribution of the i.i.d.r.v.s critically affects the calculation of the expected minimum difference. Different distributions have different properties such as variance, skewness, and kurtosis, which all influence how spread out the values are and how likely they are to be close to each other. For example, a narrower distribution like a normal distribution with a small standard deviation will typically have a smaller expected minimum difference compared to a wider one.

Are there practical applications of computing this expectation in real-world scenarios?

Yes, there are several practical applications. For instance, in optimization and operations research, understanding the minimum difference can help in configuring systems to minimize conflict or overlap. In finance, it can be used to estimate the minimum expected change in asset prices, aiding in risk assessment and investment strategies. In machine learning, this concept can help in clustering or anomaly detection where understanding close neighbors in a high-dimensional space is crucial.