Order statistics in the uniform (probability) distribution

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SUMMARY

The discussion focuses on calculating the distribution function and probability density function for the sample median Y(n+1) derived from a uniform distribution U(0, 1). The user, Mathmos6, proposes a solution involving the binomial coefficient {{2n+1}\choose{n+1}} and the expression a^n[(1-a)^n-(1-b)^n] to represent the probability of the sample median falling within a specified interval. The user questions the symmetry between the variables a and b in the derived function, suggesting a potential oversight in the formulation of the probability density function.

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  • Understanding of order statistics in probability theory
  • Familiarity with uniform distribution U(0, 1)
  • Knowledge of binomial coefficients and their applications
  • Basic concepts of probability density functions (pdf) and cumulative distribution functions (cdf)
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  • Explore the implications of symmetry in probability distributions
  • Investigate the use of binomial coefficients in statistical calculations
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Mathmos6
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Homework Statement


Hi there: I just need someone to tell me if I've made a mistake somewhere in my solution to this:

Suppose that X1 , . . . , X2n+1 are i.i.d. random variables that form a random sample
from the U (0, 1) distribution. Suppose that the values are arranged in increasing order as
Y1 ≤ Y2 ≤ . . . ≤ Y(2n+1) . Calculate expressions for the distribution function and for the probability density function of the random variable Y(n+1) (the sample median).

The Attempt at a Solution


Now if we want Y(n+1) to be in the interval [a,b] we need to have exactly n of the Xi in [0,a] and n+1 in [a,1], but ensure not all of the latter n+1 are in [b,1]. So we have {{2n+1}\choose{n+1}}a^n[(1-a)^n-(1-b)^n] where the (1-a)^n-(1-b)^n ensures not all n+1 of the latter Yi are in the [b,1] interval. But shouldn't there be some sort of symmetry in a and b with this function: shouldn't it be F(b)-F(a) where F'(x) is the pdf of the question? Because it doesn't look much like an F(b)-F(a) to me...

Thanks a lot,

Mathmos6
 
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