The characteristic function of order statistics

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Discussion Overview

The discussion revolves around finding the characteristic function of the sum of order statistics from a set of independent and identically distributed (i.i.d) random variables. Participants explore the implications of ordering these random variables on the characteristic function and the underlying probability density functions (PDFs).

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant proposes that the characteristic function of the sum of the smallest K order statistics can be expressed as an expectation involving the ordered random variables.
  • Another participant questions whether the random variables are independently chosen and then ordered, suggesting that if they are, the definition of Y remains unaffected by the order.
  • A clarification is provided that the selection involves N i.i.d random variables, and Y is defined as the sum of the K smallest of these variables.
  • A further contribution presents the joint PDF of the ordered random variables and attempts to derive the characteristic function through integration, questioning the correctness of the derived expression.

Areas of Agreement / Disagreement

Participants express differing views on the implications of ordering the random variables and the correct approach to deriving the characteristic function. The discussion remains unresolved regarding the correctness of the proposed mathematical expressions and the treatment of independence in the context of order statistics.

Contextual Notes

There are limitations regarding the assumptions made about the independence of the random variables and the conditions under which the characteristic function is derived. The discussion does not resolve these assumptions or the implications of the ordering process.

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Suppose that ##Y=\sum_{k=1}^KX_{(k)}##, where ##X_{(1)}\leq X_{(2)}\leq\cdots X_{(N)}## and (##N\geq K##). I want to find the characteristic function of ##Y## as

\phi(jvY)=E\left[e^{jvY}\right]=E\left[e^{jv\sum_{k=1}^KX_{(k)}}\right]

In the case where ##\{X\}## are i.i.d random variables, the above characteristic function will be

\phi(jvY)=\prod_{k=1}^K\phi(jvX_k)

but when the random variables are ordered, they are no longer independent. How can the characteristic function be found in this case?
 
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, Are the random variables independently chosen and then ordered? If so it doesn't matter, since the definition of Y doesn't depend on the order. . On the other hand if your question means choose a random variable, then a second variable greater than the first, etc., it is a different problem.
 
mathman said:
, Are the random variables independently chosen and then ordered? If so it doesn't matter, since the definition of Y doesn't depend on the order. . On the other hand if your question means choose a random variable, then a second variable greater than the first, etc., it is a different problem.

The selection is like this: I have ##N## i.i.d random variables. Then ##Y## is the sum of the ##K## smallest random variables of the ##N## random variables.
 
I found that the joint PDF

##f_{X_{(1)},\,X_{(2)},\ldots,\,X_{(K)}}(x_1,\,x_2,\,\dots,\,x_K)=K!\prod_{k=1}^Kf_X(x_k)##

for ##0<x_1<x_2<\cdots<x_K<\infty##, where ##f_X(x_k)## is the PDF of the original random variables before ordering. Then we can write the characteristic function as

\phi(jvY)=\int_{x_1=0}^{\infty}\int_{x_2&gt;x_1}\cdots\int_{x_K&gt;x_{K-1}}K!\prod_{k=1}^Ke^{jvx_k}f_X(x_k)\,dx_1\,dx_2\cdots\,dx_K

which is the multiplication of ##K## integrals

\phi(jvY)=\left(\int_{x_1=0}^{\infty}e^{jvx_1}f_X(x_1)\,dx_1\right)\left(\int_{x_2&gt;x_1}e^{jvx_2}f_X(x_2)\,dx_2\right)\cdots \left(\int_{x_K&gt;x_{K-1}}e^{jvx_K}f_X(x_K)\,dx_K\right)

Is this right?
 

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