The characteristic function of order statistics

  • #1
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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

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

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

[tex]\phi(jvY)=\prod_{k=1}^K\phi(jvX_k)[/tex]

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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Answers and Replies

  • #2
, 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.
 
  • #3
, 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.
 
  • #4
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

[tex]\phi(jvY)=\int_{x_1=0}^{\infty}\int_{x_2>x_1}\cdots\int_{x_K>x_{K-1}}K!\prod_{k=1}^Ke^{jvx_k}f_X(x_k)\,dx_1\,dx_2\cdots\,dx_K[/tex]

which is the multiplication of ##K## integrals

[tex]\phi(jvY)=\left(\int_{x_1=0}^{\infty}e^{jvx_1}f_X(x_1)\,dx_1\right)\left(\int_{x_2>x_1}e^{jvx_2}f_X(x_2)\,dx_2\right)\cdots \left(\int_{x_K>x_{K-1}}e^{jvx_K}f_X(x_K)\,dx_K\right)[/tex]

Is this right?
 

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