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Verify Equation (6.6), which gives the joint density of Xi and Xj.

Short Answer

Expert verified

Equation :

fx(i)xjxi,xj=n!(i-1)!(j-i-1)!(n-j)!F(xi)i-1Fxj-Fxij-i-11-Fxjn-jfxjfxi鈭赌xi<xjproved.

Step by step solution

01

Joint probability distribution :

The related probability distribution on all possible pairings of outputs is the joint probability distribution. For each given number of random variables, the joint distribution may be studied.

02

Explanation :

Equation (6.6):

fx(i)xjxi,xj=n!(i-1)!(j-i-1)!(n-j)!F(xi)i-1Fxj-Fxij-i-11-Fxjn-jfxjfxi鈭赌xi<xj

Let the denote the joint probability density function of Xiand Xjto prove the equation,

fx(i)xjxi,xjwhere 1ijn, then we have,

role="math" localid="1647420071276" fx(i)xjxi,xj=limxi0xj0PxiXrxi+xixjXrxj+xjxixj.....(1)

Now, let the event role="math" localid="1647419990583" E=xiXrxi+xixjXrxj+xjcan be written as :

1i-11j-i-11n-j1

Now,

Xixifor i-1of the X'rs,

xi<Xrxi+xifor one X(r)

xi+xi<Xrxjfor j-i-1of X'rs,

xj<Xrxj+xjfor one Xr

And Xr>xj+xjfor n-jof the X'rs,

Hence, by using multinomial probability law we get the following as :

P(E)=P[xiXrxi+xixjXrxj+xj]=n!(i-1)!1!(j-i-1)!1!(n-j)!p1i-1p2p3j-i-1p4p5n-j.......(2)

where p1=PXixi=F(xi)

p2=Pxi<Xrxi+xi=F(xi+xi)-F(xi)p3=Pxi+xi<Xrxj=F(xj)-F(xi+xi)p4=Pxj<Xrxj+xj=F(xj+xj)-F(xj)p5=PXr>xj+xj=1-PXrxj+xj=1-F(xj+xi)

03

Explanation :

Thus, substituting (2)in (1), we get,

localid="1647425314276" fx(i)xjxi,xj=n!(i-1)!(j-i-1)!(n-j)!limxi0F(xi+xi-F(xi))xilimxi0[F(xj)-F(xi+xi)]j-i-1limxj0F(xj+xj)-F(xj)xjlimxj01-F(xj+xj)n-j=n!(i-1)!(j-i-1)!(n-j)!F(xi)i-1f(xi)F(xj)-F(xi)j-i-1f(xj)1-F(xj)n-jfx(i)xjxi,xj=n!(i-1)!(j-i-1)!(n-j)!F(xi)i-1F(xj)-F(xi)j-i-11-F(xj)n-jf(xj)f(xi)鈭赌xi<xj

Hence proved.

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