Order Statistics

Suppose X1,,Xn is a random sample of size n from an infinite population with a continuous pdf. If we arrange the values in ascending order and

X(1)X(2)X(n)

X(1) = minimum, X(n) = maximum

For a random sample of size n, there are n! possible arrangements of the random variables

Distribution of minimum and maximum

Suppose X1,,Xn iid fX(x) (independently identically distributed) are continuous. Let

X(1)=min(X1,X2,,Xn)X(n)=max(X1,X2,Xn)

The cdf of X is FX(x)=P(Xx)=fx(t)dt

The pdf of X(1) (min) is

FX(1)=P(X(1)x)=1P(X(1)x)=1P(X1x,X2x,Xnx)=1(1P(X1x))(1P(Xnx))=1(1FX(x))(1FX(x))(1FX(x))=1[1FX(x)]nfX(1)=ddx(1[1FX(x)]n)=n[1FX(x)]n1fX(x)

And the pdf of X(n) (max) is

FX(n)=P(Xnx)=P(X(1)x,X(2)x,X(n))=P(X(1)x)P(X(2)x)P(X(n)x)=[FX(x)]nfX(n)=n[FX(x)]n1fX(x)

Distribution of the rth order

P(xX(r)x+Δx)=fX(r)(x)Δ(x)

The pdf of the rth order statistic X(r) is given by

fX(r)=limΔX0fX(r)Δ(x)Δ(x)

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P(xX(r)x+Δx)=P((r1) x’s in(,x), one in (x,x+Δ(x)),(nr) x’s in (x+Δx,))=n!(r1)!1!(nr)![P(Xx)]r1P(xXx+Δx)[P(Xx+Δx)]nr=n!(r1)!(nr)![FX(x)]r1[FX(x+Δx)FX(x)][1FX(x+Δx)]nr

Then from

fX(r)(x)=limΔx0P(xX(r)x+Δx)Δx=limΔx0n!(r1)!(nr)![FX(x)]r1[FX(x+Δx)FX(x)][1FX(x+Δx)]nrΔx=n!(r1)!(nr)![FX(x)]r1limΔx0FX(x+Δx)FX(x)Δx[1FX(x)]nrfX(r)(x)=n!(r1)!(nr)![FX(x)r1]fX(x)[1FX(x)]nr

From which we can derive the formula for the min (r=1) and the max (r=n)

Applying the above result we find that the sample median X~ of a random sample X1,X2n+1 (odd size) has the pdf

fX~(x)=fX(n+1)(x)=(2n+1)!n!n![FX(x)]nfX(x)[1FX(x)]nfX~(x)=fX(k)(x)=n!(k1)!(nk)![FX(x)]k1fX(x)[1FX(x)]nk

For an even number of observations:

X~=X(n)+X(n+1)2

which needs to find the joint pdf of (X(r),X(j)) using Change of Random Variables:

Y1=X(n)+X(n+1)2,Y2=X(n)