lixXix2=wixc2xc1xXixXix2c2xc1x2=c2x2c1xc3xc2xc1x2.

si287_e

For any nonnegative integer r, using Chebychev’s inequality, it can be established that

nhns+11KxXi/hnXixs=nhn1KxXi/hnXix/hnsPusKudufx.

si288_e

Noting that usKudu=0si289_e for s = 1 and s = 3 (since K is symmetric about 0), we have

c1x=hnuKudu+oP1=oPhn,c2x=hn2u2Kudu+oP1=hn2σK2+oP1,andc3x=hn3u3Kudu+oP1=oPhn3.

Plugging in these approximations for c1x, c2x, and c3x in the expression ...

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