Table 3

Examples of realized cumulants

lRealized lth cumulant
2j=1N(ΔStj)2
3j=1N((ΔStj)3+3ΔMtj(2)ΔStj)
4j=1N((ΔStj)4+6ΔMtj(2)(ΔStj)2+3(ΔMtj(2))2+4ΔMtj(3)ΔStj)
5j=1N((ΔStj)5+10ΔMtj(2)(ΔStj)3+15(ΔMtj(2))2ΔStj+10ΔMtj(3)(ΔStj)2+10ΔMtj(3)ΔMtj(2)+5ΔMtj(4)ΔStj)
6j=1N((ΔStj)6+15ΔMtj(2)(ΔStj)4+20ΔMtj(3)(ΔStj)3+45(ΔMtj(2))2(ΔStj)2+15(ΔMtj(2))3+60ΔMtj(3)ΔMtj(2)ΔStj+15ΔMtj(4)(ΔStj)2+10(ΔMtj(3))2+15ΔMtj(4)ΔMtj(2)+6ΔMtj(5)ΔStj)

Note(s): This table describes the realized lth cumulants in Bae and Lee (2021) and Fukasawa and Matsushita (2021). ΔMtj(l) in the row 2–6 are Mtj(l)Mtj1(l). Recall that St=Mt(1)

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