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if the first four moments of a distribution about the value 5 are equal to -4,22,-117 and 560. Determine (a) about the mean (b) about zero


Sai Rohit 15th Jan, 2020
Answer (1)
Mounika Sonti 29th May, 2020

Hello!!!

hope you are doing great!!!!

Given moments about an arbitary origin  5.

μ1' = -4      μ 2' = 22     μ 3' = - 117        μ4'=560

Moments about mean:

μ2 = μ2' - (μ1')^2

μ1'μ2' + 2(μ1')^3

μ3=μ3' - 32

μ4 = μ4' - 4μ1'μ3' + 6μ2'(μ1)^2  - 3(μ1')^4

substituting the values,

μ2 = 22 - (-4)^2 = 22-16=6

μ3 = -117-3(-4)(22)+2(-4)^3=-117+264-128=19

μ4 = 560-4(-4)(117)+6(22)(-4)^2-3(-4)^4

=560-1872+2112-768=32

Moments about mean are μ1=0,μ2 = 6,μ3=19,μ4=32

Moments about zero::

Let the moments about zero be denoted by v1,v2,v3,v4

First moment about zero = v1 or mean

Second moment about zero= v3=μ1 + (v1)^2

Third moment about zero = v2=μ2 + 3v1v2 - 2v1^2

fourth moment about zero = v4 = μ4 + 4v1v2 - 6(v1)^2v2 + 3v1^4

Hope it helps!!!!!

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