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
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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!!!!!
