Answer to Question #224032 in Statistics and Probability for Madhu

Question

The first four moments of a distribution about the value 5 of the variable X are 2,20,40 and 50 respectively. Calculate first four central moments, mean, variance, B1 and B2

Answer 1

$\space exact \space question \space is \space \\ The \space first \space four \space moments \space of \space a \space distribution \space about \\ \space the \space value \space 5 \space of \space a \space variable \space are \space 2 \space , \space 20 \space , \space 40 \space , \space and \space 50 \space . \\ \space calculated \space first \space four \space central \space moments \space about \space the \space mean \space and \space \ \space \beta_1 \space and \space \beta_2 \space \\----------------------------\\ solution\\ Let \space rth \space moment \space of \space a \space variable \space x \space about \space 5 \space is \space \\ \mu'_r=E(x_r-5)^r\\and \space left \space rth \space moment \space of \space x \space about \space its \space mean \space be\\ \mu_r=E(x_r-5)^r. \\ so \space \mu'_1 =2 \implies E(x_i)-5=2 \implies E(x_i) =\bar{x}=2+5=7 \\ So, \\----------------------\\first \space moment \space about \space mean \space = μ_1=E(x_i−\bar{x})=E(x_i)−\bar{x}=7−7=0...(1) \\ ----------------------\\ 2nd \space moment \space about \space mean \space =μ_2=E(x_i−\bar{x})^2=E(x_i−7)^2\\=E[(x_i−5)+(5−7)]^2=E[(x_i−5)−2]^2=E(x_i−5)^2−4E(x_i−5)+4\\=μ^′_2−4μ^′_1+4=20−4×2+4=20−8+4=16....(2)\\ ----------------------\\3rd \space moment \space about \space mean \space =μ_3=E(x_i−7)^3=E[(x_i−5)+(5−7)]^3=E[(x_i−5)−2]^3\\=E[(x_i−5)^3−3(x−5)^2×2+3(x_i−5)×2^2−2^3]=E(x_i−5)^3−6E(x_i−5)^2+12E(x_i−5)−8\\=μ^′_3−6μ^{'2}_2+12μ^′_1−8=40−6×20+12×2−8=−64 . .....(3) \\ ----------------------\\4th \space moment \space about \space mean \space =μ_4=E(x_i−7)^4=E[(x_i−5)−(7−5)]^4\\=E(x_i−5)^4−^4C_1E(x_i−5)^3×2+^4C_2E(x_i−5)^2×(2)^2−^4C_3E(x_i−5)×(2)^3+^4C_4(2)^4\\=μ^′_4−4×2μ^′_3+6×4μ^′_2−4×8μ^′_1+1×16=50−8×40+24×20−32×2+1×16\\=50−320+480−64+16=546−384=162...(4)\\----------------------\\ Skewness= \sqrt{β_1}=\sqrt{(\frac{μ^2_3}{μ^3_2})}=\sqrt{(\frac{64×64}{16×16×16})}=−1\\ (negative \space sign \space because \space sign \space of \space μ_3 \space is \space negative)\\ hence \space \beta_1 =1\\ \\----------------------\\ Kurtosis= β_2=\frac{μ_4}{μ^2_2}=\frac{162}{16×16}∼0.63$